Transcript for:
Applications of Calculus 3

calculus 3 calculus 3 takes a look at some different applications and more advanced uses of derivatives and integrals the actual calculus part of finding derivatives and finding integrals we're very comfortable with at this point we know how to do it all now we're going to see some uses of it and chapters five and six we're going to look at together and they look at these things called sequences and series we've got to lay a little bit of foundation with sequence in series first and then we'll start to see how the calculus plays into these so first to start this whole conversation off we're going to talk about sequences as we answer the question how do we work with sequences and really what a sequence is is an ordered list of numbers so for example 1 4 9 16 we say the first number is one the second number is four the third number is nine the fourth number is 16 they're in order in fact we usually label them a sub 1 is the first number a sub 2 is the second number a sub 3 is the third number and a sub 4 is the fourth number and so on well with calculus we're often interested in the infinite so let's take a look at what is called an infinite sequence the idea of an infinite sequence is the sequence really never ends it starts with the first term then you'll see the second term then you'll see the third term and will keep going on and on and on forever there's no last term there's a first term but no last term and we'll often express the infinite sequence in one of two ways we can use what's called an explicit formula which can define each term you'll often see a sub n is equal to some function that defines each term at that function so if we want the fifth term a sub 5 we would just plug 5 into the function so for example in the sequence i have up above 14916 you might notice those are all the perfect squares so we could express that as a sub n equals n squared and then when i want a sub 1 that's in the first term is 1 squared or 1. when i want the second term we plug 2 in we have 2 squared which is 4. when i want the third term we want three squared which is nine we're just plugging the subscript into the function to find out what that term is this way we could quickly find the tenth term by plugging 10 in 10 squared is a hundred that's an explicit formula and we could keep going on and on and on with those but another way to express them is what's called a recurrence relation which can define [Music] each term based on a previous term so for example we might say okay the first term is equal to one and then the nth term is equal to the previous term a sub n minus one plus 2 n minus 1. so to get the next term we have to know the previous term and so we can build this recurrence relation one term at a time based on what happened previously so here for example a sub 1 we already know is 1. so a sub 2 the second term is the previous term which is 1 plus 2 times the term number which is 2. minus one this makes the second term four if i want the third term a sub three we take the previous term which was four plus two times the term number three minus one and this becomes four plus six minus one which is nine and what you might notice is we're really building the same sequence this is the perfect squares 1 4 9 the next one is going to be 16. so there's different ways to represent the same sequence both explicitly and as a recurrence relation in terms of the previous terms now there are a couple sequences that are very important to us and so we're going to look at them separately we're going to look at what's called the arithmetic sequence and the idea of the arithmetic sequence is we have a common difference added to the previous term so for example i might have the number 5 then 8 then 11 then 14 and so on and what you see is happening in the sequence is we have a common difference of three that's being added to each term to get to the next term five plus three is eight eight plus three is eleven eleven plus three is fourteen we're adding three to get each term and we can represent an arithmetic sequence here with recursion by saying okay the first term is equal to five so the nth term is equal to the previous term and as we said above we're just going to always add three to the previous term maybe we could generalize this and say the first term is the first term whatever that is and the next term is the previous term plus that common difference although we also know how to express this with an explicit formula the explicit formula is going to be that any given term is equal to what we really have is a linear relationship we're going by three each time we really have a y equals mx plus b situation where we can take the common difference as our slope and we're going to have to say n minus 1 plus the y-intercept or the starting value of 5. now we have to do n minus 1 instead of n with this y equals mx plus b because we don't call the first term the zeroth term but the y-intercept is the zeroth term so we're kind of staggered off by one unit so what we really have is our slope or common difference and we have our y-intercept or our starting number often we call that v so maybe generally we would say a sub n is equal to the common difference times n minus 1 plus the first term that's the arithmetic sequence another important sequence though that we work with is called the geometric sequence if the arithmetic sequence has a common difference that's always added the geometric sequence has a common ratio that is multiplied so for example we've got 4 negative 12 36 negative 108 and so on what you see here is from to move from one term to the next term to the next term we have a common ratio that we're multiplying by we're multiplying each term by negative three to get the next term 4 times negative 3 is negative 12 negative 12 times negative 3 is 36 and so on well if i wanted to express this with a recurrence relationship we again will say well the first term is equal to four and then to get that oops sorry the first term not the nth term the first term is equal to four and then to get the nth term just take the previous term n minus 1 and multiply by that common difference of negative 3. maybe generally we say the first term is the first term and the nth term is that common ratio times the previous term but we can also express this with an explicit formula here we're multiplying by the same number over and over again repeated multiplication we should know is an exponential so we can take our first term times the ratio to the n minus 1 power telling us we multiply by negative 3 over and over and over again here we have to do n minus 1 because the 4 the first term a sub 1 is not multiplied by negative 3. so if we plug 1 in for the n you'll notice 1 minus 1 is 0 negative 3 to the 0 is 1 and we're just left with 4 for the first term so we've got to be careful sometimes with that n minus 1 bit so kind of generally an explicit formula for a geometric sequence is the first number times the ratio to the n minus 1 gives us that geometric sequence we might even label that the 4 is the first term and the 3 is the ratio okay so if that's a geometric sequence we're ready to actually start combining what we know about arithmetic and geometric and also pattern recognition we're going to see if we can build and find explicit formulas and as we do this our general strategy is to just look for patterns and do a little guess and check because sometimes you have to try stuff is the exponent in is it n minus 1 is it in plus 1 let's guess and check a few and see how we do so for example if i have negative one-half comma two-thirds comma negative three-fourths comma four-fifths negative five sixth we're going to break this down and see what we observe first thing i observe is it's alternating it's going from negative to positive to negative to positive to negative and the way we can generate an alternating sequence like this is we just take negative 1 raised to an exponent now i always check the first term the first term is negative so if i plug 1 in for n negative 1 to the first power is negative if the first term was positive we'd need to use n plus one so that we shift the negative sign down one so the first thing we saw is that we're alternating second thing i observe here is i just look at the numerators the numerators are counting 1 2 3 4 5 and so on well what we see is the first term is a numerator of one the second term has a numerator of two the third is three the fourth is four that's just n the term number is the numerator one two three 4 5. when i look at the denominators they're a little different the denominators go 2 3 4 5 6 2 3 4 five six i see the denominators are kind of a shift of one they don't start the first term doesn't start at one the first term starts at two it's always one more than the term number it's n plus one so the fifth term has a denominator of six one more than that and so when i put it all together what we end up with is that the nth term is alternating negative 1 to the n times the numerator of n divided by the denominator of n plus 1. and this becomes the explicit formula for our series let's try another one see if we can continue finding these patterns let's do three-fourths nine-sevenths twenty-seven over ten eighty-one over thirteen and so on if i look just at the numerators again the numerators are three nine twenty seven eighty one what i see is those are powers of three three to the first three to the second three to the third three to the fourth and that exponent is always matching the term numbers so three to the n seems to create my numerators so for the fourth term three to the fourth is eighty-one in the numerator in the denominator we've got 4 7 10 13 4 7 10 13. what i noticed there is we are adding 3 to get the next term if we're adding 3 to get the next term what we actually have is an arithmetic sequence my arithmetic sequence says i can take the common difference of three times n minus one plus the starting number of four well let's simplify if i distribute the three three n minus three plus four combining like terms we have three n plus one so when i combine this together any term the numerator is 3 to the n the denominator is 3n plus 1. and we've got an explicit formula for what's going on with this sequence let's do another example then let's try 2 over 2 i know it's not reduced but we're going to go with it negative 4 over 10 12 over 50 and negative 48 over 250 and so on again the first thing that i'll notice on something like this is we're going from positive to negative to positive to negative we're alternating again to get an alternating sequence we take negative one to some exponent the problem is is we've got our first term second term third term fourth term notice if we use the term number of 1 plugging 1 in for n negative 1 to the first power is negative 1. our first term is not negative our first term is positive we've got the wrong signs we're set up for negative positive negative positive and this one's positive negative positive negative so to shift by 1 we just need to add 1 to the exponent now when we plug 1 in we get 1 plus 1 is 2 and negative 1 squared is a positive 1 and we're set up for a positive negative alternation let's look at the numerators 2 4 12 48 to go from 2 to 4 you might notice we multiply by 2 but from 4 to 12 we multiply by 3 and from 12 to 48 we multiply by four we're multiplying by one more number to get to 48 we have four times three times two to get to twelve we have three times two notice that counting down of multiplication we should recognize that as a factorial to get the fifth term it should be five times four times three times two times one however this doesn't quite work because the first term isn't one the first term is two plugging 1 and 1 factorial is 1 how do we get a 2 well let's shift by multiplying that by 2. let's see if that works let's test that out on the third term the third term has a numerator of 12. well 3 factorial is 3 times 2 times 1 which is 6 times 2 is 12 and it looks like we've got our numerators figured out all that's left to look at then is the denominator the denominators are 2 10 50 250 2 10 50 250 and what you might see there is we seem to be multiplying by five every time we know that if we're multiplying by 5 every time that's a geometric sequence where we take our first term times the ratio to the n minus 1. and now if we put all of that together we can say the nth term is negative 1 to the n plus 1 times 2 n factorial over 2 times 5 to the n minus 1. but one thing you might notice we've got twos in both the numerator and denominator we can divide those twos out so our reduced formula then becomes negative one to the n plus one n factorial over five to the n minus one so that's kind of how we can play and observe patterns in order to find the nth term of any geometric or any sequence really whether it's geometric arithmetic or neither but sometimes we're more interested in as these terms go out as these terms get to the infinitieth term kinda what's the last term approaching we call this the limit of the sequence are the terms getting closer and closer to some number are they asymptotically getting closer to some value or what is the infinitieth term well to do that we will say if the limit as n goes to infinity of a sub n of all the terms is equal to some limit l that means the sequence is approaching l as in becomes large if this happens if it approaches a specific limit we say then the sequence is convergent or the sequence converges to a specific number kind of the opposite case then is if the limit as n goes to infinity of the terms does not approach a value l it is called a divergent sequence let's do some examples got quite a few of these to look at let's take the explicit formula of one minus one half raised to the n power well if we take the limit as n goes to infinity of one minus one half to the n power we end up with one minus and one half times one half times one half is getting smaller and smaller and smaller that's approaching zero and so one minus zero equals one therefore as n gets really large these terms are getting closer and closer to one we can say that this series converges or the sequence converges sorry converges to one let's try the sequence one plus three n well if we take the limit as n goes to infinity of 1 plus 3n well we know that becomes 1 plus 3 times as n gets huge it becomes infinity basically these numbers are just going to get bigger and bigger because we're always just adding 3 each time this sequence diverges because it goes off to infinity how about the sequence three n to the fourth minus seven n squared plus five over six minus four n to the fourth well if we took the limit as n goes to infinity of that entire thing we know from our work with limits that the largest exponent is going to take over and that largest exponent is the fourth powers and basically the fourth powers take over and we get three over negative four and this series does converge i'm sorry the sequence does converge to negative three-fourths now we have a good friend named l'hopital that can help us with these as well let's say we've got a sequence that's two to the n power over n squared we would try and take the limit as n goes to infinity of 2 to the n over n squared to see if it approaches a value but the numerator goes off to infinity and the denominator goes off to infinity so what we can do is we can apply l'hopital's rule and take the derivative of the numerator and denominator and see what limit that gives us the derivative of 2 to the n is 2 to the n times the natural log of 2 over the derivative of n squared is 2n but those still are each going off to infinity so that doesn't help us so i guess we'll take another derivative so we have the limit as n goes to infinity the derivative of 2 to the n is still 2 to the n natural log of two times the constant natural log of two all over the derivative of the denominator which is two and now what you see happening is the numerator natural log of 2 squared that's just a constant divided by 2 that's just a constant but this numerator 2 to the n goes off to infinity we have infinity divided by 2 which is still infinity so this series or this sequence diverges the sequence diverges because it goes to infinity let's do one more with our friend l'hopital let's say the nth term is five n squared plus one over e to the n again the numerator is off to infinity and the denominator is off to infinity so we can use l'hopital's rule to say that the limit as n goes to infinity is the derivative 10 n over the derivative e to the n but as n goes to infinity these are still both going off to infinity so we'll do l'hopital's rule again the limit as n goes to infinity of 10 over e to the n and now when we plug infinity in e to the infinity is infinity we're taking 10 divided by a very large number basically we end up with zero this sequence converges it's a convergent series now sometimes we can't just take the limit very clearly and see what it's approaching so we need a slightly different strategy and there is another nice strategy that we can use sometimes not always but it works out nice when it does with what are called bounded sequences and the idea of a bounded sequence is all the terms are smaller than a constant that's called bounded above because they're never going to be bigger than this constant number and or all the terms are larger than a constant and that's what we call bounded below for example a sub n equals 1 over n is bounded above by one and below by zero what you see is the first term when we plug one it is one over one which is one then it's one-half one-third one-fourth one-fifth they're gonna get smaller and smaller but they're never gonna get to zero so it's always smaller than one bounded above by one but always bigger than zero bounded below by zero and this idea of being bounded is very useful because of what we call the monotone convergence theorem if we are bounded above means i'm never larger than some number and increasing so i'm always getting larger then the sequence converges kind of visually what we're talking about is let's say we're bounded above by some constants we'll call it c my series is always increasing but it's never going to get bigger than that constant you can see it's going to kind of converge to that constant it's approaching some number and really the opposite is also true if we are bounded below and decreasing then the sequence converges and visually that's almost the same thing we've got some constant c and the series is always decreasing i'm sorry the sequence is always decreasing but it's never going to pass that c it's going to get closer and closer to that c we see it converges at that point let's do one example where we can see this work out let's say a sub n is equal to 4 to the n over n factorial okay what we'll do then is we'll look at the next term to see if it's getting smaller or getting bigger the next term oops the next term a sub n plus one is equal to four to the n plus one over n plus one factorial but we can break that up a little bit 4 to the n plus 1 is really 4 times 4 to the n over and n plus 1 factorial is really the n factorial or let's if we break it out into its terms the first term is n plus one break it out into its factors not terms times the next term is n then n minus 1 n minus 2 that's really n factorial which means we've broken this up into two fractions we have 4 over n plus 1 times 4 to the n over n factorial what we see is we're taking the previous term and multiplying it by four over n plus one when we multiply by 4 over n plus 1 with large numbers of n because we're always interested at infinity if n is large 4 over n plus one we're really multiplying by a really small number we're multiplying by four over infinity we're we're not really multiplying by much it's always going to be positive four over n plus one is getting smaller but never negative this means we are bounded below by zero and decreasing each term is smaller than the previous term because we multiply by some fraction four over a large number and it's smaller and smaller and smaller and it's decreasing decreasing decreasing but it's never going to pass zero so therefore by the monotone convergence theorem we know therefore the sequence converges and that's how we can use the monotone convergence theorem so this is a little introduction to sequences to in our next video we're going to look at what we can do with sequences in a little bit more of a useful fashion but today we're just trying to get familiar with using those sequences expressing them as an explicit formula finding out if they converge so take a look at your homework practice a few of these and we'll discuss them more in class now while sequences are very interesting to take a look at what we're more interested in with calculus is what are called series specifically infinite series so our question is going to be how do we work with infinite series and first let's define what exactly we mean by an infinite series basically a series is a sum of a sequence it is a sum of an infinite number of terms and we'll write it with this notation we'll take the sum as our n goes from 1 to infinity of each of the terms a sub n which basically means we take the first term and add the second term and add the third term and just keep adding all the way up to infinity kind of an example of this might be to say that the sum of some number of terms is equal to the sum as n goes from 1 to infinity of maybe n over n plus one so if i want the first sum that's just the sum of the first term the first term is one over one plus one which is really just saying one half or as a decimal 0.5 the second sum then says take the first term which was one half and add the second term which now is putting two in for n two over two plus one or one half plus two thirds which is equal to about 1.17 and change so if i want the third sum that means we're going to add the first three terms the first term was one half plus the second term was two-thirds plus the third term which is three over three plus one or one half plus two thirds plus three fourths now my sum is equal to about 1.92 and change and we could keep doing this but ultimately what we're interested in is if these are kind of the partial sums the first sum the second sum the third sum we're interested in kind of what happens with the entire sum so we've got these things called the kth partial sum we'll call it s sub k is equal to the sum as n goes from 1 to k instead of infinity of each of the individual terms well if the partial sum diverges goes to infinity then the whole series will diverge as well and if the partial sum converges then the whole sum or the whole series will converge as well let's take a look at what that means in an example we've been talking about the sum as n goes from one to infinity of n over n plus one and uh we found out that the sum of the first term was equal to one half the sum of the second term was one half plus two thirds the sum of the first three terms was one half plus two-thirds plus three-fourths and the sum of the fourth term is one-half plus two-thirds plus three-fourths plus four-fifths and so on and what we're interested in is if we kept going what would we know about the kth sum does it converge or diverge to a specific value well what i notice is if i look at each one of these terms they're either one half or bigger than a half one half or bigger and bigger than a half one-half and bigger and bigger and bigger than a half what i could say is that the first term is greater than equal to one-half times the number of terms which is one therefore the second one is going to be greater than or equal to one-half times the two terms because two-thirds is bigger than the one-half and then the third series is one-half times three we've got one half times the three terms but our answer is going to be bigger and similarly we're going to be bigger than one half times the fourth term which means ultimately the last k is going to be greater than or equal to one half times the k number of terms but here's the catch with that if i were to take the limit of this stuff the limit as that k term goes to infinity of one half of k we know that is equal to infinity because we're just taking one half times a large number it's still a large number therefore we can conclude that this series is unbounded and that the sum as n goes from one to infinity of n over n plus 1 is just going to keep getting larger and larger and larger ultimately this series diverges because the k-th term diverges but this is not always the case let's take a look at another example let's take a look at the sum as n goes from one to infinity of one over n times n plus one to get an idea of an example of what's happening let's take the individual first few sums the first sum is just the first term 1 over 1 times 2. which is one half the second sum is equal to that first term of one half plus one over two times three which turns out to be one-half plus 1 6 which is equal to two-thirds the third sum is the two-thirds the prior sums plus the third term which is one over three times four or two thirds plus one twelfth which is equal to three fourths and with this one you might start to notice a pattern approaching for what each term is equal to can you tell what the fourth sum is going to be 3 4 plus the 1 over four times five is equal to three fourths plus one over twenty which turns out to be four fifths did you catch the pattern the kth term is going to be equal to notice our numerators are always equal to the term number k over the denominators are one more than the term number k plus 1. so this time if i were to take the limit of that stuff as k goes to infinity as we get to the infinitieth term of k over k plus 1 we know the ks are going to get really large the plus 1 isn't going to really matter and this is ultimately going to be 1. therefore we know that the sum as n goes from one to infinity not just k of one over n times n plus one is equal to one therefore i know this series converges basically as we get more and more terms added in we're going to get closer and closer to the number one so we converge that's kind of a really brief introduction to deciding if series converge or diverge i want to talk about three special series that come up quite a bit as we work through series and the first one is one of the most famous series of calculus it is called the harmonic series and the harmonic series is the sum as n goes from one to infinity of one over n basically it's one over one or one plus one over two plus one over three plus one over four plus one over five one over six and so on and so forth we should recognize this harmonic series quickly as we continue to study in calculus here well let's take a look at what's happening with that harmonic series first let's take a look at the first sum well the first sum is one over one that's just one the second sum is one plus one over two the third sum is one plus one half plus one third and the fourth sum is one plus one half plus one third plus one fourth and now what i want to do is stop here for a moment and take a look at these last two terms one-third plus one-fourth notice one-third is bigger than one-fourth each term is bigger than the next term after it which means this is bigger than one-fourth plus one-fourth if i replace the one-third with something smaller the one-fourth we end up with something smaller well one-fourth plus one-fourth is one-half so the sum of the first terms is greater than one plus one-half and i'm going to say one-half times two because we've got the first half and then the next one's bigger than another one-half similarly if i kept going all the way down to the eighth sum the eighth sum is one plus one-half plus one-third plus one-fourth plus one-fifth plus one-sixth plus one-seventh plus one-eighth and we've already established that one-third plus one-fourth that part is greater than one-half but what's interesting is if i looked at all the other terms the next four terms what i have is that the one-fifth plus one-sixth plus one-seventh plus 1 8 1 8 being the smallest meaning all four of those are bigger than if they had all been replaced with 1 8 plus 1 8 plus 1 8 plus 1 8 and 4 8 adds up to one half so the eighth sum is greater than one plus one half times three because we've got a half another half and the next four terms add up to a half and in fact in general if we were to continue this pattern out s sub 2 to the n power it has to be a power of two notice the ones we played with were the fourth power and the eighth power the next one half would come off of the sixteenth power but it will always be greater than one plus one half times whatever that exponent is on the two so for example with the eighth sum eight is two to the third we had three one halves multiplied together which means on the sixteenth there'll be four one-halfs added to the one we said though if we can take the limit as n goes to infinity of the sum which is one plus one half of n this one gets there really slowly but as the n gets larger and larger and larger half of infinity is still infinity and if you add one you still get infinity so therefore the partial sums diverge to infinity which means the big idea in fact let's change colors we'll do this in brown therefore the harmonic series also diverges to infinity the sum is n goes from one to infinity of one over n which is one plus one half plus one-third plus one-fourth and so on is going to equal to infinity the harmonic series diverges this is one of the most important series of calculus and we should recognize that whenever we have the sum of one over n it will always go to infinity very slowly but it still diverges so that's the first important series the harmonics series the second of three important series i want you to see today are what is called the geometric series and it comes from the geometric sequence the geometric sequence was some number then some number times the ratio then some number times the ratio squared because we kept multiplying by that ratio and so the geometric series says we're going to add those all together a plus a times r plus a times r squared plus a times r cubed in other words we're taking the sum as n goes from one to infinity of a times r to the n minus one it turns out and the proof is a lot more involved but it turns out that this geometric series will diverge actually i'll write the series again here because we're going to highlight this is very important the sum as n goes from 1 to infinity of a times r to the n minus 1. it will diverge if the absolute value of r the common ratio is greater than or equal to 1. and if the absolute value of the ratio is less than one it converges to whatever a divided by one minus r is so the geometric series which we should recognize diverges if our ratio is greater than or equal to one and it converges if the ratio is less than one let's take a look at some examples let's take a look at the sum as n goes from one to infinity of negative three to the n plus one over four to the n minus one we have to be careful here because this isn't quite a geometric series yet let's take a look at what this series is doing and see if we can convert it into the right form of the geometric series so starting when n is one we end up with negative three squared over four to the zero power plus when n is 2 we have negative 3 cubed over 4 to the first power plus we've got negative 3 to the fourth over 4 to the second power and so on and so forth which means what we really have four to the zero is just one so this is really just saying negative three squared plus negative 3 that negative 3 squared gets multiplied by another negative 3 over 4. plus then that negative 3 squared is multiplied this time by a negative 3 over 4 twice giving us a total of 4 in the numerator and 2 in the denominator and so on which means really what we have is positive 9 plus 9 times negative 3 4 plus 9 times negative 3 4 squared plus and so on and what we see is the starting value for this series is nine and the ratio that we're being multiplied by is negative three-fourths the negative three-fourths squared the negative three-fourths cubed and so on because that ratio is smaller than 1 the absolute value is smaller than 1 we know it converges to the starting value over one plus or actually one minus the ratio which is negative three-fourths well let's clean up that fraction by multiplying by 4 all the way across and we get 36 over 4 plus 3 or 36 over 7. so this series as it gets closer and closer to infinity is ultimately going to converge to 36 sevenths getting closer and closer because the ratio is smaller than one or the absolute value of the ratio let's compare that to the sum as n goes from 1 to infinity of e to the 2n now i could rewrite this because that e to the 2n is really e squared to the n power as the sum as n goes from 1 to infinity of e squared raised to the n power what we see here is we're multiplying each term gets multiplied by e squared and then e squared and then times an e squared what we're really doing is we've got a ratio of e squared being multiplied over and over again the problem is e squared is greater than one whenever the ratio is greater than one we know the entire series diverges to infinity so that's the geometric series it diverges if the absolute value of the ratio is greater than one and it converges if the absolute value of the ratio is less than one one last series i want to talk about it's called the telescoping series this one doesn't come up much but when it does it's really nice it is a series where most terms subtract out what i mean is in general we've got the sum as n goes from one to infinity of one term minus the next term that comes after it in other words what you end up with is the first term minus the second term and then when n is two we get plus the second term minus the third term and then when n is three we do plus the third term minus the fourth term and so on and so forth all the way up to basically infinity and what you see happening is the second subtracts out with the next and then the next one subtracts out with the next one and so on until we end up with the kth partial sum is equal to the first term minus the last term well if the limit as k goes to infinity of b sub k converges in other words that last term is approaching a specific number not infinity or negative infinity then the partial sum also converges so if i have a telescoping series where we're subtracting out the next term it's going to basically come out to the first term minus the last term if the limit converges so for example if we're working with the sum as n goes from one to infinity of the cosine of one over n minus the cosine of one over n plus one if i recognize we've got a telescoping series that we're subtracting the cosine of the next term every two terms is going to subtract out all the way down to the kth term which is the cosine of the first term 1 over 1 which is 1 minus the cosine of the last term which is 1 over k plus 1. well if i think about the limit as k goes to infinity of the cosine of one over k plus one ultimately as k gets huge we have one divided by a huge number which is zero and the cosine of 0 is 1. so overall this series converges to the cosine of one minus the very last term at infinity is going to be one and we found the sum of this telescoping series let's do one more let's do the sum as n goes from one to infinity of e to the one over n minus e to the one over n plus one notice we are subtracting the next term n plus one so then i know the k partial sum is e to the one over one or e to the first power everything else subtracts out all the way down down to the last one of e to the one over k plus one so what we think about is what is the limit as k goes to infinity of the last term e to the 1 over k plus 1. well again 1 over k plus 1 ends up going to 1 over infinity which becomes 0 and e to the 0 is 1. and so what we find out is if that converges then the entire sum converges the sum is equal to e minus the last term here is approaching one and that gives us our telescoping series so we've taken a look at these infinite series and begun to play with them we talked about the harmonic series geometric series and telescoping series three important series as we continue moving forward and we're playing with this idea of when does it converge and when does it diverge in our future lessons we're going to talk about some tests we can do to get a better idea about when it's converging and diverging but for now take a look at your homework assignment practice those and come to class with questions as we're working with series the big question we're really interested in is does the series go to infinity or does the series add up to a finite number so that's going to be our question that drives us today how do we know if a series converges meaning goes to a specific number or diverges meaning goes to infinity and actually this is the question that's going to keep with us throughout this entire unit does the series converge or does the series diverge we're actually more interested in whether or not it converges or diverges then what it actually adds up to so the first way we can do this and this is really the first test we should always do when trying to decide if a series converges or diverge it's called the divergence test it's a quick test that will tell us if a series diverges or if we're not sure what we do is we consider the sum as n goes from 1 to infinity of a sub n we're looking at some series some sum if the limit as n goes to infinity of those individual terms is not equal to zero basically what that means is as we get up to infinity are we adding numbers are we adding zero if we're adding zero it may or may not converge but if we're adding anything else but 0 it's definitely not going to converge because it's always going to be getting bigger and bigger in fact that also works if the limit does not exist or if the limit is infinity then the sum as n goes from one to infinity of a sub n diverges so if the terms aren't ultimately going to zero then the series is not going to converge let's take a look at some examples let's first try the sum as n goes from 1 to infinity of n over three n minus one well to use the divergence test to see what's happening we can take the limit as n goes to infinity of n over three n minus 1 and we know that the higher powers the n over 3n are going to take over and that's going to ultimately equal 1 3 which is definitely not equal to 0. in other words once we get out to enough terms we're basically doing plus one-third plus one-third plus one-third plus one-third and that's going to keep getting bigger and bigger so because that does not equal zero therefore we will say this series will diverge any time the limit as n goes to infinity of the individual terms is not zero it will diverge so if we try to find the sum as n goes from one to infinity of the cosine of one over n squared we can do the same thing we'll take the limit as n goes to infinity of the individual terms cosine of 1 over n squared well if n gets huge this becomes 1 over a huge number which basically means we're taking the cosine of 0 and we know the cosine of zero is equal to one which again is not equal to zero so eventually we're just doing plus one plus one plus one plus one plus one that's never going to settle out at a number therefore this series as well let's try one more example let's take the sum as n goes from one to infinity of n plus one over n squared minus three n plus one well again our strategy for the divergence test is we're going to see what happens with the limit as n goes to infinity of n plus 1 over n squared minus 3n plus 1. here you'll notice the n squared has the highest exponent so it's going to take over which means the denominator is going to become huge and we'll have a small number divided by a huge number which is 0. so now in the end this one is eventually doing plus 0 plus 0 plus 0. but here's the catch about the divergence test if the limit equals 0 that does not mean it converges because sometimes it will converge and sometimes it will diverge so when the limit equals zero we say the test is inconclusive when we're inconclusive we're going to have to do some other testing and that's really what the rest of this chapter is going to be about but for now we're just going to say inconclusive so that's the divergence test if it the limit as n goes to infinity of the individual terms goes to anything but 0 we know it diverges if it equals 0 we don't really know anything so we have another option to test and this is what is called the integral test by the way the integral test only works if all the terms are positive if we have negative terms the integral test will not work so alternating series don't work even if there's one term that's negative the integral test won't work but if all the terms are positive and we're looking at the sum as n goes from one to infinity of some series what we'll do is we will let the function f of n be equal to the individual terms and we will take the integral from some number any number we want up to infinity of f of n dn this integral and the original sum will either both converge or both diverge so if we can take the integral they will both do the same thing so if the integral equals infinity the sum equals infinity if the integral equals a number the sum also equals a number but they are not necessarily the same number it just means that they both converge to a number so let's take a look at some examples let's take a look at the sum as n goes from one to infinity of one over n to the fourth if we use the divergence test on this the limit as n goes to infinity we see a zero so that's inconclusive so what we'll do instead is we will use the integral tests to see if it tells us something we'll take the integral from let's go from the first term 1 to infinity and we don't like to integrate fractions so i'm going to write this as n to the negative 4 dn which we know is equal to n to the negative 3 divided by negative 3 which we know let's simplify that a little bit further we'll bring the n to the bottom so we've got negative 3 into the third and we're integrating from one to infinity well if we plug infinity in we get one divided by a huge number which is zero minus if we plug one in we get a negative one-third and minus a negative one-third is positive one-third and it equals one-third the number doesn't matter so much as the fact that it equals a po a number it doesn't equal infinity because the integral converges the sum also converges because both the integral and the sum are going to do the same thing let's take a look at one more of these examples let's take the sum as n goes from 1 to infinity of n over three n squared plus one again the divergence test says that this is going to zero so it's inconclusive so let's try the integral test let's take the integral from one because that's the lower limit to infinity of n over three n squared plus one well we're kind of set up for a u substitution where u is three n squared plus one d u is six n so oops i forgot the dn so we'll multiply by six and we will multiply by one sixth and so that's going to give us 1 6 times the integral and let's go ahead and plug the limits in 3 times 1 squared plus 1 is 4 to infinity of 1 over udu which is equal to 1 6 natural log of u integrated from 4 to infinity but when we plug the infinity into the natural log the natural log always gets bigger and bigger we end up with infinity so the integral diverges to infinity and because both the integral and the series do the same thing if the integral diverges then the sum will also diverge and so we see here is a great example of the how the divergence test is inconclusive because both of these the limit was zero so we had to use a different test to decide one converged and the other diverged before we wrap up here though i want to talk about one special series and this is a series when we see it we're going to just recognize whether or not it converges or diverges it is called the p series a p series is a series of the form as the sum going from 1 to infinity of 1 over n to the p power if we were to do the integral test on this we would take the integral from one to infinity of one over n to the p d n or the integral from one to infinity of n to the negative p d n and the exponent property says we increase that exponent by one and we divide by the new exponent negative p plus one and if that's the case we're integrating from 1 to infinity when we plug that in what we really end up with is and this is going to be kind of loose notation here not quite correct we should be doing the limit as n goes to infinity but i'm just going to write infinity in for the n we get infinity raised to the negative p plus one over negative p plus one minus then when you plug one in one to any power is one over negative p plus one but the important thing to note here is infinity has an exponent of negative p plus one if this is in the numerator then the whole thing goes off to infinity but if that infinity ends up in the denominator the whole thing goes to 0 and it converges so this means it will diverge if the exponent negative p plus one is greater than 0 because if we have a positive exponent then it will diverge and if we solve this by subtracting 1 and dividing by negative 1 if p is less than 1 or really or equal to 1 because it is the harmonic series if it's equal to one we already know the harmonic series one over n diverges very slowly but it diverges and then the opposite is true that it converges if the exponent negative p plus 1 is negative which moves the infinity to the denominator or if we add 1 and divide by negative 1 if p is greater than 1. so this is kind of the important idea that the p series will diverge if p is less than one or equal to one let's go ahead and write the or equal to in there and it will converge if the exponent is greater than one and that can save us doing all the tests we just have to look at the exponent if we're actually working with a p series let's try some examples really quick to wrap up this video first let's consider the sum as n goes from one to infinity of one over n to the five thirds well this is a p series because it's one over n to some exponent and that five thirds is greater than one because it's greater than one we know it converges similarly we can look at the sum as n goes from one to infinity of one over n to the two fifths power we're looking at that exponent two-fifths is less than one because it's less than one this p series diverges so this will be an important series to add to our other series that we're familiar with the harmonic series the geometric series the p series these are ones we should recognize really quickly when they converge and when they diverge so we're going to wrap up here give you a chance to practice some of these trying to to decide using the divergence test and the integral test or recognizing a p series does this series converge to a number we don't really care what number or does it diverge to infinity take a look at the practice and we will see you in class to discuss it further when working with theories and trying to decide if they converge or diverge we have the divergence test to see if it diverges and if that's inconclusive we can try the integral test however the integral test only works if all the terms are positive and if we can take the integral some integrals are difficult or impossible to take and so we need another method to try and determine if series converge or diverge if one of those first two fail and so one idea that might come up is to use a series we know to establish something about a series we don't know the question here is going to be can known series help determine convergence of an unknown series and obviously the answer is yes because that's what we're taking a look at so let's see about this first one called the comparison test the idea of the comparison test is if i know a series converges so we'll say if the sum as n goes from one to infinity of maybe the series b converges and i've got some other series where each term is smaller than the b term but still positive still greater than one then well if every term of a is smaller than b and b converges then the series a must also converge to something smaller than whatever the b converges to but again the important question is does it converge or not and we can kind of take the opposite logic as well let's say we've got a series that diverges the series that we know diverges maybe it's the harmonic series or something like that and every term of this series a sub n is bigger than every term of the series b which we know already goes to infinity which is also bigger than zero that's important these actually should have or equal two on them then well if series a is bigger than series b which is infinity series a must also diverge as well let's take a look at some examples where we can use this logic number three let's consider the sum as n goes from one to infinity of one over n to the fourth plus three n squared plus two and with the divergence test this clearly goes to zero but the integral of this is going to be very difficult to take so let's see if we can compare it to something we know we know something about the sum as n goes from one to infinity of one over n to the fourth that's a power series where the exponent is bigger than one so it converges sometimes reciprocals are harder to think about so first i'll look at the fact that if i take n to the fourth and i add something to it and add something more to it that's going to be bigger than just the n to the fourth by itself well if we took the reciprocal of this inequality it's going to switch the direction of the inequality just going to show what we did there um off to the side here on the left 10 is bigger than 2. if we took the reciprocals 1 10 that would be smaller than the reciprocal one-half when we take reciprocals the inequality switches direction so what we've shown is our series is smaller than a known series we know that known series converges and so then our series being smaller must also converge because it can't go to infinity if it's smaller than a convergent series let's try one more example let's consider the sum as n goes from one to infinity of one over the natural log of n again as n goes to infinity this goes to zero and so what i might try and say is the natural log of n compared to some either an n or an n squared or some other p series that we know well how does the natural log of n compare to n if i think briefly about the graphs natural log comes up and moves slowly but n is a straight line that's going to grow much faster the n is bigger than the natural log of n well if i took the reciprocals 1 over the natural log of n then must be bigger than 1 over n and 1 over n we know that's the harmonic series so it's going to diverge but if we've got something bigger than infinity it must also go to infinity it must also diverge as well so if we can find a series that is bigger than our series that converges we must also converge if we find a series that's smaller than our series that diverges we must also diverge the order's important because if i found something bigger that diverges that doesn't necessarily mean i diverge and if i find something smaller that converges that doesn't necessarily mean i converge so we need to make sure the order is right to make this comparison test work and sometimes it's not so clear from our comparison test exactly how the pieces come together so maybe a more powerful test will be what we call the limit comparison test and the limit comparison test says consider a sub n and b sub n are greater than zero for all n in other words we've got all positive terms if we were to divide them if we took the limit as n goes to infinity of their quotient a sub n divided by b sub n and that actually equals a limit but not zero maybe it equals four or two or one third one half if it equals a specific limit that's not zero then both converge or both diverge they both do the same thing if the limit equals a number besides zero if the limit equals zero though then we only know something if the denominator converges if the denominator converges that tells us the numerator also converges if the denominator diverges and we equal 0 we don't know anything it's inconclusive similarly if the limit of the quotient is equal to infinity and we add to that that b sub n the denominator diverges then the numerator will also do the same thing then a sub n diverges as well so what we can do is take a known series and divide by the known series and take the limit as n goes to infinity what we hope happens is that it equals a constant that's not equal to zero if that's the case they will both do the same thing whether it's converge or diverge if it does equal to zero we hope that denominator converges because then our numerator also converges if it goes to infinity we hope the denominator diverges because the numerator also diverges but if we don't get that second half then the test also is inconclusive and you have to use another test but let's see how we can use this test with some examples let's find the sum as n goes from 1 to infinity of 3 over the square root of n minus 2. now i see that square root of the n in the denominator so i think okay let's compare it with 1 over the square root of n or 1 over n to the one half that's a p series it diverges because the exponent is smaller than one let's see what happens when we take the limit as n goes to infinity of our unknown series 3 over the square root of n minus 2 divided by our known series 1 over the square root of n well let's simplify that by multiplying by the square root of n on top and bottom when we do the square roots of n's divide out and we get the limit as n goes to infinity of 3 root n over the square root of n minus 2. and that's really nice because those square roots of n being the highest exponent are going to take over and we end up with three over one or just three because that three is not equal to zero they are going to both do the same thing so we said our comparison diverges therefore the sum as n goes from 1 to infinity of 3 over the square root of n minus 2 doing the same thing also diverges when we equal a number not equal to 0 they both do the same thing let's try another example let's consider the sum as n goes from 1 to infinity of 5 to the n plus 2 over 6 to the n divergence test the 6 to the n takes over it equals 0. the integral is difficult to take so let's try a comparison test let's try to compare it with kind of a simple form of the 5 to the n over 6 to the n which is 5 6 to the n plus 2 is probably not going to matter when we get off to infinity so it's probably going to behave similar to the 5 6. well 5 6 to the n is a geometric series where the r is less than one multiplying by 5 6 over and over and over again when the r is less than 1 we know that it converges so let's see what happens when we take the limit as n goes to infinity of what we don't know 5 to the n plus 2 over 6 to the n divided by what we do know 5 to the n over 6 to the n well uh we're dividing fractions so dividing fractions means we multiply by the reciprocal so we get 5 to the n plus 2 over 6 to the n times 6 to the n over 5 to the n 6 to the ends will reduce out and we get the limit as n goes to infinity of 5 to the n plus 2 over 5 to the n and as the ends get huge we see those 5 to the ends are going to take over ultimately the plus 2 is going to make very little difference and this will equal 1. which is not zero because it equals a number that's not zero they both numerator and denominator will do the same thing well we said it converges in the numerator in the denominator so therefore our unknown sum of five to the n plus 2 over 6 to the n also converges because our limit equals a number they both do the same thing let's try one more example let's take the sum as n goes from 1 to infinity of the natural log of n divided by n squared well the natural log of n is smaller than n so if i pick something bigger we'd be thinking like n over n squared which is one over n so let's try that one over n let's compare it to one over n 1 over n is the harmonic series it diverges so let's see what happens when we take the limit as n goes to infinity of the natural log of n over n squared divided by the known series 1 over n multiplying by the reciprocal we get the natural log of n over n squared times n the n divides out with the n squared and so we get the limit as n goes to infinity of the natural log of n divided by n well both of those go off to infinity so let's try to l'hopital it to see if that can help us figure out what the limit as n goes to infinity the derivative of natural log of n is one over n the derivative of n is one and now if n goes to infinity if n goes to infinity we got one divided by a large number which is zero divided by one which is zero so remember our rule about zero if we equal zero we need the denominator to converge uh-oh our denominator diverges so that is inconclusive that doesn't actually tell us anything so let's try a plan b uh 1 over n did not work so let's compare it maybe with one over n squared we know one over n squared converges because it's a p series where the p is greater than one so let's take the limit as n goes to infinity of the unknown natural log of n divided by n squared divided by one over n squared well this one's going to simplify quite nicely when we multiply by the reciprocal the natural log of n over n squared times n squared over 1. the n squareds divide out and we're left with the limit as n goes to infinity of the natural log of n which is infinity so we go up to our rule and if it equals infinity that's only useful if the denominator diverges well our denominator converges so this tells us nothing back to the drawing board so squared made us diverge first power made us converge let's go somewhere in the middle between one and two so we might try one over n to the three halves that's right in the middle it's still bigger than one is an exponent so it's a p series that converges so let's take i didn't cross off the last one that one was bad let's take the limit as n goes to infinity of the unknown series the natural log of n divided by n squared over the known series 1 over n to the 3 halves well if we multiply by the reciprocal we get the natural log of n over n squared times n to the three halves over one now when we reduce we're going to end up with a half left in the denominator so we have the limit as n goes to infinity of the natural log of n over n to the one half as you plug infinity in you'll notice the numerator is going to infinity and the denominator is going to infinity so we'll use l'hopital's rule which allows us to take the derivative of the numerator and denominator the derivative of natural log of n i'm just going to come down here to the next row limit as n goes to infinity the derivative of natural log of n is one over n the derivative of n to the one half pull out the one half and then subtract one from the exponent you get n to the negative one half that sticks it in the denominator and so we'll multiply by the reciprocal again one over n times two n to the one half all over one subtracting the exponents we get the limit as n goes to infinity of 2 over n to the one-half reducing the exponents and now as n gets really really huge in that denominator we get two divided by a large number which is zero so we go back to our rule if it equals zero we're happy if the denominator converges here the denominator we used converges and so we finally have a solution because the denominator converges and we equal zero therefore the natural log of n over n squared also converges so we're not as flexible when the limit equals zero or infinity when it equals zero or infinity it either needs to converge or diverge depending on the direction we're at but if it equals any other number it's really convenient to use the limit comparison test because if it equals a number other than zero they both converge or both diverge working with these takes a little bit of practice identifying which which series to use in the denominator so practice a few of these take a look at the homework assignment we'll discuss them more and practice more in class up to this point we've been mainly working with series where the terms are all positive but when the terms are negative we have what's called an alternating series where we have positive and negative terms bouncing back and forth there's another strategy we can use to determine the convergence of an alternating series so our question for the day is how do we determine the convergence of an alternating series and first we'll just really define the alternating series a little clearer oh turn 18 series there's two types and they're almost identical uh if we have the sum as n goes from one to infinity of negative one to the n plus one times some other series basically what that does is when n is one we get one plus one which is two and negative one squared is a positive one so we just end up with uh the first b term but then when we go to the next one we end up with an odd exponent so we end up subtracting the second b adding the third b subtracting the fourth b and so on and so forth alternating between positive and negative the other type and it's really the same idea the difference is that the negative one just has an n power and what that does is it allows the first term to start out negative and then we add the second term subtract the third term add the fourth term and so on those are alternating series and an interesting question that can kind of come up out of this introduction of alternating series we've talked about the harmonic series which diverges to infinity what if we made the harmonic series into an alternating harmonic series in other words we were to take a look at the sum as n goes from 1 to infinity of negative 1 to the n plus 1 over n because if we look graphically at what this does if we call this one over to the right the first term is going to be a positive one so we'll add one the second term is a negative one-half so we subtract one-half then the next term adds one-third and then we end up subtracting 1 4th and what you see is happening is we keep going back and forth but we don't go quite as far as we did last time and it looks like we're actually converging down to a point on the line and in fact it does the alternating harmonic series turns out to converge at the natural log of 2 and we can look at that a little more closely in class because there's some interesting stuff that happens there but for now what i want to point out if we change a divergent series like the harmonic into an alternating series sometimes it could turn out to be a series that actually converges but this is not always the case so we need some better rules to test the convergence of an alternating series and so we have what is called the alternating series test and the alternating series test says that an alternating series such as the sum as n goes from 1 to infinity of negative 1 to the n plus 1 times sum b sub n terms converges if two things are true first each b sub n is positive and the n plus one term is smaller than the nth term in other words every term we go to next is smaller than the one before it and that's what happened with the alternating harmonic we didn't quite go back as far and then we didn't quite go back as far and we didn't quite go back as far and eventually that's going to pull us down to a point on the line so it each term needs to be smaller than the one before it and the other thing that's important and it's just the divergence test that the limit as n goes to infinity of b sub n has to equal 0. we have to pass the divergence test with the b sub n without the alternating part if that's true if each term is getting smaller and the divergence test works on the non-alternating part then the series will converge so let's take a look at some examples where we can use this alternating series test let's take a look at the sum as n goes from 1 to infinity of negative one to the n plus one over n cubed well to do that we will consider n plus one and n ignoring the alternating part the negative 1 to the n plus 1. if we ignore the alternating part we end up with 1 over n plus 1 cubed and 1 over n cubed well if i think about the fact that n plus 1 cubed is going to be bigger than just n cubed because we've added one we've got a bigger number we're cubing the reciprocals then one over n plus one cubed must be smaller than one over n cubed in other words we've shown that the next term is smaller than the first term if that's the case the terms are getting smaller therefore we can say the series converges let's try another one let's try the sum as n goes from 1 to infinity of negative 1 to the n times n over 2 to the n to use the alternating series test we consider the n plus one and the n to see if n plus one is smaller ignoring the alternating part we just have n plus one over two to the n plus 1 and just n over 2 to the n what i notice here is n does not grow nearly as fast as the denominators the denominators grow faster two to the fifth is a lot bigger than five two to the twelfth is a whole lot bigger than twelve the denominators are growing much larger so they're going to kind of take over so when we compare those denominators 2 to the n plus 1 is going to be bigger than just 2 to the n because we've added 1 to the exponent making it bigger so the reciprocals 1 over 2 to the n plus 1 must be less than 1 over 2 to the n therefore multiplying both top and bottom by n or n plus one they're basically the same amount we can conclude that n plus one over two to the n plus one is less than n over two to the n so we found out is that the next term is smaller than the first term and when that happens we know that we will ultimately converge with our alternating series so now that we've got some idea of this uh work with convergence and divergence with alternating series we're ready to talk about what's called absolute convergence versus what's called conditional convergence the idea behind absolute convergence is that the sum as n goes from one to infinity of your terms exhibits absolute convergence if the sum is in goes from one to infinity of the absolute value converges however we have conditional convergence if the sum as in goes from one to infinity of your terms exhibits conditional convergence if the sum is then goes from one to infinity of the original series converges but the sum as n goes from one to infinity of the absolute value diverges an example of number two here would be the harmonic series when the harmonic series is positive it diverges but when the harmonic series is alternating where we've got negatives and positives it does converge so we have conditional convergence with that example conditional convergence isn't very exciting to us because we're forced to use the alternating series test but what's really interesting is the fact that absolute convergence if the positive form in other words ignoring the alternating part if that converges it implies convergence in other words if the sum as in goes from one to infinity of the absolute value of a series converges then we know for sure that the sum as n goes from one to infinity of all the terms positive or negative also converges notice the opposite isn't true if the original converges does that mean the absolute value converges not necessarily the alternating harmonic is a great example of that but if the absolute value converges then we know the entire thing converges well what that means is if we can prove the absolute value converges we can then work with alternating series by using the other convergence test looking for absolute convergence so if we have maybe the sum as n goes from 1 to infinity of the alternating negative 1 to the n over 4 n minus 1. we're going to take a look at the absolute value version of it which means we're going to really see if the sum as n goes from 1 to infinity of 1 over 4 n minus 1 converges well i notice 1 over 4 n minus 1 is really similar to 1 over n which diverges that's the harmonic series so let's use the limit comparison test to see what happens when we divide these and take the limit as n goes to infinity we take the unknown one one over four n minus one over the known one one over four n multiplying by the reciprocal we have the limit as n goes to infinity of n over four n minus one we know from experience the ends take over and we get one-fourth which is not zero or infinity therefore we conclude that they both do the same thing so because our one over n diverges therefore this one diverges absolutely which doesn't really tell us much we like it when it converges absolutely because then the original one converges but this one does not divert does not converge absolutely but it might converge conditionally so now we go to the alternating series test we will consider n plus 1 and n when we use n plus one we get one over four times n plus one minus one and one over four n minus one which if we simplify these that becomes one over four n plus four minus one is plus three and one over four n minus one well if we were to compare these denominators 4 n and 4n are the same so if i add 3 to 1 and subtract 1 from the other the one we added 3 to is going to be bigger so then the reciprocals one over four n plus three must be less than one over four n minus one and so what we've shown is that the next term is smaller than the first term if that's the case the alternating series will converge therefore it converges not absolutely because when we did the absolute value thing we diverged absolutely but we do converge conditionally similar to the harmonic series if it alternates it converges if it doesn't alternate it diverges so we've got conditional convergence here let's do another one example number two we're going to do the sum as n goes from 1 to infinity of the sine of n over n cubed this doesn't look like an alternating series at first but if you think about sine of n sine of n goes from positive to negative to positive to negative oscillating between one and negative one so we're really going to end up with alternating signs here so what we'll do is we're going to compare it by getting rid of the negative part the sine is what makes it oscillate between 1 and negative 1. we can compare it with 1 over n cubed and that we know converges because it's a p series where the exponent is bigger than one all right let's compare it to the absolute value of the sine of n over n cubed because the absolute value forces everything to become positive we're looking for absolute convergence well if we do that compare it to having n cubed in the denominator the absolute value of sine is never going to be bigger than one which means our series every term is going to be less than or equal to one over n cubed so we've got a smaller series than a known conversion series therefore our series by the comparison test that we saw in the previous video the sum as n goes from 1 to infinity of the absolute value of the sine of n over n cubed must converge if the absolute value converges then we have absolute convergence and the original series without the absolute value also converges so if we can prove the absolute value converges with either the integral test the comparison test the limit comparison test or another test then we have absolute convergence and the alternating series converges if the absolute value diverges then we check for conditional convergence by using the alternating series test so that's how we can work with alternating series take a look at practicing these again series take lots of practice to get comfortable with so practice practice practice come to class with questions and we'll talk about it more at that time as we wrap up chapter 5 we're going to ask kind of the last question which is what other tests do we have that can be used for convergence that's kind of been the theme of chapter five we started with the divergence test to see if the limit goes to anything but zero then it definitely diverges but if it goes to zero it's inconclusive so we could use the integral test if we could take the integral easily we could use a comparison test to see what's happening compared to a known series whether by dividing or being smaller or bigger than that known series we then looked at if we have an alternating series what we can do in terms of converging absolutely or if it doesn't converge absolutely does it converge conditionally because the terms are getting smaller each time and then to wrap it up we've got two more tests that i want to look at here today the first one is called the ratio test and the ratio test is particularly helpful with factorials if we have factorials in the problem you can pretty much bet the best way to get at convergence is by doing the ratio test and here's what the ratio test says we're going to let the sum as n goes from 1 to infinity of all of our terms be the series that we want to see if it converges or and we're going to define greek letter rho it kind of looks like a p but it's a greek letter rho rho is equal to the limit as n goes to infinity of the absolute value of the n plus one term divided by the n term and this is going to be useful because when we take that limit one of three things is going to happen if zero is less than or equal to rho which is less than oops not equal to just less than one if we get a number that's less than one for rho then the sum as n goes from one to infinity of the entire series converges absolutely positive or negative it will converge if rho is less than one if rho ends up being bigger than 1 then the series as n goes from 1 to infinity diverges this is because if we're dividing and the next term is bigger dividing will result in a bigger number than one however there's a problem that comes out of this test and that is if rho equals one we get no information it doesn't happen often but that's kind of the worst case scenario if rho ends up equaling one we need another test so let's see if we can do some examples with the ratio test and see how it works out let's start by doing the sum as n goes from 1 to infinity of 3 to the n over n factorial i mentioned that this test is best with factorials so we define our rho as the limit as n goes to infinity of in the numerator we replace n with n plus one we have three to the n plus one actually let's do it this way over n plus 1 factorial and then the denominator is just the a sub n term which is 3 to the n divided by n factorial well if we simplify this by multiplying by the reciprocal we get rho is equal to the limit as n goes to infinity of 3 to the n plus 1 over n plus 1 factorial times n factorial over 3 to the n which is very interesting because n plus 1 factorial represents n plus 1 times n times n minus 1 factorial multiplies by everything below it specifically i could write n plus 1 times everything below it which is n factorial which is nice because then the n factorials would divide out leaving behind an n plus one with the with the threes three to the n plus one over three to the n we can subtract the exponents and the three to the n divide out and we're just left with three so we have now the limit as n goes to infinity of three over n plus one and we know as n gets huge this becomes three divided by a huge number which is zero well if rho is zero our ratio test tells us in letter a up there that if rho is greater than or equal to zero therefore the sum as n goes from 1 to infinity of 3 to the n over n factorial converges absolutely actually we could probably just say converges because there's no negative terms in this case but we did determine from the ratio test that this first series converges let's do another example let's look at the sum as n goes from 1 to infinity let's take n to the n power over n factorial let's see what happens there well now rho is equal to the limit as n goes to infinity of the numerators the n plus one terms so it's n plus one to the n plus one power over n plus one factorial divided by the regular series n to the n power divided by n factorial well if we multiply by the reciprocal and i'm going to even break this up the n plus 1 factorial we're going to write that as n times in oops sorry we're going to write that as n plus 1 times n factorial that way it's going to match down there and the n plus 1 to the n plus 1 power i'm going to rewrite that as n plus 1 to the n times n plus 1. that way when you add the exponents you get that n plus one so now multiplying by the reciprocal we have n plus one to the n times n plus one over n plus 1 times n factorial times n factorial over n to the n power and that's nice because one of the n plus ones can divide out with the denominator the n factorials can divide out and we're just left with n plus one to the n limit as n goes to infinity n plus 1 to the n over n to the n power which if we play with this we get n to the infinity both of those are to the n powers so we have n plus 1 over n to the n power which is equal to the limit as n goes to infinity if i divide both terms by n we get 1 plus 1 over n raised to the n power but this guy the limit as n goes to infinity of 1 plus 1 over n to the n power should look familiar way back when we talked about limits we found out that the limit as n goes to infinity of 1 plus 1 over n to the n power that goes to e that's one way that we can define e is that limit right there so the whole thing's equal to e which is definitely greater than one it's almost three a little less than 3 and we know from the ratio test that if we're greater than 1 the series has to diverge so therefore the sum as n goes from one to infinity of n to the n over n factorial diverges to infinity so a lot of these we massage algebraically to reduce so we can figure out the limit let's do one that's an alternating series before we step away from the ratio test let's consider the sum as n goes from 1 to infinity of negative 1 to the n times n factorial squared over 2 n factorial well row technically row we've been leaving it off on the past two but technically row says it's the absolute value of the quotient so because it's the absolute value of the quotient we don't really need to worry about the negative one to the n part because the absolute value will just always make it positive so really our numerator we've just got n plus one factorial squared over 2 times n plus 1 factorial all over n factorial squared divided by 2n factorial well as i simplify this the 2 times n plus 1 is really 2n plus 2 factorial so if we break that up counting down so that it will divide out with a 2n factorial that's really 2n plus 2 times 2n plus 1 times 2n factorial as it counts down similarly the n plus 1 factorial squared is an n plus 1 squared times n factorial squared just counting down and putting the squared on both parts because the whole thing is squared so when we multiply by the reciprocal oops i forgot the limit as n goes to infinity the limit as n goes to infinity of we have n plus 1 squared times n factorial squared divided by n factorial squared oops i'm getting ahead of myself finish out the first fraction divided by two n plus two times two n plus one times two in factorial and then multiply by the reciprocal which is 2n factorial over n factorial squared and let's see what we can divide out here the n factorial squareds are gone the two n factorials are gone and that's it so we've got the limit as n goes to infinity of and let's go ahead and multiply out to that numerator becomes n squared plus two n plus one multiplying out the denominator two n plus two times two n plus one is four n squared plus six n plus two and we know that the larger exponents are going to take over so it's going to become one-fourth which is less than one and so the ratio test tells us that if we're less than one it's going to converge absolutely and we know if it converges absolutely it also converges for the actual series so the sum as n goes from 1 to infinity of negative 1 to the n times n factorial squared over 2 n factorial converges absolutely so that is the ratio test really helpful with factorials because it allows us generally to reduce out the factorials by just breaking up maybe the n plus one from the n factorial maybe up to a couple terms and then if our row is less than one it converges absolutely if it's greater than one it diverges and we hope it doesn't equal one because then we have no information and we need another test a second test that's useful as we wrap up the chapter though is what's called the root test and the root test is really helpful with powers of n the reason it's helpful with powers of n is because it ultimately is going to take the nth root and get rid of that power of n so let's let the sum as n goes from 1 to infinity of a sub n be what we're looking for trying to decide does it converge or diverge we're going to also define a row in this case but here rho is going to be equal to the limit as n goes to infinity it's a terrible infinity as n goes to infinity of the nth root of the absolute value of the a sub n term if rho equals that one of three things will happen very similar to the ratio test if zero is less than or equal to rho which is less than one then the series converges absolutely and if that row is greater than one then the series diverges and just like last time if rho is equal to 1 we have no information so we'll have to use some other tests to decide if it converges or diverges again this root test is really helpful with powers of n so let's take a look at a couple of examples starting with taking the sum as n goes from one to infinity of three n cubed minus seven n squared plus one raised to the n power all over four n cubed plus two n squared minus one all raised to the n power well when i see those powers of n we're going to use the root test with rho equal to the limit as n goes to infinity of the nth root of our terms three n cubed minus seven n squared plus one to the n over four n cubed plus two n squared minus one to the n and what's nice about that is the nth root takes care of those nth exponents as we take the nth root of the numerator and denominator so we're left with the limit of three n cubed minus seven n squared plus one over four n cubed plus two n squared minus one and we know the largest exponents take over which is going to leave this with just three-fourths three-fourths which is less than one and the root test just like the ratio test tells us that if we end up with a limit a row that is less than one the whole series converges absolutely so therefore the sum as n goes from one to infinity of three n cubed minus seven n squared plus one to the n over 4 n cubed plus 2 n squared minus 1 to the n converges absolutely whether positive or negative it converges let's try one more example of the root test let's take the sum as n goes from 1 to infinity of n to the n power over the log or let's do the natural log the natural log of n raised to the n power again rho is going to be equal to the limit as n goes to infinity because we have n powers we're thinking the root test where we take the nth root of n to the n over the natural log of n to the n power well the nth root and the nth exponents will be inverses of each other leaving us with just the limit as n goes to infinity of n over the natural log of n the problem here is that as n goes to infinity the numerator goes to infinity and the denominator also goes to infinity so what we'll have to do is use l'hopital's rule which says we can take the derivative of the top and bottom the derivative of n is one the derivative of natural log of n is one over n which means we have the limit as n goes to infinity multiplying by the reciprocal is just n and as n goes to infinity n goes to infinity which is much larger than one and so the root test just like the ratio test tells us if rho is greater than one then the series as n goes from one to infinity of n to the n over the natural log of n to the n power diverges so we have two tests today to wrap up our work with sequences and series try to decide if series converge or diverge we have the root test which is good for powers of n where rho is the limit as n goes to infinity of the nth root of the absolute value and we have the ratio test where rho is the limit as n goes to infinity of the n plus one over the nth term and that one's helpful with factorials once we do that and calculate our row the process is the same if rho is less than 1 we converge absolutely if rho is greater than 1 we diverge if row equals one we have no information take a look at the homework assignment and try a few of these and we'll discuss them in class further and answer any questions that you might have now that we've gotten comfortable with various series and decide if a series converges or diverges we're going to look at some specific series that have some important properties that actually make calculus a lot easier to work with when the equations are complex our is going to be when does a power series converge so first we got to make sure we understand what we're talking about when i say power series a power series is a very special series it can be either centered at zero or any other value so first kind of the simple version of a power series is a power series centered at x equals zero is of the form the sum as n goes from zero to infinity of some constant and that constant varies based on the term times x to the n power in other words if i wrote it out it's basically a polynomial the first constant plus another constant times x plus another constant times x squared plus another constant times x cubed and so on and so on and this is when the power series is centered at zero we can shift it and actually do a power series centered at x equals a some other number which is of the form the sum as n goes from zero to infinity of those same constants but now instead of x it's going to be x minus a to the n power which means if we were to write this out when n is 0 the x part goes away and so we get the first constant or c sub 0 plus c 1 times x minus a to the first power plus the next constant times x minus a to the second power plus the next constant times x minus a to the third power and so on and so forth and so let's take a look at some examples of power series so we can get an idea of what we're talking about an example of a power series would be something like 1 plus x plus x squared plus x cubed and so on and so forth that's a power series because we're looking at all the different powers of x we could even write that as equal to the sum as n goes from zero to infinity of x to the n power in fact we can make our power series as interesting as we want maybe we could make it 1 plus x plus x squared over 2 factorial plus x cubed over 3 factorial plus and so on where we're dividing by the exponent factorial you know you can kind of see the first one we're dividing by 1 factorial and 0 factorial as well but both those are equal to one so we don't need that so this power series then is really the sum as n goes from zero to infinity of x to an exponent and then we divide by that exponent factorial let's look at one more example of a power series we're going to do 1 plus the x minus 2 divided by 2 times 3 plus x minus 2 squared divided by 3 times 3 squared plus x minus 2 cubed divided by 4 times 3 cubed and so on and so forth this is also a power series because we see that x minus 2 squared cubed fourth power and so on so what we end up with is the sum as n goes from 0 to infinity of the x minus 2 raised to the n power divided by and what we notice is if we look in the denominators we've got two which is one more than the exponent of one three which is one more than the exponent of two four which is one more than the exponent of three so we're really dealing with x plus one times 3 to the and you notice the exponent is always one no the exponent's always exactly the exponent so the exponent on 3 is the same as the exponent on the numerator and so we end up with this power series in fact we go one step further to classify these power series you notice the first two don't have that x minus a property the first two are centered at zero while the last power series because it's x minus two is centered two so that's what a power series is now that we've defined what a power series is the real question is when does that power series converge we're going to take a look at the convergence of a power series and as it turns out the power series the sum as n goes from zero to infinity of those constants times x minus a raised to an exponent we'll always satisfy one of the following it will either a converge only when x equals a or only where it's centered or b it will converge for all real numbers x maybe it converges everywhere or c and this one's a little more involved it will converge if the absolute value of x minus a is less than some value r for some r that represents the radius of convergence i'll underline that term because we're going to use that a lot today and what's really interesting though is while when x minus a the absolute value is less than r what we don't know is when the absolute value of x minus a equals r it may or may not converge kind of to visually see these three options option a if i've got a number line of all possible values for x and right in the middle is a it could converge only on that value a and then diverge everywhere else so that's the case where it converges only at a or same number line we might have a value a in the middle where it's centered and it really converges everywhere it converges for all real numbers or c that same number line and then off to the right we're going to add the radius and off to the left we subtract the radius and what that does is give us a range of values where it converges and outside of those values it diverges but what's important to note in this case we have a radius of convergence it is not immediately clear what happens at those edges it may converge and it may diverge so we'll have to test those edges to get an idea of exactly what's happened but only converges on the radius of convergence where the absolute value of x minus a is less than r and then we have to really decide what happens exactly at r so if these are the three cases for x when the whole series will converge or diverge what we're ultimately interested in finding then is what is the radius of convergence on what interval does the series converge and diverge to find the interval and radius of convergence the easiest thing to do probably is to use the ratio test which we know if rho is less than one it converges so let's take a look at an example of a power series and see if we can determine where it converges or where it diverges we're going to look at the power series as n goes from 0 to infinity of x to the n over n factorial and see if it converges using the ratio test then we're looking for rho which is equal to the limit as n goes to infinity careful not to confuse your variables we're always taking n to infinity of the absolute value of the next term x to the n plus 1 divided by n plus 1 factorial divided by the current term x to the n over n factorial so if we simplify that we get the limit as n goes to infinity of x to the n plus one over n plus one factorial we know is n plus 1 times n factorial times the reciprocal n factorial over x to the n which is nice because it reduces with the n factorials dividing out the x to the n divides out with the x to the n so we're left with the limit as n goes to infinity of x to the first power over n plus one and really it doesn't matter what x equals as n gets huge we're taking x divided by a huge number which means this is going to equal zero in other words rho is equal to zero which is less than one for all real numbers for x whatever number i pick for x rho will always be less than one it's always going to be zero so in this case what that tells us is the sum as n goes from 0 to infinity of x to the n over n factorial this power series converges for all x in other words the radius of convergence is infinity it will converge no matter what x equals if i wanted to express it as an interval it converges from negative infinity all the way up to positive infinity because rho is always less than one this power series will always converge but that doesn't always happen let's take a look at the power series the sum as n goes from 0 to infinity of n factorial times x to the n let's run through the same process with this and see when it converges well we have rho is equal to the limit as n goes to infinity of the next term which is n plus one factorial times x to the n plus one divided by the current term n factorial x to the n using our ratio test and we simplify that we get the limit as n goes to infinity of n plus 1 is n plus 1 times n factorial x to the n plus 1 over n factorial x to the n which is nice because the n factorials divide out the x to the n divides out and we're just left with the limit as n goes to infinity of n plus 1 times our x well in this case regardless of what x is actually equal to when n goes to infinity this whole thing is going to be infinity times our x and it comes out to be infinity to converge we need this to be less than one but it is never less than one because we're never less than one we don't really have a radius of convergence in other words what we can conclude therefore the sum as n goes from 0 to infinity of the power series n factorial times x to the n it will diverge [Music] everywhere except one place recall that we always converge at least on the center at least when x is equal to a well this power series is centered at 0 because we're not doing x minus anything so it diverges for all x not equal to zero in other words the radius of convergence because we're only on one point the radius is zero in the interval of convergence there's not really an interval it's just the value at the center which happens to also be zero so that's the second case where it diverges everywhere except the center which in this case happens to be zero so when do we end up with an actual radius of convergence well let's try that other series we played with earlier let's try the power series as n goes from zero to infinity of x minus two to the n divided by n plus 1 times 3 to the n using the ratio test rho is equal to the limit as n goes to infinity of the next term which is x minus two to the n plus one divided by n plus one plus one is n plus two times three to the n plus 1 divided by the current term which is x minus 2 to the n divided by n plus 1 times 3 to the n and when i multiply by the reciprocal we get x plus two to the n plus one divided by n plus 2 3 to the n plus 1 times the reciprocal n plus 1 times 3 to the n oops i put x plus 2 that should be x minus 2. careful not to change the signs x minus 2 to the n in the denominator reducing here things become really nice the x minus 2 to the n divides up with the n leaving 1 behind the 3 to the n divides out with the n leaving 1 3 behind and so what's left is the limit as n goes to infinity of an x minus 2 times an n plus 1 over a 3 times n plus two not forgetting the absolute values that are around everything because that's important here well as n goes to infinity we know the n term is going to take over but notice it's the same in on numerator and denominator which means we're going to end up with the coefficients of n which is the absolute value of x minus 2 divided by the absolute value of 3 which is just 3. well that's interesting because the first time when we took the limit we ended up with a value that was always smaller than one so we said it was all real numbers the second time we took the limit in example number two we end up with a value that's never less than one so we said it's basically has no radius but here we got x minus two's absolute value over three which we want to be less than one which means what we've ended up with is an equation that we can solve for x we've got the absolute value of x minus 2 over 3 is less than one multiplying both sides by 3 the absolute value of x minus 2 is less than 3. remove the absolute values puts x minus 2 between negative 3 and positive 3. and adding 2 to both sides puts x between 5 and negative 1. what we have is a radius of convergence the radius is what the absolute value is less than it's less than 3 which means it's going to have a radius of convergence of 3 wide 3 to the left and 3 to the right from the center and it will converge on the interval from negative 1 negative 1 to 5. however we don't know what happens at negative 1 and at 5. recall we have to actually check those values so let's go back and check those values specifically let's see what happens at negative one actually i'll say at x equals negative 1. when x is equal to negative 1 our original series we're going back to the beginning here this original series that we're working with if i plug negative 1 in for x i get the sum as n goes from 0 to infinity of negative 3 to the n over n plus one times three to the n which is nice because the three to the n divide out leaving behind a negative 1 to the n and so we have the sum as n goes from 0 to infinity of negative 1 to the n divided by n plus one notice those denominators are just counting up one uh when n is zero it starts at one two three four we have one half one third one-fourth one-fifth one-sixth we should recognize that as the harmonic series but because we have negative one to the n in the numerator this is actually the alternating harmonic which we know converges so on our interval when x equals negative 1 we can put a square bracket because we know the alternating harmonic series which we end up with at negative 1 does converge but we still need to see what happens at x equals 5 the other edge if x equals 5 we have the sum as n goes from 0 to infinity of 5 minus 2 which is 3 to the n over n plus 1 times 3 to the n which when the three to the ends divide out leaving one in the numerator is the sum as n goes from zero to infinity of one over n plus one and so when we start plugging in values of n we get one over one plus one over two plus one over three plus one over four plus one over five we should recognize that as the harmonic series and the harmonic series which is not alternating we've shown diverges so at five it actually diverges so we're going to do a curved bracket which means we have a power series with a radius of convergence of 3 meaning x can be 3 away from its center x can the whole series will converge when x ranges from negative one including negative one to five excluding the five [Music] so that's how we can decide if a power series converges we'll use the ratio test and find out does it converge only at the center for all real numbers or within some radius of convergence and we can figure out what that radius is and exactly what interval works for x so that the series converges but why are power series important power series are important to us because quite often we can represent functions as power series and usually the power series is easier to do calculus with than the original function so if we can represent the function as a power series we'll do the calculus on the power series and save us a lot of grief for example we know the sum as n goes from 0 to infinity of a times r to the n that's the geometric series the geometric series is equal to the starting value a divided by one minus the ratio given the ratio is less than one because if the ratio is greater than one it ends up diverging so if i have the function f of x equals 1 over 1 plus x cubed i could represent that as a power series if i can make it look like a over 1 minus r well we're not too far away from a over 1 minus r because we've already got a 1 over 1 but instead of plus if i make it a minus and i think what could i subtract that would be the same as adding x cubed well if we subtract a negative x cubed then we're in the same form where r is equal to the negative x cubed so we can go to our power series as n goes from zero to infinity of a a is the numerator the starting value of 1 times r which is what we're subtracting which is negative x cubed to the n power or cleaning that up we have the sum as n goes from zero to infinity one times anything is just that and i'm going to separate out the negative signs so we can see it's an alternating series negative 1 to the n times x to the 3n this power series then comes out to be the same value as one over one plus x cubed or we could write it out as the individual terms when n is one everything comes out to just a positive one minus because we're alternating i'm sorry starting out when n is zero now when n is one we get a negative x to the third plus when n is two we get x to the sixth minus when x is 3 we get x to the ninth plus and so on i want to take a look visually at this function coming together all right what i've done here on desmos is i've graphed the function we were working with 1 over 1 plus x cubed and then i'm going to start to build the power series that we found the power series that we found was 1 minus x cubed is the first two terms and notice that the blue line starts to kind of follow the red line now look what happens when we add x to the sixth it gets a little closer minus x to the ninth it gets a little closer plus x to the twelfth it gets a little closer minus x to the fifteenth it gets closer and as i continue to work through this power series the graph will start to get closer and closer to the graph of one over x one over one plus x cubed eventually leveling out at exactly the right value so that's how we can find a power series that gives us the exact same graph let's do one more example before we wrap up let's look at f of x equals x squared over 4 minus x squared now if we want to be in power series form we're ok with the numerator but we do need a 1 minus right now we have a 4 minus so what i'm going to do is i'm going to factor that 4 out of the denominator leaving behind a 1 minus and now it's x squared over 4. but we don't like to have a constant in front of the denominator so i'm going to pop that up to the numerator which is going to give us x squared over 4 in the numerator divided by one minus another x squared over four now we're in power series form of a geometric series where we've got the sum as n goes from 0 to infinity of the first term which is the numerator x squared over 4 times r which is what's subtracted in the denominator x squared over 4 raised to the n power well if i want to clean that up a little bit we'll have the sum as n goes from 0 to infinity of if i multiply the exponents we've got x to the 2n times x squared adding those exponents we get x to the 2 plus 2 n over 4 to the we've got 1 4 and putting the n on the denominator as well 4 to the 1 plus n power and we end up with a power series that's equal to the original function so we looked at several things today in this video we defined a power series we're going to continue to work with them over the next couple videos and then we looked at when the power series converges by identifying the radius of convergence in the interval where it converges it'll either converge at the center everywhere over a specific interval and we use the ratio test to decide when that is and then we concluded it looking at why power series are important by converting a known function into a power series specifically looking at the geometric series later we'll look at other series that aren't in the geometric series form that we can still convert into power series but this is where we're going to start with for now so take a look at the homework practice a few of these and we will see you in class now that we've taken a look at what a power series is and how to work with a power series we're ready to actually do some calculus with power series the question is going to be how can power series make derivatives and integrals easier and the brief answer to this is that often calculus is easier with a series than the function for example if the function f of x is equal to some power series from n to 0 to infinity of the constants times x minus its center a raised to the n power then we know that the derivative of the function would be equal to the derivative of a power series well what's nice about power series is it's a sum of many terms that have the exact same form so we can take the derivative term by term or really we can take the derivative of the entire sum as n goes from 0 to infinity and this c sub n is a constant and the exponent we know comes out front and then x minus a stays the same and then the exponent is decreased by 1. just using our exponent property we've taken the derivative term by term and we can do the same thing with the integral the integral of f of x dx is equal to the integral of the power series well going to the original power series as n goes from zero to infinity this time we keep the same constant out front and we know that we increase the exponent by 1 and divide by the new exponent now there's a couple adjustments that we need to make to these definitions though first let's look at the blue one the derivative when we take a derivative remember the first term in a power series is a constant and the derivative of a constant is always zero so actually it doesn't go from 1 to infin 0 to infinity it actually goes from 1 to infinity because that first term's derivative is 0 so we don't need to include it in our derivative with the integral we make up for that first term by having an extra constant now i'm going to add the constant before the sum because i don't want that constant to look like it's part of the sum but the power series anti-derivative does have a constant plus the sum of the anti-derivative and that kind of sticks that extra term back in there so minor adjustments to the intuitive formula just a couple things to be aware of so if that's true that we can just take the derivative term by term and the antiderivative term by term what we can do is we can use what we know to find derivatives and antiderivatives we know that for the function f of x equals 1 over 1 minus x we know that's a geometric series sum as n goes from 0 to infinity where the first term is 1 and we take the x to the n power based on what we know we're going to find a power series for the function g of x equals 1 over 1 minus x squared which is really equal to one minus x to the negative two well you might notice that that's very similar to the function that we know that we started with at the beginning so we know that the function f of x is equal to 1 minus x to the negative 1. and if i take the derivative of f f prime of x we would bring the exponent out front bring the exponent down by one times the derivative of the inside the derivative of negative one is positive one i'm sorry the derivative of negative x is negative one a negative and a negative makes a positive which means our derivative is one over one minus x squared which is what we're trying to find a power series for so we know the power series for the first original function is the sum as n goes from zero to infinity of x to the n and we know the second function is its derivative so we'll take the derivative term by term remember we have to have n equals one because the first term goes to zero to infinity bring the n out front and we have x to the n minus 1. and we have just found a power series for an unknown function because it's the derivative of a known function so if we can make that connection with something we know we can use calculus to find the new power series here's another one what if we want to find a power series for h of x equals the negative natural log of one minus x well again the function we know is 1 over 1 minus x we also know that the antiderivative of f of x dx would be the natural log of the stuff 1 minus x and then we have to divide by the derivative of the inside which is negative 1. and so the antiderivative of our known function is the negative natural log of 1 minus x well if we know the power series for the original function is the sum as n goes from 0 to infinity of x to the n we're just going to integrate that to get our new power series don't forget we need a constant plus the sum as n goes from zero to infinity of x to the n plus one divided by n plus one one little thing that remains though is what is that constant c so we make sure we're equal to the original function well to figure out what's going on with that constant we'll take the zeroth term and see how much it is off by so if we plug zero in for our x we get negative natural log of one minus zero which is the negative natural log of one which is just 0. so our constant in this case is equal to 0. so we don't need the constant here it's equal to 0 we're just going to say it's equal to the sum as n goes from zero to infinity of x to the n plus one over n plus one and actually all that n plus one stuff kind of seems silly to me that we're shifting one over in the numerator and shifting 1 over to the denominator instead of shifting 1 over in the formula let's shift one of our ends over starting with n equals 1 to infinity and now it becomes just x to the n over n so that's how we can use calculus to move from a known series to a power series for an unknown series if we can connect the known series to a derivative or anti-derivative of the known series we can just take the derivatives term by term and we end up with our new power series another place that calculus comes in handy is with partial fractions and power series so let's take a look at that really quick partial fractions and power series and we're actually not going to take the antiderivative but we're going to use the idea behind partial fractions to build an unknown power series let's say f of x is equal to one over one minus x times x minus two and we wanna build a power series for this well if we were to break it up into two fractions a over 1 minus x plus b over x minus 2. we could use what we know about the form of a geometric series to change these into gmet2 geometric series but first we have to figure out what a and b is so if we multiply across we get 1 equals a times x minus 2 plus b times 1 minus x if we let x equal 2 we end up with one equals b times negative one so b is negative one if we let x equal one we end up with one equals negative one times a so a also equals negative 1. and so what we end up with then is that f of x is negative 1 over 1 minus x plus negative 1 over x minus 2. okay let's see if we can massage these to get them in the right format where we have a geometric series form remember a geometric series form is going to be a over 1 minus the ratio the first one is actually already there the second one we're going to do a little bit of work with one thing we can do on the second one is we're going to switch the order so we have negative 2 plus x that way the variable is on the right side and then like before we can divide out a negative two so that we just have one minus x over two and then that negative two needs to move up to the numerator so that we're in power series form so we finally end up with is negative one over one minus x plus negative one over negative two is one half over one minus x over two which means we're dealing with a power series two power series added together we've got the sum as n goes from zero to infinity of the first term is the numerator so that's the negative one times the ratio which we're just subtracting x in the denominator so the ratio must just be x to the n plus the second power series which has a first term of one half because it's in the numerator times what's subtracted in the denominator x over two to the n power and maybe i want to clean that up a bit we've got the sum as n goes from 0 to infinity i want to factor out ultimately an x to the n so i'm putting that in onto both top and bottom because it's not a power series unless we're multiplying by x to the n everything else becomes that constant out front and so we factored out that x to the n we're left with negative 1 plus a 1 over in the denominator we've got a 2 and a 2 to the n so combine that together we get 2 to the n plus 1 is our coefficient and now we ended up with a power series that represents this function we were able to do that because our partial fractions allowed us to break it into two geometric series looking parts and we could make each part into a geometric series and factor out the x to the n to end up with our final partial fraction power series one last point then to wrap up this video let's take a look at how we can take a power series and find a function doing the exact opposite process we were doing before before we started with the function and we worked to find the power series now we're going to start with the function which is the sum as n goes from 1 to infinity of 1 over x plus 4 to the 2 n minus 1 and we're going to see if we can build this into a function that's equal to the sum well one thing i notice is we're trying to make it into a geometric geometrics like to be start at n equals zero so what we're going to do is we're going to shift this power series down to n equals zero which means we need to account for an extra in if you will so where we see the n we're gonna call it n plus one and that'll shift to the zero over one so if we replace that n with n plus one we get one over x minus four to the two times n plus one minus one and a little bit of algebra in that exponent gives us one over x minus four to the two n plus two minus one gives us a plus one well i'm going to keep trying to massage this as we work to figure out where our geometric series form might be when we see that addition inside an exponent we can break that into a product so let's see what that gives us the sum as n goes from 0 to infinity of 1 over and i'm going to do the first power first x minus 4 times 1 over x minus 4 to the 2n well notice we're getting really really close oops it's a plus 4 isn't it i've been writing this wrong make sure we're not changing our signs halfway through the problem sorry about that we're getting really close to that geometric form because the geometric form says we've got our first term times our second term which needs to be raised to the n power so if we could write that second part as something to the n power we'd be in business and if we go as n goes from zero to infinity of one over x plus four times what's nice about a numerator of 1 is it can be written as 1 to the n so if i pull out an n power we're left with x plus 4 squared in the denominator in fact we could even write that as 1 over x plus 4 the entire thing squared because the one squared doesn't matter and now what we've got is we've got our first term which becomes the numerator and the ratio which is subtracted in the denominator recall that the sum as n goes from zero to infinity of a times r to the n is equal to the numerator a over 1 minus the r so in our case the numerator a is the 1 over x plus 4. over the denominator one minus the ratio that's multiplied is one over x plus four squared but this is really an ugly format so to get rid of the fractions and fractions remember we've got x plus 4 squared in the denominator so we're going to multiply by x plus 4 squared on top and bottom it's going to distribute through and so what we'll end up with is a single x plus 4 in the numerator and x plus 4 squared minus and the x plus 4 is divide out so it's just minus 1. and we end up with this final function that is equal to the original sum that we started with so sometimes we can kind of parse this power series apart to find the pieces of the geometric sum if we are in fact in a geometric form if it's not a geometric series we're going to need a new game plan and that's what we're going to take a look at in our next video but for now take a look at how we can take derivatives and anti-derivatives of known power series and unknown power series and then also how the partial fractions can help us find power series and break up a power series to find a function good luck and we will see you in class thus far we focused our attention of power series on geometric series everything relates to a geometric series and can be written in the form of a geometric series but that's actually not the case in fact most functions cannot be written in the form of a geometric series so the question that we're going to try and address today is how do we find a power series for other functions and the answer to that is what are called a taylor series and mclaren series and taylor and mclaren developed these processes independent of each other there's a slight difference between the two of them taylor's is probably the most well-known because it's the most versatile taylor said a series for a function at the point a is the function at a plus the first derivative at a divided by 1 factorial times x minus a to the first power plus the second derivative at a over 2 factorial times x minus a to the second power plus the third derivative and we'll just put a little 3 at a divided by three factorial times x minus a to the third power and so on and so forth ultimately we're adding the nth derivative at a divided by n factorial times x minus a to the nth power on and on and on to infinity in other words it is the sum as n goes from zero to infinity of the nth derivative at a divided by n factorial times x minus a to the n this is what is called a taylor series now mclaren wasn't quite as versatile when he came up with his version of this same result mclaren said basically the same thing a series at but mclaren kept it at a equals 0 and didn't expand it to any a so if we plug 0 in for all the a's we basically get what's called a mclaren series which is f of 0 plus the first derivative at zero divided by one factorial times x to the first plus the second derivative at zero divided by two factorial times x to the second plus the third derivative at zero divided by three factorial times x to the third and so on and so forth and so what you see is we're ultimately taking the nth derivative at zero dividing by n factorial times x to the n and we keep going all the way up to infinity in other words it can be summarized as the sum as n goes from zero to infinity of f to the n at 0 divided by n factorial times x to the n and we get a mclaren series and really a mclaren series is a special case of the taylor series because it's just when a equals zero turns out the algebra is a lot nicer when a is zero for a mclaren series but the process is exactly the same the idea is if we can identify the generic form of all of the derivatives at the point we can write it as a mclaren or taylor series let's take a look at how we can write any function as a power series we're going to find either the [Music] well let's start with maclaren actually let's start with the mclaren series sometimes they're called mclaren polynomials a maclaren polynomial is not infinite it stops at a certain point but if we make it into a series going all the way to infinity we can find a mclaren series for example for f of x equals e to the x so we're going to start to look for a pattern that's going to occur with the derivatives so f of x actually start with f of x we know f of x is e to the x the first derivative is also e to the x the second derivative is also e to the x and so we can see pretty quick that any derivative is just e to the x and since this is a mclaren series we're actually going to plug 0 into all of these well e to the zero is one the first derivative at zero is also one the second derivative at zero is also one and so on and so forth we see that any derivative at zero is e to the 0 which is equal to 1. so we can build this mclaren series by saying okay e to the x is equal to first the function at 0 which is 1 plus the first derivative which is one over one factorial times x to the first plus the second derivative which is one over two factorial times x squared plus and so on continuing with this pattern until we have the nth derivative which is 1 over n factorial times x to that n power and so on well if we've seen this pattern building essentially we have the sum as n goes from zero to infinity of all of the x to the n's divided by n factorial this series then is equal to the e to the x function in fact in certain context as we continue our study of mathematics it's more convenient instead of working with e to the x to work with the power series the sum of x to the n over n factorial this one wasn't too interesting though because all of the zero points ended up being equal to one so let's see if we can find a mclaren series for something a little more interesting how about f of x equals the sine of x well if we start building our derivatives f of x is the sine of x f prime the derivative of sine is cosine the second derivative is negative sine the third derivative is negative cosine and the fourth derivative then becomes positive sign and so we're starting to cycle back through sine cosine negative sine negative cosine okay let's plug 0 in because that's what mclaren wants us to do the sine of 0 is equal to 0. the first derivative at 0 is the cosine of 0 which is 1. the second derivative at zero is the negative sign well negative sine of zero is zero the third derivative at zero is negative cosine which is 1 so negative 1 and then the fourth derivative at 0 is going to start to cycle back through our answers and so we end up with coefficients of 0 1 0 negative 1 0 1 0 negative 1 which means all of the odd terms the first term the third term the fifth term and so on are all going to be 0 because we're multiplying by that derivative so we don't really have any odd terms the first term 0 or the 0th term is 0. and now for the first term we'll take x to the first power divided by one factorial times the derivative of one the second derivative is zero so we have zero when we do the third derivatives the third derivative is negative one times x to the third over that three factorial the fourth derivative is zero but the fifth goes back to x to the fifth over five factorial then we're going to do minus and you start to see the pattern forming the sixth power goes to zero but the seventh power is negative x to the seventh over seven factorial and so on so we need a way to represent this pattern so we can convert it into a series well one thing that i notice is we are alternating positive negative positive negative so we've got a negative 1 to the n to give us that alternating series notice when n is 1 actually we're going to start at 0. so when n is 0 negative 1 to the 0 is a positive 1 which matches our first term so that's good then we've got x raised to an exponent but the exponents on x are all the odd numbers so we can't just say n what we can say though if 2n is the odd numbers 2n plus 1 i'm sorry 2n is the even numbers adding 1 2n plus 1 becomes all the odd numbers so as you can see when n is 0 the first term has x to the first when n is 1 plugging 1 in we end up with x to the third when n is two the second term remember that we're starting at zero zero one two we end up with two times two plus one which is five so those exponents work out and we're dividing by that same exponent two n plus one factorial and that's going to continue forever as we end up taking the sum as n goes from zero to infinity of that exact thing we just found negative one to the n x to the two n plus one over two n plus one factorial this is our mclaren series our power series that is exactly the same as the sine of x i want to take a look at this on desmos because i really think this is amazing when you see the graph work out okay what i've done on here is i've graphed sine of x we know sine of x starting at zero going up and down period of two pi amplitude of one that's that's not very exciting but what we've just found out is that the mclaren series which starts with x over one factorial and then it's going to be alternating with the odd exponents should be exactly the same as this graph and this is really neat to watch it build when we subtract x cubed over three factorial it starts to bend with the graph then we add x to the fifth over five factorial and it continues to bend with the graph subtract x to the seventh over seven factorial and it gets even better add x to the ninth over nine factorial and it gets even better and if we keep going we're gonna get more and more of the graph building up as it stretches out and out and eventually when you get to an infinite number of terms you've got the entire sine x curve from this power series so what we've done is we've changed the sine of x into a polynomial polynomials are much easier to work with as we get to more complex operations so that's how we can build a mclaren series we find all the derivatives we plug 0 into all the derivatives and we look for patterns in this mclaren series format where we're taking the first term then times x over 1 factorial times x squared over 2 factorial times x cubed over 3 factorial all the way up and ultimately we're just trying to find that generic form so that we can change it into a series that's mclaren let's take a look at taylor because taylor's a little bit more versatile we're going to find some taylor polynomials and series a polynomial is finite a certain number of terms a series goes off to infinity and we're going to start simply with f of x equals 1 over x and we're going to find this taylor series at a equals 1. we're going to plug 1 into all the derivatives and then we're going to have x minus 1 squared cubed fourth power fifth power and so on so first f of x our function is one over x or let's actually write it as x to the negative one we can convert it so that we can see what it is our first derivative then is negative x to the negative 2 or negative 1 over x squared our second derivative is positive 2 x to the negative 3. so 2 over x cubed and our third derivative notice if we pull the negative 3 out we end up with negative 3 times 2 times x to the negative four or negative three times two over x to the fourth and the reason i write it like that is you can sort of start to see a factorial forming in this numerator you can see next time it's going to be 4 times 3 times 2 then 5 times 3 4 times 3 times 2. it's going to count down so we're ending up with those factorials in the numerator that's going to be really helpful to us as we come up with a generic form all right we're doing it at one so we're gonna plug one into each of these one divided by one is one the derivative at one is negative one over one which is one the second derivative at one well the denominator is just one so we end up with two the third derivative at one gives us negative three times two and you can kind of see we're building towards the nth derivative at one is going to have an n factorial as we said next time it's gonna be four times three times two times one we just have to worry about the fact that it's an alternating series we want it to start out positive on the zeroth term on the zeroth derivative so we'll just do the n power and in much the same way we can use the taylor formula to start building then that one over x is equal to and let's see if we can jump right to the generic form the sum as n goes from zero to infinity of the derivative pattern notice our derivative pattern was negative 1 to the n times n factorial then we divide by an n factorial times x minus our a which is 1 raised to the n power but this one's really interesting because the n factorials actually divide out so what we really have is the sum as n goes from 0 to infinity of negative 1 to the n times x minus 1 to the n power and we have ended up with a power series that is equal to the function one over x let's try one more finding another taylor series we did cosine last time let's do i'm sorry we did sine last time let's do cosine this time and let's do it at a is equal to pi pi is a nice number to stick into cosine and so in much the same way we're going to run through the derivatives the zeroth derivative is cosine x that's the regular function first derivative is negative sine x second derivative is negative cosine x the third derivative is positive sine x and the fourth derivative we start to see it cycling back to cosine this time because it's a taylor series though we're not plugging 0 in we're plugging pi in well the cosine at pi is negative 1. negative sign at pi the first derivative is zero the second derivative negative cosine well cosine at pi is negative one the opposite of negative one is positive one and then the third derivative at pi is also zero and then you can start to see it the fourth derivative it's going to start to cycle around might be a little trickier to see the pattern of the derivatives off the bat here so we're going to go ahead and build a couple terms and see if that helps us see the pattern the first term is or the zeroth term i should say is negative one so cosine x equals negative one then our first derivative is going to be a zero so that all goes to zero for the second derivative we end up with one times the x minus pi to the second power divided by two factorial then a zero then another negative one x minus pi this is now the fourth derivative fourth power divided by four factorial and you start to see the pattern this time it's the even exponents times x minus pi the next one is going to be to the sixth power over six factorial and so on can we come up with a generic formula well it's an alternating series but this time the first term is negative so we need to stagger negative 1 to the n will make the first term positive negative 1 to the 0 is positive 1. so if we do n plus 1 that'll stagger us by 1. and then all we have left is the x minus pi to the oh what power now this time we end up with the zeroth term the second term the fourth term the sixth term this time we have the even exponents the even terms so to get even numbers we'll do 2 times n and we'll divide by that 2 n factorial as it keeps going so what we end up with is that the cosine of x is really equal to the taylor series as n goes from 0 to infinity of negative 1 to the n plus 1 times x minus pi to the 2n over the 2n factorial we should have that 2n in parentheses and this one is centered at pi because we did at pi at a equals pi what does that mean centered at pi as opposed to the mclaren series which was centered at zero well let's go back to the graph to discuss that notice when we started to build the sine of x as a mclaren series it was centered at zero and if i start deleting these terms off you see ultimately where the function started building was from the center of zero but if i do a cosine x i have to move the graph over a bit and now if i start building negative 1 plus x minus pi squared over two you start to see where it's building is right at pi if i do x equals pi and put a line here that's the center and now if i go minus x minus pi to the 4th power over 4 factorial that 2 should be factorial as well but that doesn't really make a difference you see it's starting to build that cosine but it starts building from the center of pi and if we keep going we'll get closer and closer to the actual function over eight factorial and we've almost got a full period now and if we keep going off to infinity you can see how we'd end up eventually with the entire graph forming from a center of pi so taylor allows us to start building the graph from any point mclaren always builds the graph from zero but let's take give you a chance to look at the homework assignment try and build some of these taylor and mclaren series get comfortable with what they are and how they work and then we will see you in class to discuss them more after looking at the conic sections what we realize is that we have non-functions that are relationships between x and y that we use quite often it would be nice if there's a better way to to express those relationships that are not functions and so what we're going to take a look at in the next couple sections is some different ways to express relationships between our variables we're answering the question is there a better way to describe relations that are not functions we're used to this whole xy relationship and x determines y and that's what allows us to graph our points and our function to represent all the solutions however there might be another way to represent functions what if instead of just having x dependent on y what if x and y were both defined to be dependent on another variable and this is where we get what are called parametric equations and the idea of a parametric equation is the x value comes from some function x of t and the y value comes from some function y of t where t now is the independent variable and what's really important is that independent variable moves over an interval in other words we're graphing x and y with relationship to time as time passes we have this movement of our x and y values and so the time can quickly determine the graph let's take a look at an example let's say we have a relationship where x of t is equal to three t plus two sometimes you just see x equals and y of t is equal to t squared minus 1 and negative 3 is the lowest value for t and 2 is the highest value for t what if we wanted to graph this parametric equation well what we can do is we can make a value a table of values for our t values going from negative 3 to 2 negative 3 negative 2 negative 1 0 1 and 2 and we can look at what happens to our x and y values so that we can graph what happens at each of these moments in time as the time progresses so when t is negative 3 we plug negative 3 into the x equation and so we end up with 3 times negative 3 which is negative 9 plus 2 is negative seven then we plug negative three into the y function when we plug negative three into the y function we get negative three squared which is nine minus one is eight then we plug the negative two in three times negative two plus two is negative four negative two squared minus one is three plugging the negative one in three times negative one plus two is negative one negative one squared is one minus one is zero plugging zero in we get two and negative one plugging one in we get five and zero and plugging two in we get eight and three and so what this does is give us the coordinate points that we can graph to see our function popping up some graph paper to help us graph this x starts at negative seven comma eight then it's negative four three then it's negative one zero then it's 2 negative 1 then it's 5 0 and then it's 8 3 and what we end up with is this graph turns out what we have is the forming of a parabola but what's really important to note is that the graph also has direction and that's something that's unique about parametric curves is you end up with direction to your graphs so this is our function x of t is three t plus 2 y of t is t squared minus 1. now sometimes it might be useful to convert our parametric equation and eliminate the parameter so that we can see the relationship between x and y a little more straight forward and really the way we can eliminate the parameter is we can solve one equation for the variable for the parameter t and substitute into the other function so let's take a look at some examples starting with x is equal to 2 plus 3 over t and y is equal to t minus 1. what would this look like as a regular function well the y equation is really easy to solve for t by adding 1 to both sides and then we can take that y plus 1 and replace the t with that y plus one when we do we end up with x equals two plus three over the y plus one and then we might wanna solve this for y because we're used to seeing functions solve for y so if we subtract 2 from both sides we can multiply both sides by y plus 1 and divide both sides by x minus 2 and subtract 1 from both sides to get y is equal to 3 over x minus 2 minus 1. and this would give us the exact same function but with the parameter eliminated let's take a look at a little bit more of an interesting function let's look at the parametric equation x equals 4 plus 2 cosine of t and y equals negative 1 minus 3 sine of t this is actually a very familiar graph to us let's see what happens when we work to solve it if it becomes something more familiar we can start by working with the x equation i'm going to subtract 4 from both sides and then divide by 2 so we've solved for the cosine of t and the same thing with the y i'm going to solve it for the sine of t by adding 1 and dividing by negative 3. one way we could attack this is to take the cosine inverse and then plug it into the sine that's going to be a lot more work than we need to do though because we know that sine squared of t plus cosine squared of t equals one so let's do that we're going to take the sine of t which is y plus 1 over negative 3 squared plus the cosine of t which is the x minus 4 over 2 squared must equal one and if we put that squared on top and bottom we end up with y plus one squared over nine plus x minus four squared over four equals one and we should recognize that function that is an ellipse that is centered at four comma negative one with a vertical major axis equal to the square root of nine is three double that at six and a horizontal minor axis equal to the square root of 4 is 2 and double that we get 4. so this is an ellipse centered at 4 negative 1 with a vertical major axis of 6 and a horizontal minor axis of 4 that is what the graph looks like but what we've done because this is a parametric equation is x and y depend on t the independent variable now the ellipse is defined as a function of t this is one advantage of parametric equations is we can start working with non-functions and define them in terms of this parametric variable this parameter t today's assignment is designed to have you get familiar with parametric graphs and converting between parametric and rectangular coordinates with the x's and y's take a look at practicing those and then in our next video we'll look at how we can do calculus when we're working with parametric equations now that we've taken a look at parametric curves and how to work with uh parameterized equations we're ready to take that up to the level of calculus and ask ourselves the question how do we do calculus with parametric equations and really calculus is very similar in parametric as it is in rectangular coordinates but there are a few adjustments that we're going to make first we're going to take a look at derivatives with parametric equations and the big thing with derivatives is normally when we find a derivative we call it d y d x well d y and d x are now separate equations so we have to take their derivatives separately so we'll take the derivative of the y equation with respect to t and divide it by the derivative of the x equation with respect to t in other words we're taking y prime of t divided by x prime of t to get d y d t so some examples of doing this would include if x of t is equal to t squared minus 3 and y of t equals 2t minus 1. so if we want to find the derivative of this the derivative is d y over d x so first we need to find d y well y prime the derivative of the y function with respect to t we know is 2. over the derivative of x with respect to t well x prime of t is equal to 2t so our denominator is 2t and when we simplify by reducing out the twos we get the derivative d y d x being one over t and this becomes the function of the slope of the tangent line at any point on this parametric curve with respect to t let's do one more example let's try x of t equals 5 cosine squared of t and y of t is equal to 5 sine squared of t well again we need the individual derivatives first we need y's derivative which we know is 5 times bring the 2 out front sine of t times the derivative of the sine which is the cosine of t using the chain rule there so really that's tan sine cosine and now the derivative of x with respect to t is we have the 5 times bring the 2 out front 2 cosine of t times the derivative of the inside which is negative sine of t and so if we clean that up we get negative 10 cosine t sine t and so when we want to find the derivative which is d y dx we simply plug those in the derivative of y was tan sine t cosine t over the derivative of x with respect to t which was negative 10 cosine t sine t and this time we get a lot of reducing that happens leaving just the negative in the denominator really we have one over negative one which is just negative one which means the slope of this function is negative one now when we were back in calculus one we didn't just want to find the derivatives because the derivative was the slope of the tangent line but we actually wanted to find that tangent line quite often through a given point so let's see if we can do that as well let's find the tangent line what you remember is y minus y1 equals m times x minus x1 where m is equal to the slope dydx at the point x one comma y one so if we have the function x of t is equal to t squared minus four t and y of t equals 2t cubed minus 6t and we want the equation of the tangent line at the point t equals 5. we can build this in much the same way first we're going to find the slope y prime of t is equal to six t squared minus six and x prime at t is equal to two t minus four and then the derivative d y over dx is equal to the y six t squared minus six over the x two t minus four and this represents the slope of the tangent line to any point we're specifically interested in the slope at t equals five so if we plug five into that we have six times five squared minus six over two times five minus four six times 25 minus six is 144 over six and dividing we get exactly 24. now we've got the slope of our tangent line we still need to go back and find what point that is actually the slope at so if we go back to our x of t equation and find out what's happening at x of 5 we'll have 5 squared minus 5 times 4 or 4 times 5 same thing which is 25 minus 20 which is 5. and y at that point of 5 is 2 times 5 cubed minus 6 times 5 which is 125 times 2 is 250 minus 30 is 220 so we also have this point then of 5 comma 220. now we have a point and we have a slope we're ready to make the equation of our tangent line which is y minus the y coordinate of 220 equals the slope of 24 times x minus the x coordinate of 5 and we now have the equation of the tangent line through this curve when time is equal to five let's do one more example let's say x of t is equal to the sine of t and y of t is equal to the cosine of t and we want the tangent line at t equals pi over 3. well the derivative of y with respect to time is negative sine of t and the derivative of x with respect to t is cosine of t so our derivative d y d x is equal to the negative sine of t over the cosine of t which we should recognize sine over cosine is tangent so we get the negative tangent of t and we want to know specifically what happens with that slope at t equals pi over three so what we're really working on is negative tangent of pi over three and the tangent of pi over three is the square root of three so we have negative square root of three is the slope of our tangent line at pi over three now we just need our point if we plug pi over 3 into the x equation careful not the derivative in the equation x is equal to sine of t so we have the sine of pi over 3 and the sine of pi over 3 is one-half plugging pi over 3 into the y function y is going to be cosine of pi over three which is root three over two which gives us the point one half comma root three over two to go with our slope of negative root 3 and make our tangent line which is y minus one-half the y-coordinate equals the slope negative square root of 3 times x minus the x-coordinate square root of three over two and we found the equation of our tangent line so it's kind of a brief overview of doing derivatives with parametric equations we can also do anti-derivatives with parametric equations and they just need a little bit of a tweak to what we've done before so let's take a look at the antiderivatives these would be our integrals and remember with anti-derivatives or integrals the fundamental theorem of calculus told us what we're really finding is the area under a curve and it turns out that if we want to find the area under a curve in parametric equations we still take the integral from a to b but we need to account for the fact that our dx's is split up into an x function and a y function with respect to t so we've got the y function that we're going to integrate with but when we use the chain rule with the dx we end up with an x prime of t dt also it's important to note that a is the left endpoint because sometimes with parametric equations the graph ends up being graphed from right to left in which case we need to switch the order of the integration or just take the opposite of the final integral we'll take a look at what that means when it comes up but for now we'll say if we want the area under the curve we take the anti-derivative from a to b the integral from a to b of the y function times the derivative of the x function dt here's an example of what that looks like if x of t is equal to 2 cosine of t plus the cosine of 2t and y of t is equal to 2 sine of t minus the sine of 2t we're going to find the area under the curve between 0 and pi well the first thing we need to know if we're going to use this function this operation to find the area under the curve is we need to know the derivative of the x function so first the x derivative the antiderivative of two cosine is negative two sine t and the anti-derivative of cosine of 2t using the chain rule is negative 2 sine of 2t so when we plug that into our integral function the area under the curve from 0 to pi of the y function which is 2 sine t minus the sine of 2t times the derivative of the x function which is negative 2 sine t minus 2 sine of 2t dt and then all we have to do is evaluate that integral to find the area under that curve this integral is going to take a little bit of work to work out so let's give ourselves a little bit more space to work with and let's see what happens first thing i'd notice is that i've got a gcf of negative 2 inside this last factor i can factor out that negative 2 and pull it all the way out of the integral so let's do that negative 2 times the integral from 0 to pi of 2 sine t minus the sine of 2 t times now it's positive sine t plus the sine of 2t dt and then to keep integrating let's take let's foil this out yeah let's boil this all out so we've got two sine squared of t plus 2 sine of t times the sine of 2t minus the sine of t times the sine of 2t minus the sine squared of 2t dt let's go ahead and combine the like terms in the middle so we have negative 2 times the integral from 0 to pi of 2 sine squared of t plus sine of t sine of 2t minus the sine squared of 2t dt a couple things that i notice here we each piece of this integral we need to do some work with using some trig formulas in order to simplify we see the sine squared of t and we remember that the sine squared of t is equal to one half times or minus one half times the cosine of double the angle two t so we're going to replace that sine squared with this property and what's nice when that 2 distributes through it's going to clear out those one halves so we end up with negative 2 times the integral from 0 to pi when the 2 distributes through we have 1 minus the cosine of 2t plus another thing i notice is we've got a sine of 2t we have a a formula from trig that says the sine of 2t is equal to 2 sine of t cosine of t and when we multiply that by the sine we'll end up with 2 sine squared of t times the cosine of t and that's set up now for a nice easy substitution where u is sine and du is cosine minus and again we see the sine squared so we'll use the sine squared formula again to say one half plus because we have to distribute the negative through one half cosine and we double the angle we get four t dt let's go ahead and combine there's not much for like terms but we do have one minus a half so now we have negative two times the integral from zero to pi of one-half minus the cosine of two t plus 2 sine squared t cosine t plus 1 half cosine of 4t dt and after all that algebra we finally have something we can integrate we have negative two times one half of t minus the antiderivative of cosine is sine of two t but we need a one half to take care of the two on the inside plus sine squared comes from a sine cubed but we need to divide by three to take care of that and then using the chain rule gives us the cosine you can check that using substitution plus anti-derivative of cosine is sine of 4t but we need to divide by a 4 which means we have a total of divide by 1 8. and we're integrating this whole thing from 0 to pi what's nice is when we plug pi into sine pi or zero both pi and zero of sine go to zero which means all the terms go away except for the very first one so we end up with negative two times one half of pi minus zero because when we plug in half of zero is zero and then when we multiply this together negative two times pi over two is negative pi but wait why is it negative we made a small error at the beginning and that we didn't take the time to notice if a was a left endpoint or a right endpoint if we start integrating at 0 we're claiming that 0 is the left endpoint let's take a look at this graph really quick what i've done here is i've asked desmos to graph this parametric curve for us as t grows from 0 to some value a and a is going to be a video that's going to scroll from 0 up to pi and i want to notice when this graph is drawn it's going to start here at 3 0 and as a gets bigger it goes off to the left this curve is actually drawn backwards until our time is equal to pi and this then becomes the curve that we're trying to find the area underneath down to the x-axis but because it was drawn from right to left the left-most endpoint was actually pi not zero which means we should have integrated not from 0 to pi but from pi to 0 to make it integrate from left to right that way our dt is positive well we could fix that easily at the end because if we switch the order of integration that would have stuck a negative out front well let's just make that adjustment at the end and say this is drawn backwards which means the area under the curve is actually equal to a positive pi so if it's drawn backwards you'll end up with a negative area and that should be your hint oh this is probably drawn from right to left we need to take the absolute value and say it's equal to the positive value because area can never be negative i want to take a look at one more use of integration of calculus with parametric curves we've talked about the area under a curve but let's actually talk about another interesting thing and that is i think this is number two arc length arc length was generally one of the most difficult formulas to work with in rectangular coordinates and there are some situations where arc length is actually easier to work with in parametric equations we just need a slightly different formula to account for the fact that we have an x and a y function to find the arc length it's going to be the integral from a to b of the square root of and instead of 1 plus the derivative squared it's going to be dx dt squared plus d y dt squared it really comes from the pythagorean theorem the change in the x plus the change in the y squared dt one thing we need to be careful of as we use this formula and it's a property from precalculus that we don't spend a lot of time on but it does become important here is when we take the square root of something that's been squared it's not actually equal to that something it's actually equal to the absolute value of that something and so sometimes if you get a weird answer we might need to go back and say let's take the absolute value of what was under the square root instead of what was what we ended up with it doesn't become an issue often but when it when you get a weird answer like an arc length of zero something like that that's probably what happened all right let's try an example let's say x of t is equal to 3t squared and y of t is equal to 2t cubed and we're going to find the arc length as t runs from 1 2 3. well to use the formula we need to take both derivatives and square them so let's take the derivative of x which is 6 t and when we square the derivative of x we get 36 t squared with the y's the derivative of y is 6 t squared and when we take that derivative and square it we'll get 36 t to the fourth so using our formula we're going to take the integral as t runs from 1 to 3 of the square root of the x derivative squared which was 36 t squared plus the y's derivative squared which is 36 t to the fourth dt and if we can integrate this guy we'll end up with our arc length from one to three let's do a little simplifying with this we've got a common factor of 36 t squared that can come out that's going to leave behind one plus t squared dt and we can take the square root of 36 t squared it's 6t times the square root of 1 plus t squared dt and we're set up then for a nice substitution where u is equal to one plus t squared d u is equal to two t dt well to get that six to be a two we're going to divide it by three so we'll divide by three inside and multiply by three outside so we end up with three times the integral make sure we plug our limits of integration into the u one plus one squared is two one plus three squared is ten and we're left with just the square root of u or u to the one half d u which is equal to three times u to the three halves times the reciprocal which is two thirds integrated from two to ten the threes divide out and so we're just left with two times u which will start by plugging in 10 to the three halves minus plug the two in two to the three halves and that's not going to be any pretty numbers so let's just leave our arc length as that value 2 times 10 to the 3 halves minus 2 to the 3 halves and that's how we can find arc length in parametric curves so we've taken a look at both derivatives and anti-derivatives specifically in the context of tangent lines and arc lengths and area under curves take a look at the homework assignment to practice a few of these and we will talk about them more in class see you then now that we've taken a look at parametric equation there's an alternate way to graph x and y coordinates using two different functions we're going to look at another option for graphing in a completely different graphing system the question is going to be what other graphing systems can we use normally when we've graphed a point up to now we've said we're going to graph a point really with an x y coordinate where x is how far we go to the right and y is how far we go up however we could graph using a different version of reference points and this leads to what are called polar coordinates the idea behind polar coordinates is we've got the coordinate system and instead of giving an x y coordinate what we're going to do is we're going to have some distance which we will call r that is going to be some angle theta away from the polar axis or the what we traditionally have called the x axis to get to our point so a point is now made up of an ordered pair of r comma theta where r represents the radius or distance from the center and theta represents the angle that we open up to now what's interesting about polar coordinates then is unlike the x y coordinate system where there's only one way to get the to the point four comma seven with polar coordinates there are multiple ways to get to the same point they are not unique coordinates to represent the same point let's say we want to look at a point over here and if i were to draw it out let's say that radius has a distance of 2 and the angle is a distance of pi over 4 a 45 degree angle if that's the case then our radius is 2 and our angle is pi over 4 and that's going to represent that point but that's not the only way to get to that point i could have taken my angle and gone around a full circle and a little bit more 2 pi plus another pi over 4 would be 9 pi over 4. and so we end up with the same radius of 2 but an angle of 9 pi over 4 representing the exact same point in fact there's a third way to represent to that exact same point and that is to take an angle that goes around to the distance exactly opposite it that angle would be 5 pi over 4 but then our radius is going to come out the other direction and to show the radius coming the other direction we'll say the radius is negative and so we end up with this point negative 2 comma 5 pi over 4 and we've got three ways to represent the exact same point in fact there are infinite number of ways that we could represent this exact same point including negative angles and angles that are several times around the entire unit circle so polar coordinates are not unique but they do provide a way to do a lot of graphs in a much more convenient way but before we do that let's take a look at how we can do conversions between polar and rectangular coordinates and to set these conversions up let's take a look at a point that is drawn by some radius r and some angle theta and if we were to drop a triangle on that we end up with an x distance to the right and a y distance up well we know that the cosine of theta the cosine of the angle is the adjacent x over r and multiplying by r tells us that the x coordinate is equal to r cosine theta similarly if we took the sine of the angle it would be the opposite over r or y over r and multiplying both sides by r we get r sine theta is equal to the y coordinate we also have ways to convert the other direction we know from the pythagorean theorem on this exact triangle that x squared plus y squared is equal to r squared and so if we know x and y we should be able to calculate r also if we know x and y we can use the tangent of theta to equal the opposite over the adjacent or y over x and so tangent of theta is equal to the ratio of our coordinates let's see if we can use these conversion properties to help us convert maybe the point 3 comma negative 4 to polar coordinates well we've got the x and y so we know that r squared is equal to x squared plus y squared so r squared is equal to 3 squared which is 9 plus 4 squared which is 16. so r squared is 25 and our radius must be 5. we also know that the tangent of the angle is equal to the ratio of y over x negative 4 over 3. and if we take the tangent inverse of both sides it's not a pretty number but we get negative 0.9273 approximately and so we end up with the polar coordinate of 5 comma negative 0.9273 for our polar coordinate that's equal to 3 comma negative 4. let's go the other direction let's convert the polar coordinate four comma two pi over 3 to a rectangular point well to do that we know x is equal to r cosine theta and we know that y is equal to r sine theta from our formulas up above so now we just have to plug in what we know that x is equal to the radius of four times the cosine of two pi over three or x is equal to four times cosine of two pi over three is negative one half so the x coordinate is negative two for the y coordinate y is equal to the radius of four times the sine of two pi over three so y is equal to four times the sine of two pi over three is root three over two which reduces down to two root two and we end up with the rectangular coordinate of negative two comma 2 root 2 being the exact same as the polar point 4 comma 2 pi over 3. now that we've taken a look at how we can convert between rectangular and polar coordinates we're ready for our next task which is to see if we can actually plot points and we're going to plot two points on this graph we're going to plot the point 4 comma 5 pi over 3 and we're going to plot the point negative 3 comma negative 3 pi over 2. now when we're graphing in polar coordinates we can convert them to rectangular coordinates and figure out where they should be or we can just graph them using our angles in what's called a polar grid this here is our polar grid and each circle represents a radius of one so we can kind of count out just like we count on our x-axis of one two three four five and so on and we can see the growing radii coming out of the center here on this polar grid every pi over six has been labeled so we've got zero pi over six two pi over six which is pi over three then we end up with pi over two then two pi over 3 and 5 pi over 6 and then pi and then 7 pi over 6 and then 4 pi over 3 and then 2 pi or 3 pi over 2 and then 5 pi over 3 and 11 pi over six and then we're back around to two pi and so you can see you're working yourself around this one does not label the pi over fours but those would be just kind of in the middle so if you wanted to you needed a pi over four you could stick it right up in the middle and you could label the pi over fours in this graph as well just kind of like a unit circle but with a different radius established for each coordinate point so when we want to plot point a as four comma five pi over three i'll go over to an angle of five pi over three and i'll count out four radii one two three 4 and that fourth radius out from 5 pi over 3 is point a similarly with negative 3 negative 3 pi over 2 negative angles count backwards so we're going to go the opposite direction i'll do this one in green because we did the first one in blue going the opposite direction we've got negative 3 pi over 2 if you count 1 2 3 negative 3 pi over 2 is up here at the top but instead of coming vertically up towards negative 3 pi over 2 because the radius is negative we're going to count backwards to 3 points and that gives us a negative 3 radius because we're going backwards away from the negative 3 pi over 2 and there becomes point b so if that's how we plot points then we're one step away from being able to actually graph polar equations let's take a look at that let's take a look at polar graphs the first polar graph we're going to look at is r equals 4 plus 4 cosine of theta and to set this up we're going to take a bit of time with this so we can get good comfort with how these polar graphs work we're going to look at all the common angles theta and see what radius they give us and see if we can graph what shape this is going to give us our common angles that we're going to look at is we're going to count every pi over six it does skip the pi over fours and we could include the pi over fours if we wanted to but uh for the sake of time i'm just going to do zero one pi over six pi over three pi over 2 2 pi over 3 5 pi over 6 pi 7 pi over 6 4 pi over 3 three pi over two five pi over three and eleven pi over six and then 2 pi would be coming back around to the same point so i'm going to use my calculator to take a look at these so let's pull up our calculator on your calculator in order to graph polar equations we need to change the mode that we're in if we click mode down a couple spots you see function is highlighted next to function is par that's for parametric equations if you want to use your calculator for those and the next is the polar equations let's go down to polar equations and hit enter to highlight polar equations and now when i hit the y equals instead of saying y equals it's actually going to say r equals and we want r to equal 4 plus 4 cosine of theta and now when we hit the x button the x t theta n we're going to end up with a theta for our variable and then i'm going to hit second table and i'm going to delete all of these values out that i don't want and now i can start entering in my values i enter in 0 and i see that the radius should be 8. if i enter in pi over 6 we see the radius should be 7.5 approximately and then i can enter in pi over 3 and i see the radius should be approximately six and so i can go down and finish filling in to figure out what values i want to fill into my table so when the rate when the angle is zero the radius ends up being eight pi over six is approximately 7.5 pi over 3 is approximately 6 pi over 2 is 4 2 pi over 3 comes out to 2 5 pi over 6 comes out to 0.5 pi is zero seven pi over six is point five four pi over three is two three pi over two is four you start to see it's kind of looking symmetrical five pi over 3 is 6 and 11 pi over 6 is 7.5 and you can see it's going to go to 8 next so let's see what this graph looks like when the angle is 0 the radius is eight so i'll count out one two three four five six seven eight on a zero radius then at pi over six that's the first line it's 7.5 pi over three it's six pi over two is four two pi over three is two five pi over six is one half and then at pi it actually comes down and hits zero and then one half and two and 4 and then 5 pi over 3 it goes out to 6 and 11 pi over 3 it comes out to 7.5 and then you can see it's gonna go back around to eight and so what we end up with is this interesting little curve that comes in it almost looks like a squished heart actually i'm going to make that line a little bit thicker so we can get a real good c on what what that graph looks like here it curves in and then balances out and that's our polar graph that is the function r equals 4 plus 4 cosine of theta and actually we can end up with some pretty gnarly looking graphs in polar coordinates to save us some time we're going to cheat and we're going to look at some graphs on desmos and kind of graph what these things look like first we're going to graph theta equals pi over 3. actually this pi over 3 option let's see if we can kind of talk through what it's going to do pi over 3 means the angle is going to stay consistent at pi over 3 regardless of what the radius is equal to theta equals pi over 3 is just a line so let's see if we can find something a little more interesting than that let's do r equals 2 cosine theta minus 3 sine of theta when we graph that r equals 2 cosine theta minus 3 sine theta what we end up with is a kind of off-centered circle this is the equation of a circle kind of was off-centered but it was a circle nonetheless let's take a look at another equation this one's kind of gnarly let's take a look at r equals theta over three r equals theta over three gives you a gnarly spiral and as you scroll out you see this only goes up to 12 pi yeah the spiral only stops here because theta has been programmed to only go to 12 pi but if theta keeps going the spiral is going to keep going out and out and out so theta equals pi over 3 is actually equal to a graph of a spiral which is kind of cool how about this one r equals 3 plus 3 cosine of theta r equals 3 plus 3 cosine theta that's kind of similar to the one that we saw earlier it looks like a squished heart something like that how about this graph r equals 2 plus 4 sine of theta this one's interesting because it ends up with a little squiggly in the center a little tiny circle on the inside so that one's kind of neat in that it comes around and then it loops around before it finishes in the center kind of the classic polar graph is what's called the rows r equals 3 sine of 2 theta let's take a look at this graph this one gives us a nice little rose with four petals on it we can actually play with the number of petals by changing what's in front of theta turns out if the number's even you end up with twice as many petals so if i do six theta i've got a 12 petal rose but if it's odd we end up with exactly that many petals so if i do 5 theta we get a 5 petal rose so you can play with that and get some gnarly looking graphs in polar coordinates now it might be useful though to change a polar graph into a more familiar rectangular coordinates and that takes a little bit of creative spark so let's take a look at how we can do one last thing and that is letter c to transform polar to rectangular functions let's start by doing theta is equal to pi over 3. well we know this is probably going to be a line based on our little experiment with functions up above in desmos where we said if that angle is going to stay constant and the radius is going to change you're just going to end up with a straight line but to find out what that line is what we know about theta is that the tangent of theta is equal to y over x by definition so we can take the tangent of the angle pi over 3 and that must be equal to y over x and the tangent of pi over 3 is the square root of 3 equal to y over x and so if we multiply both sides by x we end up with x square root of 3 equals y and that's a line through the origin with a slope of the square root of three that one's not too exciting let's try another one let's do r equals three well if the radius is going to stay consistent and the angle is going to change you might expect this to be a circle because we keep the same radius and go all the way around the unit circle we also know we've got a property that x squared plus y squared equals r squared so plugging in what we know x squared plus y squared equals three squared or nine and that really doesn't need any simplifying we recognize that from our conic sections as a circle with radius of 3. so let's try a more interesting graph than just r equals or theta equals let's try r equals 6 cosine theta minus 8 sine of theta now this one we're going to have to be a little more creative with a nice little trick that can help us quite often is we're going to multiply both sides by r and when i do that we end up with r squared equals and when i distribute the r r cosine theta minus eight r sine of theta and the reason that's nice is we know that r cosine theta is equal to x and we know that r sine of theta is equal to y and we know that r squared is equal to x squared plus y squared and so if we put that all together we get x squared plus y squared equals 6x minus 8y now it might not be clear what shape this is exactly so let's move all the x and y's to the same side so we have x squared minus six x plus y squared plus eight y equals zero and then if i try and complete the square on the x's by adding nine half of 6 squared and adding 16 half of 8 squared to both sides the x's are a perfect square x minus 3 squared plus the y's are a perfect square y plus 4 squared equals 0 plus 9 plus 16 is 25. you recognize this as a conic section as well it is a circle centered at 3 negative 4 with a radius of five let's do one last function let's consider r equals secant theta tangent theta well we don't really have any formulas with secant and tangent but we do know that secant is one over the cosine and tangent is sine over cosine and if i multiply both sides of this equation by the cosine squared of theta to clear out the denominators we get r cosine theta squared equals the sine of theta and where this gets interesting is if we multiply both sides by r that's kind of our favorite trick of polar coordinates to multiply both sides by r we get r squared cosine theta equals r sine theta and we should recognize that r sine theta is just y and r cosine theta is x and if it's all squared that must be x squared so we really have is x squared equals y is the same function as r equals secant theta times tangent theta this is a parabola so this video was a quick introduction to this new coordinate system of polar coordinates where we can graph things using a radius and an angle theta take a look at the homework assignment to practice some of these and good luck to you let me know if you have any questions and we will see you in class now that we've taken a look at using polar coordinates we're ready to do some calculus with polar coordinates as we answer the question how do we find areas in polar coordinates or polar curves would be more accurate before we actually get to working with area in polar coordinates i want to go over some tips on how we can use the ti 83 or 84 calculator to help us with finding the correct areas in the polar curves because it's really easy to find the incorrect area the wrong part of the graph so the first thing that we want to do is we need to make sure that the mode is set to polar so you'll hit the mode button and make sure polar is selected and what's nice about that as we found out in our previous video is now when you hit the y equals button it will give us r equals equations when we're going to graph these though a little tip is when you graph a polar graph i would not recommend that you do not hit the graph button it's a button for the first drawing because some of these polar graphs are too large or too small you're not going to see much in hitting just the graph button so we're going to say rather we're going to hit the zoom button and select zoom fit and what zoom fit will do is it will center the graph as large as possible on the screen so we'll use zoom fit to get a good view now sometimes it skews this graph when we do this maybe the x's will count by twos and the y's will count by sevens and so we end up with a significant skew but it does provide us a really good view of what the graph looks like something else to be aware of is the window button because the window button not only gives you a chance to set the minimum and maximum values for x and y the window buttons will also allow you to select a range for theta for the angles and then if you adjust that to reset it we will use 0 to 2 pi that's kind of the default so whenever we change it we need to remember to change it back to 0 to 2 pi otherwise some weird things might happen so let's take a look to see if we can correctly use the calculator to graph one petal of r equals three sine of two theta three sine 2 theta is your typical rows we're going to graph it first so i'll pull up my calculator here first we'll go to mode to check to see if it is in polar coordinates which it's not right now so i'm going to scroll over and select polar coordinates second quick to go back home now when i hit the y equals button i have r equals equations let's delete out the old equation and we want to graph 3 sine of 2 theta 3 sine of 2 theta and when we close the parentheses that will give us the graph but instead of just hitting graph because i'm probably going to get a bad view of it this first time i draw it i'm going to click the zoom button and i will scroll down to find zoom fit it's number zero on my calculator depending on the version of your calculator you it might be a different number so if you hit zoom and then that number it'll automatically zoom you don't have to scroll so it might be worth memorizing what number it is on your calculator so i can hit enter and it'll quickly graph its four petaled rows our task is to try and graph only one petal let's try and graph this first petal what i notice is it starts and ends at the origin at that point regardless of the angle the radius i know is zero that's what pulls it in towards the center so we need to figure out what angles give us a radius of zero so we can just graph one petal well to do that we'll go back to our equation and we'll set that radius to zero equals three sine of 2 theta and solving this for theta will tell me what angles give me that center point dividing by 3 gives me 0 equals the sine of 2 theta and i know from my unit circle that the sine is equal to 0 at the left and right points so 2 theta is equal to maybe 0 1 pi 2 pi 3 pi and so on and so forth dividing everything by 2 then my theta my angles are equal to 0 pi over 2 pi 3 pi over 2 and so on and so forth we just want one petal so let's just connect from 0 to pi over 2. we're going to take theta and stick it in between those two values and hopefully that will give us exactly one petal of the rose when i hit window i see the x min and or the theta min and the theta max these are our range for our theta so we're going to go from 0 enter to pi over 2 and if i type in pi and divide by 2 and hit enter it's going to give me a decimal approximation for that and now i can hit graph because it's already been graphed for me and what i notice is that does give me exactly one petal of my rose like we were hoping for so all that to give you kind of an orientation to how to get what we want on the calculator let's see now if we can use that to help us actually do what we wanted to do which was to find areas to find an area in polar coordinates is similar to rectangular coordinates in that we're going to take an integral from a to b but a slight difference is we're going to take half the integral from a to b of the radius squared d theta that is going to be the big formula for today half the integral from a to b of r squared d theta so going back to the example we were just looking at let's see if we can find the area of one petal of r equals 3 sine of 2 theta well to do that we'll take one half times the integral and we need to usually do some work to find our a to b however we just found out in our previous example that that theta is going to go from 0 to pi over 2 to give us one petal of the radius squared so 3 squared is 9 sine squared of 2 theta d theta this integral then should give us the entire area of one petal i'm going to pull that 9 out because it's a constant gives us 9 halves times the integral from 0 to pi over 2. then we're integrating sine squared and we know we can't really integrate sine squared but we have a nice property that says that's one half minus one half of the cosine of double the angle four theta d theta and now i can integrate that to get nine halves times half of a theta minus the antiderivative of cosine is sine but we have to divide by that 4 so we actually end up with 1 8 sine of 4 theta and that's going to be integrated from 0 to pi over 2. so we have 9 halves times one half of theta which is pi over two minus one eighth of the sine of four times theta four times pi over two is two pi minus one half times zero plus one eighth times the sine of 4 times 0 which is 0. and when we simplify all that we get 9 halves times pi over four and all the rest of the pieces actually go to zero so for our final area of one petal we get nine pi over eight so that's how we can find area with these polar coordinates let's do one more example to make sure we've got it let's find the area inside r equals 1 minus the cosine of theta well first you get an idea of what's going on we'll go back to our calculator we'll clear out the old function we're doing 1 minus the cosine of theta under window remember we've got to reset our min and max otherwise weird things happen we want to go from 0 to 2 pi as a default and then we'll zoom on my calculator hitting zero will give me a zoom fit your calculator might have a different number and this is kind of the shape we're trying to find the area inside of well what we see is it seems to start and end again where the radius is 0. so again we're going to figure out when that radius is 0 so 0 equals 1 minus cosine of theta if i add cosine of theta to both sides we want to know where the cosine is equal to one and if i think about my unit circle cosine is equal to one over here on the right so theta is equal to 0 2 pi 4 pi and so on but we just want to get one loop of it which goes from zero to two pi so we want theta to range from zero to two pi and that's where we set the defaults so we don't need to adjust it at all we're ready now to go to our formula that says the area is one half times the integral from a to b which we just found out was zero to two pi of the radius squared one minus cosine theta squared d theta and it might be worth on this one to go ahead and square that out 1 minus 2 cosine theta plus cosine squared of theta d theta we recognize that cosine squared we should be really familiar with what to do with that one from 0 to 2 pi of 1 minus 2 cosine theta plus let's break this up to one half plus one half cosine of two theta d theta we could combine like terms if we wanted to but there's not much combining that happens so let's go ahead and take the antiderivative which is going to be one half times a theta minus two sine theta plus one half of a theta plus the antiderivative of cosine is sine of two theta but to account for that 2 theta we have to divide by 2 which gives us 1 4 and we're integrating from 0 to 2 pi plugging that in then we get one half times theta which is 2 pi notice when we plug 2 pi into sine we'll get 0. actually i shouldn't cross it off just in case we get something on the subtraction step so we have two pi plus one half of theta one half of two pi is just one pi and one fourth sine of two two times two pi which is the sine of four pi which is also zero and then when we plug zero in everything goes off to zero the sine of zero is zero theta is zero half of zero is zero so we end up with a total of three pi over two is the area inside that curve one minus the cosine of theta so as long as we've got this important formula down one half integral from a to b of r squared d theta the integrals the area inside these polar curves is not too difficult to find we can do one twist on it though and instead find the area between curves and just like we found the area between curves with rectangular coordinates by subtracting the top from the bottom we're going to do much the same thing to find the area between curves we want to find the area outside of r equals 2 plus 2 sine theta and inside r equals 6 sine theta well first we need to get an idea of what shape we're looking at so we're going to use the calculator to help us with that clear out the old function the first function that we want to be outside of is 2 plus 2 sine of theta and we want to be inside 6 sine of theta as i graph this i'm going to take very careful note of which graph graphs first because i want to be outside of the 2 plus 2 sine theta i want to be outside of the first curve and inside the second curve the 6 sine theta so as it graphs i need to be very aware of outside the first inside the second let's go ahead and hit zoom fit which is zero on my operating system outside that first one inside the second one so i want to be outside the first curve and inside the second one it's kind of this sideways half moon on top is what i want to be inside of let's kind of draw a quick picture of it so one function was this half circle the other function was kind of this upside down heart looking thing but the top function the blue function is the six sine theta i want to be inside that one and outside the 2 plus 2 sine theta which means the area we're finding is this kind of sideways half moon on top which means we really need to know what are the two points of intersection so i know where to integrate from and to well those lines intersect where the radii are equal so we'll set the two functions equal the first is two plus two sine theta the second is six sine theta subtract 2 sine theta from both sides to get 4 sine theta divide by 4 and one half equals the sine of theta i think about my unit circle sine the y coordinate is one half at pi over six and five pi over six so theta is equal to pi over six and five pi over 6. to test that let's set our minimum and maximum on our calculators hitting window we're going to go from pi over 6 to 5 pi over 6 and now when it graphs it's going to graph the exact shape that we want to find the area of we're ready to integrate we know that the area is found by taking one half times the integral from our limits of pi over six to five pi over 6 and we know from our prior study that we take the outside function minus the inside function so the outside function is 6 sine theta but remember in polar coordinates we have to square them so we have 36 sine squared of theta and we'll subtract the other function which is 2 plus 2 sine of theta squared d theta we now have the integral that we can use to find the area between these curves a little simplifying as we square out the binomial so one half from pi over 6 to 5 pi over 6 of 36 sine squared theta minus that minus has to distribute through 2 squared is 4 minus 8 sine theta minus 4 sine squared of theta d theta i'm gonna combine like terms so we have one half times the integral of pi over six to five pi over six 36 minus four is 32 sine squared of theta minus 4 minus 8 sine of theta d theta noticing here that all my terms in that integral are even i'm going to go ahead and distribute that one half through the integral so i don't have to worry about it anymore pi over 6 to 5 pi over 6 of 16 sine squared theta minus 2 minus 4 sine theta d theta now i'm going to break up that sine squared and we're going to multiply by 16. so when we have the one half sixteen times a half is eight minus sixteen times a half is eight cosine of 2 theta minus 2 minus 4 sine theta d theta i'm going to combine like terms again pi over 6 to 5 pi over 6 8 minus 2 is 6 minus 8 cosine of 2 theta minus 4 sine of theta d theta and now we're finally ready to integrate we end up with six theta minus the antiderivative cosine is sine of two theta when we divide by two we get four minus the antiderivative of sine is negative cosine so we have 4 cosine theta integrated from pi over 6 to 5 pi over 6. so let's plug it all in we have six times the five pi over six minus four sine of two theta which is five pi over three plus four cosine of theta which is five pi over six minus six times the pi over six plus 4 times the sine of 2 times pi over 6 which makes it pi over 3 minus 4 times the cosine of pi over three oops sorry pi over six so to help us out we might want to think about our unit circle we need to know a 5 pi over 3 which is going to be down in the fourth quadrant this is 5 pi over three it has coordinates of one half comma negative root three over two we're also going to need to know about five pi over six which is over here five pi over six that has coordinates of negative root three over two comma one half we need to know about pi over three which has coordinates of one-half comma root 3 over 2 and pi over 6 is another one we need a lot of them on here that one's pi over six has coordinates of root three over two comma one half all right plugging it all in then we have the let's switch colors six is divide out so we have 5 pi minus 4 times the sine of pi over 5 pi over 3 assigns the y coordinate which is negative making it positive root 3 over 2 plus 4 times the cosine of 5 pi over 6 cosine's the x coordinate making it negative root 3 over 2 minus the 6's divide out just pi plus 4 times the sine of pi over 3 sine being the y coordinate root 3 over 2 minus 4 times the cosine of pi over 6 cosine being the x coordinate root 3 over 2 and this simplifies quite nicely because we've got a couple opposites in there 4 root 3 over 2 and negative 4 root 3 over 2 is 0. 4 root 3 over 2 and negative 4 root 3 over 2 is 0. so we're just left with 5 pi minus pi leaves us with 4 pi the area that we're looking for on that curve beneath the 6 sine theta and above the 2 plus 2 sine theta area there is four pi let's do one last example make sure we've got a really good grip on this number two we're gonna find the area inside the curve r equals 4 cosine theta and outside the curve r equals 2. let's take a look at our calculator all right first thing we need to do is plug this formula in so let's delete out these old formulas from the previous problem our first one is 4 cosine theta which means the first graph we want to be inside of so we'll take note of the first graph the second graph of 2 we want to be outside of for our window we have to reset it to 0 to 2 pi and then i'll zoom fit so we get a good picture we want to be inside this first curve on the right outside the second curve on the left so we're really looking for the right area here shaded so if we draw a quick picture of what we're working with the first function was the circle on the right the second function was the circle in the center and if you can excuse my crude drawing you see we want the area the right part of that which means again so we know our limits of integration we need to find where those points intersect that's where our functions are equal so we will say four cosine theta equals two divide by two cosine theta is one half so if i think about my unit circle where is the cosine one-half the x-coordinate is one-half up on top and near the bottom that happens at pi over three and 5 pi over 3 so we're going to say theta is probably going to go from pi over 3 to 5 pi over 3 but just to make sure we'll go back to our calculator and check to make sure that's going to give us the area on the right that we're looking for we'll click window for the minimum we want to go from pi over 3 to a maximum of 5 pi over 3 and when we hit graph what you notice is something unexpected happens it does not give us this section to the right we want it gave us everything else that we didn't want so we were going from pi over 3 to 5 pi over 3 and that didn't work maybe what we want to do is go the other direction and get that other grouping well we can't go from 5 pi over 3 to pi over 3 but what we can do is continue around the circle and after 2 pi the next top point 6 pi over 3 one more pi over 3 is 7 pi over 3 because remember our trig formulas have an infinite number of solutions the next point we could try is 7 pi over 3 and we're probably going to be between those values so let's take a look at the graph and see if going from a window of 5 pi over 3 to the next lap around at 7 pi over 3 does that give me the graph i want hitting graph you see it does give us that half moon shape that we were looking for so these now are our limits of integration for this problem now that we have our limits of integration we're ready to use our formula that the area is one-half times the integral from 5 pi over 3 to 7 pi over 3 of the outside function which was the blue one before cosine theta squared makes it 16 cosine squared theta minus the outside function that we crossed out which is just 2 squared becomes 4 d theta and now we have the integral we need to solve to find the area between these two curves again one thing i notice is all the terms are even so i'm going to go ahead and distribute that one half through so we have the integral from 5 pi over 3 to 7 pi over 3 of 8 cosine squared theta minus 2 d theta and it seems like the trig formula that we use the most anymore is that cosine squared or sine squared so we're going to integrate from 5 pi over 3 to 7 pi over 3. cosine squared gives us a one-half but when multiplied by 8 that leaves us with 4 plus 1 half times 8 is 4 cosine of 2 theta minus the 2 still d theta let's go ahead and combine the like terms of four minus two so we're integrating five pi over three to seven pi over three of two plus four cosine of two theta d theta and now we're ready to actually integrate when we integrate we get 2 theta plus the antiderivative of cosine is the sine of two theta we have to divide by the two which just leaves us with two sine of two theta integrated from five pi over three to seven pi over three so when we plug that in we get two times our theta which is seven pi over three plus two times the sine of two theta which is fourteen pi over three minus two times theta which is five pi over three minus two times the sine of two theta which is ten pi over three let's think about our unit circle to help us with those strange ones we've got to figure out 14 pi over 3 and 10 pi over 3. that's going to loop us around the circle a couple times so to help us out i'm going to count in three pies so we've got zero three pie over three six pie over three nine pie over three and twelve pi over 3 looping us around the circle over and over again uh that should help us see that 14 pi over 3 is right here 14 pi over 3 is going to be the same as 2 pi over 3. so the coordinates there are negative one-half comma root 3 over 2. and later when we get to the 10 pi over 3 10 pi is a little more than 9 so it's going to be down here 10 pi over 3 turns out to be the same as 4 pi over 3. so the coordinates there are negative one-half negative root 3 over 2. so when we simplify to bring it all together we get fourteen pi over three plus two times the sine of fourteen pi over three sines the y coordinate root three over two times two gives us just a root three minus 2 times 5 is 10 pi over 3 minus 2 times the sine of 10 pi over 3 sine being the y coordinate is going to be a negative negative times a negative is a positive root three over two times two is just root three combining like terms fourteen minus ten is four pi over three plus root 3 and root 3 is 2 square roots of 3 which means the area between these two circles is going to be 4 pi over 3 plus 2 square roots of 3. so now it's your turn to take a look at some of these on the homework assignment to practice them get really good the key formula here for the area we've been using it all the way through is that the area is one half times the integral from a to b of the radius squared d theta try a few of these and we'll look at them more in class this video is going to start a chapter that goes a completely different direction than the sequences and series that we've been focusing on up to this point we're going to start by looking at these things called conic sections answering the question what are conic sections and we're going to just barely touch the surface of what a conic section is each one of these different conic sections we could probably spend a whole day or more on and instead we're going to do them all in one day so we're just going to hit the highlights there are several conic sections the first one is the circle and we actually have a very precise definition for the circle it is all points that are equidistant from a point so if i have a point here in the center and i have some set distance we'll call that distance r no matter where that set distance goes that distance is going to be the same and if we collect all of the points that are exactly a distance of r away from that point we should get a perfect circle with radius r the equation that builds a circle like this is x minus h squared plus y minus k squared equals r squared where the point h comma k is the center of the circle and r is equal to the radius of the circle so if i wanted to find the equation of a circle centered at five comma negative two with radius 3 we could build that knowing that the center is h comma k and the radius is that radius of 3. so my equation becomes open a parenthesis x minus h which is the x coordinate of the center 5 squared plus y minus k which is the y coordinate of the center subtracting a negative 2 is the same as adding 2 squared equals the radius squared the radius is three three squared is nine and this gives us the equation of the circle centered at five negative two with a radius of three now a circle is the most basic of the conic sections and we don't actually spend much time on it because of how basic it is so let's go to the next most interesting conic section and that is the parabola a parabola has a very specific definition also that has to do with distance as well a parabola is all the points with a distance from a point and we call this other point the focus equal to the distance from a line and that line we call the directrix so what this means is we've got this parabola there is a focus inside the parabola and there is a directrix outside of the parabola and the idea of a parabola is if i take the distance from the focus to the line let's call that a it's going to be the same as from the parabola to the directrix that distance is going to be the same now it's going to be a different distance if i go maybe more diagonal and slightly to the right we'll call that b but then we'll get the same distance as you drop down to the directrix b so wherever this line is drawn c that length is the same as the length to the directrix c those lengths are always the same now this drawing is not to scale and not a perfect parabola so it those don't look quite the same but if it were a perfect parabola those distances from the focus to the graph would be the same as the distance from the graph to the directrix for the equation of the parabola it is based on the direction of the parabola the most common type of parabola that we are familiar with is the parabola that opens upwards the parabola that opens up where it has a vertex we'll call that vertex at the point h comma k and then up above the vertex is the focus and then below it is the line that is the directrix the equation of a parabola like this is y equals 1 over 4p times x minus h squared plus k where h comma k is that vertex point also it's important to note that this letter p p is the distance to either the directrix or focus depending on which direction we're going from the vertex and we already said the vertex h comma k is the vertex coordinates and the equation y equals since we drop from the vertex down to the directrix we can take that k coordinate of the vertex and subtract the p to get the directrix also of note just a little side note if the parabola is upside down the only difference is we make the whole thing negative so we would have y equals negative 1 over 4p times x minus h squared plus k exactly the same but the negative in front would make the parabola go the other direction so that's the first direction of a parabola the second direction we have a parabola actually goes the other direction it goes opening to the right with a vertex still at our point h comma k but now the focus is off to the right and the directrix is vertical this equation is x equals 1 over 4p [Music] times y minus k squared plus h and p on here is still the distance to either the directrix or the focus from the vertex so again that distance between them is p just like it was before h comma k is still the center notice the k is always with the y so it's y minus k here but that is the vertex and the directrix equation then is an x equals equation it's x equals h minus that p that distance from the vertex to the directrix and again we have a little side note that if the parabola opens off to the right that just means we have a negative in front of everything so it's x equals negative 1 over 4p times x minus k i'm sorry y minus k squared plus h so with the parabola the equation is based on the direction so with this in mind let's see if we can find some equations of parabola first let's say we want a parabola that has a focus at three comma two and directrix at y equals 8. i always try and draw a rough sketch of what's happening to help me decide what type of equation i'm using if i look at the point 3 comma 2 for the focus 3 comma 2 is the focus and the directrix is a y equals equation but it's y equals eight y equals is a horizontal line the parabola then needs to go between them and the only way we can get between them is to have our parabola come up kind of like this with our vertex in the center so the focus is at three comma two the directrix is at y equals eight we need to know first what is our p that p distance that separates the focus from the vertex and the vertex from the directrix well because we're going horizontally with our p distance we look at the y values of 8 and 2. the distance from eight to two eight minus two is six and that covers notice the p is there twice so six is equal to two p we can see that p value is going to be equal to three so the vertex coordinate then we're still going to the right the same three but now we're going to be off by a distance of p which is three well the focus had a y coordinate of two three more would give it a y coordinate of five so the vertex has a coordinates of 3 comma 5 and we know the p is equal to 3 so we're ready to make our equation it's a y equals equation because it's a vertical parabola but it opens downwards so we know it needs to be negative negative 1 over 4 times p which we found out was 3 times x minus the x coordinate of 3 squared plus the y coordinate which we found out to be 5. a little simplifying in that denominator we find out that y is equal to negative 1 12 times x minus 3 squared plus 5. and so in this way we can really use our sketch of the parabola it doesn't have to be to scale or perfect but our sketch really helps us build the pieces we need the vertex and that distance p between the vertex and the focus or the vertex and the directrix let's try one more example let's find the equation of the parabola that has a focus of negative 4 comma negative 1 and directrix of x equals negative eight so this time if we were to draw our rough sketch the focus is at negative four negative 1 which is the focus and the directrix is an x equals equation which means it's a vertical line x equals negative 8 which is even further to the left so that tells us our parabola has to come in through the center we need to find our vertex coordinates the focus is negative four negative one the directrix is x equals negative eight so as we go after our p value remember p is the distance between the vertex and the focus or between the vertex and the directrix so we have two p's of total distance between the vertex and the directrix the directrix has an x of negative eight the focus has an x of negative four if we subtract those we find the distance between them is four units so p here must be equal to two so if we have a distance of two between them we know that the x coordinate of the vertex is two away from our vertex i'm sorry from our focus and the directrix two away from negative four is negative six and we keep the same y-coordinate of negative one and we now know our vertex is at negative six negative one well since our parabola opens to the right it's an x equals it's positive because it's open the correct direction one over four 4p which we found out was 2 times y minus the k the y coordinate of the vertex which is negative 1 so plus 1 squared plus the x coordinate which is negative six cleaning up that denominator we have x equals one eighth times y plus one squared minus six and we now have the equation of our parabola so the parabola is our second conic section we've had circles and parabolas our third conic section is kind of a variation on the circle and it is what is called the ellipse the ellipse is also defined in terms of distance the ellipse is all the points where the sum of the distance from any point to two foci which is the plural of focus is constant an ellipse is traditionally called the oval shape and the idea of the ellipse if i had drawn it perfectly is we have two foci in the ellipse and if i were to pick any point on the ellipse and look at the distance to both foci we'll call them maybe a1 and a2 and then i picked a different point on the ellipse and took its distance to the two foci we'll call that b1 and b2 and it doesn't matter which points i pick i could pick a point that's way off on the side and i end up with c1 and c2 connecting to the two foci what happens is the sum of the distances a1 plus a2 is always constant so from the beep from the point with the b's b1 plus b2 is exactly the same distance same thing for the c1 plus c2 no matter what point i pick if i go from that point to the two vertices the total distance is always constant that's what makes an ellipse and we have two equations four ellipses and it really depends on which direction the ellipse is going the first is for the horizontal ellipse where we have x minus h squared over a squared plus y minus k squared over b squared equals one or if the ellipse opens vertically we'll put the y's first y minus k squared over still a squared plus x minus h squared over v squared equals one where the a is always bigger than the b that's important the a is always bigger than b the first denominator is always the biggest so whichever x or y has the bigger denominator we have that one come first and we have this other variable called c where c squared is equal to a squared minus b squared that's going to be an important relationship for us as we attempt to define some of the properties first the major axis which is the distance across we'll call it the long distance across it could be horizontal or vertical depending on which way the ellipse opens up but the major axis has a distance of 2 a and the minor axis which is the other direction depending on its vertical or horizontal it's the shorter distance across the middle that shorter distance is to be we also know the distance between the foci plural of focus is to c and that's where the c comes in identifying those focus or foci and of course we're getting used to seeing in all of these conic sections that the center is always the point h comma k so if all this information is true about our ellipse we should be able to find some equations let's find the equation of the ellipse that has end points of the major axes at negative three comma five and negative three comma negative seven that also has the foci at negative three comma one and negative three comma negative three notice the major axis and the foci all have an x coordinate of negative 3. the endpoints what moves is actually the y-coordinate so that tells me the y-coordinate is where we get the tall and narrow shape where we've got a center which we don't know where the center is yet but we've got two foci one above and one below the higher one has a higher y coordinate so the higher one must be negative three one the lower one must be negative three negative three and the end points of the major axes means if we draw the long line through the center the major axis is the longest line possible so the endpoints must be at negative three comma five and negative three comma negative seven we need to find a b c and the center h comma k let's do the center first that's probably the easiest to do you can see all these are in a line on negative 3 for the x coordinate the y coordinate is going to be in the middle of the foci so the folk i go from one to negative three the distance between them if we subtract one minus negative three is four so the distance between the foci let's go ahead and label that the distance between the foci is 4 and remember that is equal to 2c so we're also figuring out here at the same time that 2 equals c when we divide both sides by 2. so if the total distance is 4 half of that is 2 so to get to the center from one of the foci we'll either add or subtract 2 when we figure out the y-coordinate of the center right in the middle is at negative 1. so we've got our center we also know that the distance between the endpoints of the major axes is 2a the distance from 5 to negative 7 if we subtract is a total distance of 12 which means if we divide by 2 a is equal to 6. we still need to figure out what b is though we don't know b from this drawing but we do know that c squared is equal to a squared minus b squared and c we know is 2 2 squared is 4 equals a we know is 6 6 squared is 36 minus b squared if i add b squared to both sides and subtract 4 i find out that b squared is 32 and that's actually all we need is b squared because b squared is in the formula our formula then becomes notice this opens vertically so the y's come first y minus the y coordinate of the center is negative 1 so plus 1 squared divided by a squared 6 squared is 36 plus x minus the x coordinate which is negative 3 on the center squared divided by b squared which we just found out was 32 equals 1 and this then is the equation of the ellipse that has endpoints of the major axis at negative three five and negative three seven it also puts the foci at negative three one and negative three three let's do one more ellipse [Music] leave that on the screen let's find the ellipse that has the end points of the minor axis this is the smaller one at one negative two and one negative four also has foci at negative two negative three and three negative three notice the foci the y coordinate is consistent which means it must open up around the x's so this is going to be a horizontal ellipse we know the foci are at negative 2 negative 3. and three negative three we're going to need to find the center we're also told the end points of the minor axes the minor axis goes the short distance through the center the endpoints of the minor axes are at one negative two and one negative four well the center we know is right in the middle of those so we obviously have a y coordinate of one and right in the middle of negative two and negative four we see is negative three we also can use our distances to help us conclude a few things the distance between the foci we're looking at the x coordinate here from negative 2 to negative 3 is 5. and the distance between the foci is 2c so c equals 5 halves we also know the distance between the minor axis the minor axis gives us two b's the minor axis goes from negative two to negative four that's a total distance of two so b must be equal to one we can then find our c because c squared equals a squared minus b squared well c is five halves that's 25 fourths is equal to a squared we don't know minus b squared which is one add one to both sides that's four fourths gives us 29 fourths is equal to a squared and we're ready to make our equation [Music] because this one opens up horizontally on the x's we have the x's come first x minus the x coordinate of the center which is negative 3 plus 3 squared divided by a squared which is 29 4 plus now the y coordinate y minus the y coordinate is 1 on the center squared divided by b squared which is 1 equals 1. and even though it's a little ugly this is probably the preferred format of the ellipse because it shows all of that information we want to see it helps us identify how big the minor axis is the major axis is where the center is so we'll go ahead and leave it in this form and so that's our third conic section the ellipse there is one last conic section conic section number four we'll call it d here it is called the hyperbola and the hyperbola is defined really similar to how we define the ellipse it is all points where the only difference is the ellipse was a sum the hyperbola is the difference of the distance from any point to the two foci is constant and the idea here is a hyperbola is going to look something like this where there's going to be two vertices and we take a look at the distance if i pick a point on one hyperbola the two distances the longer one we'll call the first point the shorter one we'll call the second point no matter what point i pick on whichever one the longer one i'll call the first point the shorter one the second point no matter which points i pick when i subtract the a1 minus a2 i'll get the exact same value as if i took the b1 minus the b2 or wherever i want to draw these lines the hyperbola has that distance constant just like with the ellipse there are two equations that we use for the hyperbolas the first equation is used if it opens kind of the traditional way off to the left and right this is where x minus h squared over a squared minus y minus k squared over b squared equals 1. and here it doesn't matter which one's bigger a or b what's important here is the x is positive and the y squared is negative because if we switch that and we have the y minus k squared over a squared minus the x minus h squared over b squared equals one then the hyperbola is open vertically we also still have this relationship with c c squared this time is equal to the sum of a squared plus b squared [Music] and with that we end up with these important properties the distance between the foci is 2c and the distance between the vertices that's the corners of the hyperbolas is 2a we still need to find b to figure out how wide everything's going to be and then of course we still have our center at h comma k so using this let's see if we can find some equations starting with a hyperbola that has vertices at negative four comma three and two comma three [Music] and also has foci at negative seven comma three and five comma three so again i like to do a rough sketch just so i have an idea of what's happening with this i see the vertices the x-coordinate is changing and the y-coordinate is constant so the x-coordinate is going to change left and right well the y-coordinate is going to remain constant so we have vertices at negative 4 comma 3. and 2 comma 3. we also have foci out here at negative 7 comma 3 and 5 comma 3. we need to know where the center is and also find our a b and c well the center is going to be right in the middle of the vertices our vertices the x's are changing from negative 4 to 2 and if we subtract that's a total distance of 6. we know the distance between the vertices is 2a so that tells us that a is equal to three the distance to the center is also equal to three so if we back up three or add three depending on which point we're going from we'll end up with negative 1 comma 3 as the coordinates of our center point we now have our h and k we also have our a the only other thing we can pull off of the graph is c which we get from the distance between the foci the foci go from x of negative 7 to 5. that's a total distance of 12 which we know is 2 c so c must be equal to six now we can go after the b because we know c squared is a squared plus b squared c was six squared is thirty six a was 3 squared is 9 and subtracting 9 from both sides b squared is going to be equal to 27. so go into our equation then the equation is because it opens sideways we start with the x x minus h h it comes from the center so three i'm sorry negative one so we'll add one squared divided by our a squared three squared is nine minus y minus the y coordinate of the center which is 3 squared divided by b squared which we just found out is 27 equals 1. let's try one last hyperbola before we step away from them let's find the hyperbola that has vertices at two comma three and two comma negative one it's also going to have foci at two comma five and two comma negative three this time i see the x's staying consistent which means the hyperbola is going to go up and down around the y's so the vertices are at two comma three and two comma negative one the foci are at two comma five and two comma negative three we need to find the center and a b and c well first we know that a comes from the distance between the vertices the x's are the same so we'll focus on the y's the distance from three to negative one is four so four must equal two a therefore a equals two so the distance from the center to the edge is two x coordinates stays the same the y coordinate is two off from the vertices so three minus two is one the center must be at two comma one the vertices are two c apart so from five to negative three is eight that's two c's so c must equal four and then we're ready to go after our b because we know a squared plus b squared equals c squared a is two two squared is four plus b squared equals c which is 4 squared is 16. subtract 4 from both sides and b squared equals 12. now we have all the pieces we need to build our equation this one opens vertically so it's going to be a y first y minus the y coordinate of the center which is 1 squared over a squared 2 squared is 4 minus x minus the x coordinate is 2 squared over b squared which is 12 equals 1 and we have our hyperbola so this video was a quick overview of all of the conic sections that we're going to work with in this course we talked about hyperbolas we talked about ellipses we talked about parabolas and we talked about circles four common conic sections that we're going to talk about further in class so take a look at the homework assignment try a few of these there's a lot of pieces to keep track of but they're not too hard as a whole so take a look at them and we will see you in class this next chapter is going to take a look at vectors and how vectors can help us in calculus and in other contexts so the question we're going to do as we set up this new study of vectors is what exactly are they what are vectors and quite simply a vector is a quantity with magnitude which is a fancy word for size and direction and an example of a vector we might represent a vector from a point out to another point with direction that would be a vector we might label it vector v and what's interesting about vectors is they're not necessarily unique let's say we've got a vector u and we use the little kind of vector looking symbol above it to let us know that's a vector vector u goes from the origin to the point three comma two and vector v goes from negative 3 negative 1 to 0 comma 1. i want to notice really quick a bit of notation i have this little arrow looking thing above the vector that i'm going to write to represent a vector the textbook uses bold letters to represent vectors it's really the same thing it's just hard to type by handwriting a bold u and a bold v so my notes are going to be slightly different than the textbook but that line above means we're talking about a vector so let's look at what these vectors look like the first vector it says goes from the origin to the point three comma two so the vector goes from the origin to the point three comma two that's vector u the second vector goes from negative three comma one to zero comma one and you notice that both vectors are going the same direction and both vectors are the same size because both vectors have the same magnitude in the same direction we can say that those are equivalent vectors they both represent the same vector the same size and the same direction now we need a good way to represent vectors and one way we can represent vectors it's what's called component form which basically means it is the size and direction from the origin and usually we represent the vector with square brackets the x-coordinate and the y-coordinate from the origin now vectors aren't always set up from the origin so to find the component form what we can do is we can subtract the terminal point minus the initial point so for example if we were asked to find the component form of a vector with initial point of negative 3 negative 1 and terminal point of negative 2 comma three we could do that by finding the x component and the y component by subtracting the terminal point minus the initial point so the terminal point for x starts at negative two minus the initial point of negative three so negative two minus the negative 3 is going to equal 1. the x component is 1. for the y component we take the terminal point of three and we subtract the negative one and we end up with four and we take the x and y components together to represent our vector v as the x component comma the y component and this is our component form of the vector again i want to make sure i emphasize that we use pointed brackets when talking about a vector pointed brackets and that vector of 1 4 tells us that if this vector was centered at the origin it would go to the point 1 comma 4 with that same magnitude and direction speaking of magnitude quite often we're going to be interested in how exactly we can find that magnitude and the way we represent the magnitude of a vector is it looks like our double absolute values and if i think about a vector if i want to know the total distance of that vector the magnitude of the vector we could look as it added as set up with an x component and a y component and you can see the pythagorean theorem could be used to find the size of that vector x squared plus y squared equals the magnitude squared or if we take the square root of both sides the magnitude is the square root of the x component squared plus the y component squared and this gives us a nice little formula to find the magnitude of any vector so for example if i wanted to find the magnitude of a vector v equal to negative five comma four we would say that the magnitude of the vector is equal to the square root of the x component squared 5 squared is 25 plus the y component squared 4 squared is 16. and 25 plus 16 is 41 and so the magnitude the size of this vector is the square root of 41. let's try one more example let's find the magnitude of let's do the vector w is equal to four-fifths comma three-fifths well the magnitude of vector w is the square root of the x component squared which is 16 over 25 plus the y component squared 9 over 25 which is the square root of 25 over 25 or the square root of 1 which is just 1. and this is actually kind of interesting to us whenever a vector has magnitude of 1 we call that a unit vector a unit vector has magnitude of one in fact there are some special unit vectors that we are particularly interested in we call these the standard unit vectors there are two standard unit vectors that we're going to talk about in this video and there's a third one that we're going to talk about in the next video the first vector is vector i which has an x component of 1 and the other component is 0. other one is j j has an x component of zero and a y component of one basically i is a horizontal vector with magnitude of one and j is a vertical vector of magnitude of one and what's important is the vectors i and j meet at a ninety degree angle because we have these two special vectors i and j that are universally accepted we can write our vectors can be written as a linear combination of i and j vectors whatever i is multiplied by represents the x component and whatever j is multiplied by represents the y component so for example we talked about this vector an earlier example of negative 5 comma 4. this same vector could be expressed as standard unit vectors as negative 5 times the i vector because negative 5 is the x component plus 4 times the j vector which represents the y component another example would be the vector 2 comma negative 1. because 2 is the x component we could write that as 2i and negative 1 being the y component can be written as negative j this even gives us a way to write vectors that have zero for one of the components maybe zero comma three there's zero i components but there are three j vectors combined together similarly negative 2 comma 0 would just use the x component so we have negative 2 i so in this way every vector could actually be represented in two different ways it can be represented as standard unit vectors or it can be represented as a component vector as we go through this video we're going to express the vectors in both ways so we get used to seeing them both ways but make sure as you're working on the assignments and ultimately quizzes and tests you take the time to be aware of what format the question's asking you to express the answer in all right now that we've taken the time to define vectors let's see if we can do a little bit of work with vectors working with vectors it's really what we're going to do this entire chapter but looking at some basic operations with vectors one thing we can do with vectors is we can combine them together we can combine vectors let's say vector v is equal to x1 y1 and vector w is equal to x2 y2 two operations that are very common with vectors one is called scalar multiplication and this is the idea if k is just a number and we wanted to multiply it times the vector v what that really means is take vector v which had an x component of x 1 and multiply the x component by the k and multiply the y component by the k it kind of feels like we're distributing as we multiply by the scalar or multiply by the number k vectors can also be combined with other vectors though with what is called vector addition where we end up with vector v plus vector w and when we add vectors together what that really means we want to do is add their components together x1 plus x2 to get the new x component y1 plus y2 to get the new y component and similarly using the idea of an inverse we could also do vector subtraction where we subtract the components and so we end up with addition and subtraction of vectors and also a multiplication by a scalar in future videos we'll take a look at how we can multiply a vector times a vector there's actually two different ways to do that depending on what we're looking for specifically so for now we'll stick with vector addition and scalar multiplication so let's let the vector a be the vector seven comma one and the vector b is going to have an initial point of 3 comma 2 and terminal point of negative 1 comma negative 1. i'm going to pause here because before we get too far before we can do any linear combinations of vectors with each other we need to know what vector b actually is in component form so we already know vector a is 7 comma 1. to get vector b we need to subtract the terminal point minus the initial point so for the x component negative one minus three is negative four comma negative one minus two is negative three so we have these two vectors first problem we're going to attempt to solve is figure out what negative 2 times vector a is this is scalar multiplication because we're multiplying by a scalar or a constant negative two times vector a which is seven one and similar to distributing we end up with the vector negative 14 negative 2 or if you prefer an i j form negative 14 i minus 2 j let's look at vector a plus vector b here we're adding two vectors together vector a is seven comma one plus vector b is negative four negative three and for vector addition we just add the individual components together seven minus four is three and one plus negative three is negative two and in component form our sum is 3 negative 2 with i j we have 3i minus 2 j let's subtract the vectors a minus b well vector a is seven one minus vector b which is negative four negative three subtracting component by component seven minus a negative four becomes eleven one minus a negative three becomes four and we have our component form of a minus b which is equal to 11 i plus 4 j we want to make sure we always have that vector notation above i and j one last one of these let's find 3a minus 4b to do this we have three times the a vector the a vector is seven comma one minus four times the b vector negative four comma negative three well when we do that operation initially we have scalar multiplication which gives us 21 comma 3 plus i'm going to go ahead and take that negative in negative 4 times negative 4 is 16 negative 4 times negative 3 is 12 and now we're ready to add these vectors together 21 plus 16 is 37 3 plus 12 is 15 and we have our component form solution which is equal to 37 i plus 15 j and in this way you can see how we can do lots of different combinations of our a and b vectors another thing we can do with vectors is we can find a described vector using our trig properties and to kind of set this up we'll think about the fact that if i have a vector v the x coordinate of that vector using trig we're used to saying that x coordinate is the cosine of the theta angle that is formed and the y coordinate is the sine of theta but that comes off the unit circle if the vector is not a unit vector we need to multiply by the magnitude of the vector so we have the magnitude of the vector times cosine gives the x component and the magnitude of the vector times sine gives the y component and in this way we can now find the vector with magnitude 10 that forms a 120 degree angle because the x component is the magnitude of 10 times the cosine of the angle and the y component is the magnitude of 10 times the sine of the angle magnitude of 10 sorry well we need to know what the sine and cosine of 120 is so if we draw our little unit circle on here 120 degree angle is right up here it's the same as 2 pi over 3. so the x coordinate is negative one half and the y coordinates root three over two so ten cosine of one twenty would be ten times the x coordinate of negative one half our x components negative five for the y coordinate the sine of 120 is the y coordinate of root three over two which gives us 5 root 3 and so for our final vector that has a magnitude of 10 and is going to form 120 degree angle it's going to be negative 5 comma 5 root 3. or negative 5 i plus 5 root 3 j let's try one more and do this next one in radians instead of degrees let's find a vector with magnitude for forming a seven pi over four angle if i think about my unit circle 7 pi over 4 is down here with coordinates root 2 over two comma negative root two over two so when we want the x component it's the magnitude of four times the cosine of seven pi over four which is 4 times cosine the x coordinate root 2 over 2 or 2 root 2. to get the y component exactly the same we're just going to take the sine instead of 7 pi over 4 which is 4 times negative root 2 over 2 which gives us negative 2 root 2. and so our vector with magnitude of 4 that forms a 7 pi over 4 angle is 2 root 2 comma negative 2 root 2 in component form or in terms of i and j it's going to be 2 root 2 i minus 2 root 2 j so trig can be very helpful as we work with vectors we're going to do quite a bit of trig as we do different angles with vectors throughout this course but i want to do one other thing with vectors as we finish out our introduction to vectors and that is looking at this concept of finding unit vectors in the same direction and the trick here to get a unit vector is we're going to use a scalar multiplication basically we want to divide the vector down so it just has a magnitude of one so its magnitude is currently eight we need to divide by eight if its magnitude is 12 we need to divide by 12. we are going to take 1 over the magnitude of the vector times the vector to give us a unit vector in the same direction so for example let's say we've got the vector v is equal to 9 comma 2 and we want to find a unit vector in the same direction well first we need to know the magnitude of the vector so we know how much we have to shrink it by the magnitude is the square root of the x component plus the y component squared each so here we end up with the square root of 85 as the magnitude which means we can take 1 over the square root of 85 times the vector which will keep the vector going in the same direction but it will shrink its magnitude down so it's a unit vector this gives us our new vector of 9 over the square root of 85 comma 2 over the square root of 85 or if you prefer 9 over the square root of 85 i plus 2 over the square root of 85 j that is a unit vector in the same direction as the original vector 9 2 by multiplying by the reciprocal of the magnitude one nice thing about knowing the unit vector is it makes it very easy to get a vector of any size in the same direction to get a vector of magnitude k in the same direction we not only take one over the magnitude of the vector times the vector but we're also going to multiply by our desired magnitude of k so for example if we want a vector let's stick with 9 2 we're going to find a vector in the same direction but instead of a unit vector we want it to be of magnitude five well we already found out in the previous example that the magnitude of the vector is equal to the square root of 85. we want to get a magnitude of 5 so we're not only going to divide by the square root of 85 but we're also going to multiply by 5 and we use that as our scalar times the vector 9 2. and that gives us 45 over the square root of 85 for the x component comma 10 over the square root of 85 for the y component or in terms of i and j 45 over root 85 i plus 10 over root 85 j and that now is a vector of the same direction but now with magnitude 5. we haven't really done any calculus yet we've just been kind of exploring with this concept of a vector using scalar multiplication vector addition i k j vectors calculating the magnitude of a vector so it's been a really brief overview of vectors in the plane two-dimensional vectors but it's really important we're comfortable with these before we get to too many complex topics that these are built on so i want you to spend a day practicing with these vectors on the homework assignment we'll talk about them more in class and answer any questions that you may have now that we've taken a look at how vectors work in two space we're ready to answer the question about how do vectors work in three d in three dimensions in space but before we do that let's take a quick look at 3d graphing because we do not have as much experience with 3d graphing as we do with 2d graphing and we can think about 3d graphing as graphing in space where it's kind of like the corner in a house where one wall coming out becomes your x-axis the other wall coming out becomes your y-axis uh intersecting with the floor and then coming up vertically is what we will call the z-axis and in this way we end up with our x y z space and you can also go negative y and negative x and negative z so we get the origin is just kind of a fixed point out in space and so then if we were asked to actually sketch a point let's sketch the point one comma two comma three this becomes quite a tricky task to do in two dimensions but we'll do the best we can here uh we'll have our z-axis our y-axis and the x-axis forming the corner of the house and we can almost think about these as counting out on the y-axis one two three counting out on the x-axis one two three and on the z-axis one two three and it might be easier to kind of visualize where the point one two three is if we graph a three-dimensional box that sits in this corner where the box has dimensions of one by two by three so the box is gonna have a dimension of one down the x-axis two across the y-axis so we kind of end up with this shape and then vertically we're going to go up three units and so you almost end up with this three dimensional box sitting in the corner and that front corner closest to us becomes the point one two three because it goes a distance of one down the x-axis a distance of two down the y-axis and then vertically up a distance of three and so that's kind of how we can sketch a point in three dimensions and where we're ultimately going is from the origin we're going to have a vector that goes out to that point diagonally across the box but before we can get to that idea of vectors which have direction and magnitude to calculate the magnitude we need to know how distance works in three dimensions now in two dimensions we're familiar with the pythagorean theorem so we could take the x-coordinate squared plus the y-coordinate squared and take the square root to get that distance or the x distance squared plus the y distance squared the square root of that equals the distance we can extend that same logic out into three dimensions so that the distance is the square root of the x distance squared plus the y distance squared plus the z distance squared so a really quick example of that would be to find the distance between the point one comma negative five comma four and the point 4 comma negative 1 comma negative 1. well if we look at our x's the distance the x's travel from 1 to 4 that's a distance of 3 when we subtract them when we look at our y's the distance the y's travel from negative 5 to negative 1 that's a distance of 4. and the z's traveling from 4 to negative 1 they cover a distance of 5 when we subtract those points so then using our distance formula the distance between them is the square root of the x distance 3 squared plus the y distance squared plus the z distance squared which gives us the square root of 9 plus 16 plus 25 which is the square root of 50. and if we want to simplify that that's 5 root 2. so distance the pythagorean theorem in three dimensions works very much the same as it does in two dimensions we just add the extra z squared to our distance formula all right that was a very brief introduction to graphing in 3d because our main interest is not graphing in 3d our main interest is in 3d vectors that go through the space and basically 3d vectors are very similar to 2d vectors we just have to add the third component to get x y z or we could add a third unit vector which is the vector k in our i j k vectors where k is equal to the vector 0 0 1 giving us one unit in the z direction so for example we could have the vector v could be equal to 1 negative 2 3. and in terms of i j k vectors then that would be i minus 2 j plus 3 k or if i had the vector u equal to 0 1 negative 2. because 0 is the i vector we don't really have an i so we just have 1j minus 2 k and that tells us that this is a vector that goes from the origin diagonally to those points either 1 negative 2 3 or 0 1 negative 2. let's see if we can find a vector with initial point of 3 8 2 and terminal point of 2 negative 1 3. well our vector we know from two-dimensional vectors that we can take the terminal point and subtract the initial point and that'll give us the distance that vector's traveling so looking at the x component 2 minus 3 gives us negative 1 looking at the y component negative 1 minus 8 gives us negative 9 and looking at the z components 3 minus 2 is equal to a positive 1 or if we prefer we could say that's negative i minus 9 j plus a single k and this is that same vector and what's nice about vectors in three space once we're comfortable with vectors in two space is that we can do all the same operations from those two d vectors that we worked with in the prior video so for example if vector v is equal to negative one four three and vector w is equal to two comma zero comma negative three we could do things like find 2v minus 3w scalar multiplication and vector addition or subtraction 2 and 3 are scalars that's just 2 times the v vector of negative 1 4 3 and then a negative 3 times the vector w of 2 0 negative 3. so when we multiply those scalars through because multiplication comes first in order of operations we get negative 2 8 6 plus let's take the negative through negative 6 0 9 when those are multiplied through and finally vector addition adds those together to get negative 2 plus negative 6 is negative 8 comma 8 plus 0 comma 15. or if we do those as i j k's we have negative 8 i plus 8 j plus 15 k let's keep with these same two vectors and let's find the magnitude of vector v well magnitude is just the distance formula where we take the square root of the x component squared plus the y component squared plus the z component squared it's that pythagorean theorem we saw earlier so vector v negative 1 squared is 1 plus 4 squared is 16 plus 3 squared is 9 and we end up with the square root of 26 is the magnitude or size of vector v and once we know the magnitude we're ready to find a unit vector in the direction of v and just like with two dimensional vectors to make a vector into a unit vector in the same direction we scale or multiply by the reciprocal 1 over the square root of 26 the reciprocal of the magnitude times the vector v negative 1 4 3 which gives us the vector negative 1 over the square root of 26 4 over the square root of 26 3 over the square root of 26 or if we prefer with i j k vectors negative 1 over the square root of 26 i plus 4 over the square root of 26 j plus 3 over the square root of 26 k so as you can see working with vectors in 3d is very similar to working with vectors in two dimensions there's an extra component to work with but we follow the exact same patterns so i'll give you a chance to practice some of these on your own take a look at them and we will answer any questions you have in class working with vectors we discussed how we can add and subtract vectors together we did a multiplication with vectors called scalar multiplication where we multiplied a constant or a scalar by a vector but we haven't talked about how we can multiply two vectors together there's actually two different ways to multiply vectors together and so we specify the difference by not calling it multiplication but we call them a dot product and a cross product the difference is one gives us a scalar as a product and one gives us a vector as a product and they both have different uses depending on what we're trying to accomplish in this video we're going to look at the dot product as we answer the question how do we multiply vectors to get a scalar or a constant in other words when we multiply these vectors together using the dot product the answer is going to be a scalar not a vector so let's define this dot product let's say we've got a vector u and that vector u is x1 y1 z1 and there's another vector v that's going to be our second vector so we'll call it x2 y2 z2 as its components the way we define the product u dot v the dot product is we're going to multiply the components together and add the products so we'll take the x coordinates and multiply them together x1 x2 plus then we take the y components and multiply them together y1 times y2 plus and then we take the z components and multiply them together z1 times z2 and when we add those together we'll get some scalar so this is just a new operation we really just need to do a little bit of practice with to learn and then we'll talk about how it's useful to us when talking about vectors so for an example here we're going to do vector u is equal to 2 9 negative 1. and vector v is equal to negative 3 1 negative 4 and we're going to calculate u dot v well to do u dot v we're just going to multiply the common components together we take the x's 2 times negative 3 is negative 6. then we take the y's nine times one is a positive nine and then we take the z's negative one times negative four is a positive four and when we add those together negative six plus nine plus four we get seven so our dot product of u and v is equal to seven let's try one more example let's say vector u is equal to three i plus five j plus two k and vector v is equal to negative i plus 3k if we want the dot product u dot v we'll start by multiplying the i components together three times negative one is negative three the j components together five times and you notice there's no j component on v since there's no j component that really means there's 0 j's so 5 times 0 is plus 0. and finally on the k's 2 times 3 is a positive 6 and so when we add those together we end up with three and that's the dot product it's a very simple straightforward multiplication of vectors where we multiply the components together and add the individual products what is this useful for though well one common use of the dot product is the dot product can be used to tell us the angle between two vectors we have an important formula that says that the cosine of the angle between the vectors is equal to u dot v divided by the magnitude of u times the magnitude of v this is going to be an important formula for us to know you'll probably have it memorized by the time you finish the homework assignment because we use it a lot as we try and find the angle between two vectors so let's do some examples where we find the angle between two vectors let's start with the vector a is equal to one to zero and b is the vector 2 4 1. i like to find the individual pieces of the formula before i plug them in just makes a little more straight forward so let's find the numerator first the a dot b which is going to be 1 times 2 is 2 plus 2 times 4 is 8 plus 0 times 1 is 0. so a dot b is ten next we need the magnitude of both vectors so the magnitude of a is the square root of one squared plus two squared plus 0 squared or the square root of 5. we also need to know the magnitude of vector b vector b is the square root of 2 squared plus 4 squared plus 1 squared which is the square root of 21. now we put it all together and know that the cosine of the angle between the vectors is the dot product 10 divided by the product of the magnitudes and we multiply square roots we multiply underneath five times twenty one now we just go to our calculator to figure out what the cosine inverse of that fraction is and we can type it in just like it is cosine inverse of 10 divided by the square root of 5 times 21. make sure we close the parenthesis on the cosine inverse and this tells us that the angle is about point 22 radians so theta is equal to 0.22 radians let's do one more example finding the angle between let's say a is equal to i minus 2 j plus k and b is the vector i plus j minus 2k again i'm going to find the pieces of the formula first and then we'll go to the formula first we need to know what a dot b is 1 times 1 is 1 negative 2 times 1 is negative 2 and 1 times negative 2 is negative 2 gives us a total of negative 3 for the dot product then we need the magnitudes the magnitude of a is the square root of one plus four plus one or the square root of six the magnitude of b is the square root of one plus one plus four which is also the square root of six and so when we put that all together we know the cosine of theta is equal to the dot product negative 3 over the square root of six times six which is actually thirty-six and the square root of thirty-six is six which means this reduces to negative one-half well here we don't have to go to the calculator because negative one-half is one of our key points on the unit circle the x-coordinate is negative one-half at the point two pi over three so theta must equal two pi over three this is our angle between the vectors as we're discussing the angle between vectors one important relationship comes out of this we'll go ahead and call this number three and it's this concept of what are called orthogonal vectors and orthogonal is just a big fancy word that means perpendicular or right angles basically two vectors that meet at a right angle and to set this up let's say if u dot v turned out to be zero then as we're trying to find the angle between the vectors we would say the cosine of theta is equal to the dot product divided by the magnitude of u times the magnitude of v but zero divided by anything is zero and we know from our unit circle that the x coordinate is 0 at pi over 2 so therefore the angle theta is pi over 2. we have a right angle formed between these two vectors those vectors are orthogonal if u dot v is equal to zero the vectors are orthogonal so that can save us a little bit of work if we end up with orthogonal vectors if i asked you to find the angle between vector p which is at 1 0 5 and vector q which is at 10 3 negative 2. we would start going through that same process that we had before we would first find the dot product p dot q and then we'd find the magnitudes and set up our formula but look what happens when we do p dot q we end up with 10 plus zero minus 10 to give us a dot product of 0. as soon as i see that dot product is 0 i know the vectors are orthogonal which means the angle between them has to be pi over 2 and i can stop there saves us a little bit of time in fact we can even go a step further and we can find components to make vectors orthogonal we can find maybe an x so that vector p equal to 2 8 negative 1 and vector q equal to some missing component x negative 1 2 such that those two vectors are orthogonal well if i want these vectors to be orthogonal p dot q has to equal 0. so let's calculate p dot q multiplying the x components we get 2x the y components negative 8 the z components negative 2 and to be orthogonal that has to equal 0. well this is a real simple algebra equation by adding 10 to both sides and dividing by 2 we find out if that first coordinate is a 5 p and q will be orthogonal vectors so in this video we've taken a quick look at the dot product the dot product is a way to multiply two vectors together to end up with a scalar and that dot product is commonly used to find the angle between two vectors go ahead and take a look at the homework assignment to practice these and then in our next video we'll take a look at how we can multiply vectors together to get a vector we'll see in class the second type of multiplying vectors is called the cross product and the cross product has a very specific purpose it's to answer the question how do we find a vector orthogonal to two vectors and as i said the answer to this is the cross product it's a product where we multiply two vectors together and end up with a another vector that is orthogonal or perpendicular to the other two vectors in fact let's write that down always orthogonal to the other two vectors and unlike the dot product the cross product only works in three dimensions so let's say i've got a vector u that has component form of a b c and another vector v that has component form d e f if i want to calculate the cross product of u crossed with v the way we calculate that is we take b f minus c e to get the x component c d minus a f to get the y component and ae minus bd to get the z component now while this technically is the formula for the cross product and if you really like memorizing formulas you can memorize this formula i find it very difficult to memorize the cross product formula so instead i look at a thing that comes from linear algebra that's quite simple called the determinant and here we're not going to go into detail with the determinant we're just going to look at how a determinant works with what's called the 2 by 2 matrix if i want the determinant of a b c d the way the determinant is calculated is we take the first diagonal a d and we subtract the second diagonal cb or actually let's put it alphabetically let's call it bc and if i can remember this pattern of multiplying one diagonal and subtracting the other diagonal it's actually really easy to find the cross product the cross product is the determinant of the other rows actually overcome other columns with a little caveat that the middle component we do backwards here's what i mean by that let's say we'll take the same u and v vectors and put them on top of each other u is a b c and v we said was d e f to combine these to get u crossed with v first we look at the x component when we do the x component we ignore the x column and find the determinant of what's left which is b f minus the other diagonal c e then when we do the middle component we ignore the middle column and the middle works backwards so we're going to do the backwards diagonal first so we're going to diagonal and do cd minus the af diagonal then for our third component we can ignore the third column and we've got ae minus bd and you notice that gives us the exact same formula we had up above that i highlighted in yellow but by using this determinant strategy we don't really have to memorize the formula we just have to remember ignore the column we're working on find the determinant in the middle works backwards so let's try a few examples let's say vector p is equal to 5 1 2 and vector q is equal to negative 2 3 1. if i want to do p crossed with q we can take the 5 1 2 above and the negative 2 3 1 below and we can cross them for the first component we ignore the first column and we find the determinant 1 times 1 is 1. minus two times three is six for the second component we'll ignore the second column but we have to work backwards two times negative two is negative four five times one is five and we subtract so minus five and then for the third component we ignore the third column five times three is fifteen minus one times negative two is negative two so it becomes a plus two and when we put that all together we get our cross product which is negative five negative nine seventeen and what we've done is we have found a vector that is perpendicular orthogonal to both p and q vectors now here's an interesting question though with multiplication including the dot product order does not matter does order matter with the cross product let's take a look at it let's do b here and we're going to cross them in the opposite order we're going to do q crossed with p and when we do that now we've got the q vector first negative 2 3 1 and the p vector second 5 1 2 and when we cross those let's see what we get ignoring the first column 3 times 2 is 6 minus 1 times 1 is 1 comma ignoring the second column and working backwards in the middle one times five is five minus negative two times two is negative four so plus four comma ignoring the third column negative two times one is negative 2 minus 3 times 5 is 15. and now what we end up with is 5 9 and negative 17. how does that compare with the answer we got before when we did peak.q notice every positive is negative and every negative is positive they are not the same but they're very similar because they're both perpendicular to our vectors they just need to go in opposite directions which kind of starts to lead to some properties of our cross product p cross q and q cross p are very similar they're just opposite directions let's list some properties of the cross product so first we'll do some general properties and we'll start with the one we just talked about that if we do u cross v that's going to be the opposite of v cross u another interesting thing about our properties is if you have a constant or a scalar that's multiplied by a vector it doesn't matter when or where you multiply that scalar you can take that scalar times the cross product of uv or we can do the scalar times u and then cross it with v or we can do the u crossed with the scalar times the v it doesn't matter where that scalar appears with the u with the v or in front of the cross product you'll get the same result either way another interesting result is if we take a vector and we cross it with the zero vector zero comma zero comma zero you can imagine the multiplication working out there you're ultimately doing a lot of multiplying by zero you end up with the zero vector the zero vector is just a vector filled with zeros and if we took a vector when we crossed it with itself if you imagine what's happening there is when we do that determinant step you're going to be subtracting the same thing that you multiplied and so you'll also end up with the zero vector so those are some general properties of the cross product some specific properties though with our ikj vectors that are interesting if i is crossed with j we end up with k and if j is crossed with k we end up with i and if k is crossed with i we end up with j so it kind of looks like if we cross we just end up with the other vector but one thing we have to be careful of is with cross product the order matters that means if we were to actually calculate j crossed with i we can't just say it's the missing k it's actually the opposite of k and k crossed with j is the opposite of i and i crossed with k is the opposite of j so the order does matter there with those properties so let's see if we can use some of these properties to help us simplify some of our cross products let's leave those ijk properties on the screen as we go to number three using the properties let's take a look at doing 2i crossed with 3j and the answer crossed with j well one thing we know about scalars is we can pull those scalars out front and we still get the same result so we're really talking about is 6 times i crossed with j crossed with j well we know that i j crosses to k so we have 6 k crossed with j we also know that k crossed with j notice that's backwards so we're going to get a negative out of it negative 6 i and so we end up with our final solution using the property of negative 6 i let's do another one let's do i crossed with k cross that with k crossed with j well i k crosses to j and kj actually i k crosses to negative j because the order's backwards j k or k j cross to negative i well we know these scalars these negative scalars can pull out front so negative one times negative one is a positive one we're really just doing um j crossed with i and j crossed with i we know is negative k and so in this way we can simplify using some of our cross product properties rather than having to go through all the work of the determinant to find our final cross product that's what we're looking at today though is the cross product you can either memorize the formula or remember the cross product pattern remember the center is the opposite order but the other two follow that determinant pattern we'll take a look at these in class practice some of these on the homework and we'll answer your questions when we see you vectors provide us a great transition into talking about equations of surfaces in three-dimensional space we're going to start here in this section talking specifically about lines and planes so we're going to answer the question how do we find equations of lines and planes in space in 3d graphing and we'll start with the first one being the equation of a line and similar to how we do equations of lines in two space y equals mx plus b in three space we also get a slope intercept feel to a formula but we're going to set it up with vectors we're going to let r naught be a vector we're going to call it x naught y naught z naught be a vector to a point on the line we're going to let v which is going to be represented with abc be a parallel vector to the line in other words vector v is going in the same direction then actually let's look at a visual of what we have here so we've got our three-dimensional space and let's say there's some line floating out here in three dimensional space we're going to take a point on the line and that point we're going to call x naught y naught z naught and so the vector that points at that is going to be vector r naught and then we're going to have another vector that's parallel to the blue line we're going to call that vector v so that's kind of visually what pieces are if we combine those pieces together the line is the vector r equal to the r naught vector plus the parallel vector times t which is going to create a parametric equation in terms of t so the line is the vector r naught plus the vector times t where we kind of have a slope intercept field where we've got a slope of the vector which is parallel and the intercept or a starting point is that r naught point so it kind of feels like that y equals mx plus b format now we can break this vector out r is the vector x y z is equal to the r naught which is the x naught y naught z naught plus the parallel vector which is a b c times the parametric variable t this is the vector equation of the line and then if we take that vector equation of the line and simplify to get a parametric equation where the vector x y z you can see if we multiply the scalar t through the abc we get a t b t c t and then we add the components together and we get x naught plus a t comma y naught plus b t comma z naught plus ct and if we take this apart component by component we end up with the parametric equations x is equal to the x component x naught plus a t y is equal to the y components y naught plus b t and z is equal to the z component z naught plus c t and this becomes the parametric equation for a line where x not y naught z naught are on the line and abc is a parallel vector and in fact we can go one step further and solve each of these equations for t to get what we call the symmetric equation for a line solving the first equation we would subtract x naught and divide by a that equals t but also the second equation is y minus y naught over b is equal to t and z minus z naught over c is equal to t and these become the symmetric equations for the line so if these are the equations for a line let's see if we can make one let's do an example where we find the equation of a line through two points we're going to find the equation of the line through the points one negative three two and five negative two eight any two points in space can be connected with a line and we can find that line first we need to know a parallel vector and a point on the line so for the parallel vector that's just going to be the vector that connects these two points together let's take the 1 negative 3 2 as the initial point and the 5 negative 2 8 as the final or terminal point let's call that t for terminal point and we know to get the vector that connects them we subtract the coordinates 5 minus 1 is 4 negative 2 minus negative 3 is positive 1 and eight minus two is six so now we know four one six is a parallel vector in the same direction as this line we can use either one of these points so see we've got our parallel vector let's do our uh r naught vector being just the first point we could have picked either point but one negative three two and from this we should be able to get a vector equation for the line the vector equation for the line is x y z is equal to the initial vector 1 negative 3 2 plus the parallel vector 4 1 6 times t and once i have the vector equation we can find the parametric equations for the line by simplifying the x coordinate is equal to a 1 plus 4 t the y component is equal to the y's which is negative three plus one t and the z component is equal to the z's which is two plus six t and this becomes the parametric equations that will build this three-dimensional line that connects the two points we can then solve each of these equations for t to find the symmetric equation form of each of these lines when we do that we get x minus 1 over 4 equals t is equal to y plus 3 over 1 which we don't need to do the divide 1 which is equal to z minus 2 over 6 which are all equal to t so we have three different ways to represent this same line either in vector form parametric form or symmetric form but this gives us the equation of the line that connects the two points any two points we can find a line that connects them if we have three points though three points can be used to define what's called a plane let's look at finding the equation of a plane and a plane is just a big flat surface kind of like a piece of paper or your screen but it goes on forever just like a line goes on forever but it's incredibly thin the flat surface is this big flat space that spreads out forever and we're going to set this up in a similar but slightly different way than we set up the line we're going to let r naught be a vector to a point on the plane and just like before we can represent r naught as x naught y naught z naught and we're going to have another vector we call it the vector v before because it was parallel but this time we need a normal vector or an orthogonal vector that's perpendicular so because it's normal we're going to call it vector in be a normal which means perpendicular or orthogonal vector to the plane so visually what we have if i were to draw a picture somewhere out here is this big flat space that goes on forever and we've got a point on the plane that's our point x naught y naught z naught the vector that goes to it that's our r naught and then there's this other vector that's going to come straight out of it it's going to be perpendicular to the plane that is our normal vector what we know about vectors that are normal or orthogonal is the dot product of the vectors must equal 0. so if i pick any random point on here we'll call that vector r and i look at the vector that connects r to r naught it's going to be perpendicular to vector n well the vector that connects r to r naught is the vector r minus r naught and if i dot that vector with something that's normal to it or orthogonal to it we know orthogonal vectors have a dot product of zero so we can say that the plane is the product the normal vector with r minus r naught similar to before that normal vector we said uh let's call that normal vector just like before is going to be the vector abc so the vector a b c is dotted with the difference between our general vector at any x y z point minus the specific point on the plane x naught y naught z naught we know that product is going to be zero which means if we simplify we've got a b c dotted with x minus x naught y minus y naught and z minus z naught equals zero which means if i keep going this simplifies to what we call the scalar equation of a plane the dot product says we multiply the components a times x minus x naught plus b times y minus y naught plus c times z minus z naught zero and remember abc is the components of the normal vector and x naught y naught z naught are the coordinates of a point on the plane this is the scalar equation of a plane good one two no now i could go through a step further and i could distribute and combine like terms to get rid of the parentheses this gives what we call the general form of a plane and so when we distribute we get an ax we get a b y we get a c z and then we get a lot of constants that we're going to say all those constants are going to add together to a d equals 0. and that becomes the general form of a plane so we looked at an example where we connected two points to get a line let's do an example where we connect three points which always define a plane i guess there's one exception if those three points are linear they don't define a plane but let's say these three points are not collinear let's say our points are 3 negative 1 2 q is the point 4 negative 2 1 and r is the point negative five three negative one well the first thing we need to find the plane that connects these three points is we need the normal vector a plane is built with a normal vector and any point so we need a normal vector and the best way to get a normal vector or an orthogonal vector is to use a cross product of two vectors on the plane so we can connect p q and r in any two ways there's three or six if you count backwards ways to connect these we need to connect two of them let's connect p and q in a vector which is going to be the vector four minus three is one negative two minus negative 1 is negative 1 and 1 minus 2 is negative 1. then let's connect the vectors p and r together i could have done q and r we get the same final result any two but p to r negative five minus three is negative eight three minus a negative one is four and negative one minus two is negative three to get a vector that's normal to both of these because both these vectors actually lie on the plane connecting points that are on the plane we want a normal vector we're going to cross them and when we cross ignoring the first column negative 1 times negative 3 is 3 minus four times negative one is negative four so three plus four comma ignoring the middle column going backwards negative one times negative eight is eight minus one times negative three is positive 3 comma ignoring the third column 1 times 4 is 4 minus negative 1 times negative 8 is positive 8. and so our normal vector then is equal to seven eleven negative four we can pick any point on the plane a vector that points to any point so let's just use the first point it doesn't matter which one we use 3 negative 1 2 and now that we have a normal vector and a point we should be able to build the scalar equation and you recall the scalar equation is a times x minus x naught plus b times y minus y naught plus c times z minus z naught equals zero where a b and c come off the normal vector so a is 7 times x minus the x point which is 3 plus b which is 11 times y minus the y point which is negative 1 plus c which is negative 4 times z minus two equals zero this then is the scalar equation form of the plane that connects the three points we can do a little algebra to get the general equation form of the plane and we can get that general equation by distributing 7x minus 21 plus 11y plus 11 minus 4z plus 8 equals 0. combining like terms we get 7x plus 11y minus 4z and then we combine the negative 21 plus 11 plus 8 to get negative 2 equals 0. what i've done here is try to give you a visual of what we just graphed the 7x plus 11y minus 4z minus 2. and as you can see we end up with this flat plane that we can see from all angles in three dimensions and if we were to try and find the point on here of negative or if positive 3 on the x negative 1 on the y it should also put us at about z of 2 and this graph doesn't quite get us there exactly but it should be around here you can see 4 on x negative 2 on y it's approximately a 001 again the graph doesn't quite hit it perfectly because it's in between these points and you can also see negative 5 on x 3 on the y it's going to be approximately negative 1 on the z if i could get in there perfectly but this graph only lets me hit on the grids which are off slightly but you can see we've made this plane that connects our three points using our vectors so that's what we're looking at today the equation of a plane is found by using a normal vector and a point on the plane the equation of the line is found using a parallel vector and a point on the line take a look at the homework assignment to practice some of these and we will see you to work on these further in class now that we've taken a look at the graphs of lines and planes in 3d we're ready to extend that conversation as we ask how do we graph other shapes or other curves or surfaces in space and we're going to start with kind of the basic graph and that is going to be the graph of a cylinder and a cylinder in 3-space graphing is not what we think of with a normal cylinder normally we think of a cylinder like a soup can but actually a cylinder is going to be defined here as any 3d shape made of parallel curves and the most basic form of that and there are others but the most basic form of that is when one variable is missing from the equation what we'll do in that case is we will graph in the other planes and copy down the missing variable in other words if x is missing we'll graph it in two dimensions on the y z plane and then we'll copy that y z graph all the way up and down the x graph let's take a look at some examples so we can see that happen first graph we're going to do is z equals x cubed first thing i'll observe with z equals x cubed is that the y variable is missing so if the y variable is missing we're going to graph it on the x z plane so we'll graph it two dimensionally but we'll graph it on x z and we know that the graph of anything cubed is going to be this graph that comes up from the negative levels off at zero and then kind of takes off after that that graph on the x y plus on the x z plane then is going to be copied down the y-axis so now we try and transition that two-dimensional graph into a three-dimensional graph where we've got x y and z and we've already graphed on the x z plane let's go ahead and draw the negatives also on this coordinate system so we can kind of see in three dimensions there so the xz plane we've already got this graph that comes up on the xz levels off and then takes off where kind of y is zero but that same graph is going to be able to slide up and down the y-axis and so we'll see it levels off at zero and takes off or back down the y-axis where it comes up levels off and takes off and so what we end up with is this kind of three-dimensional sheet with a little bit of a curve to it maybe emphasize that the peaks and the valleys line up here this three-dimensional shape it's a cylinder of all of these parallel graphs these parallel lines made of y of z equals x cubed all the way up and down the line let's take a look at another example let's graph z equals y squared because we just have z and y we're going to graph it on the y z plane so we'll start with our two dimensional graph so we know what that looks like on the y z plane and we know that the squared variable gives us that parabola graph and so we're going to copy that graph down the other variable down the x-axis so let's see if we can do that we've got x y and z let's include the negatives even though we don't really need them in this graph what do we do on the x and now with the z y plane that's almost the actual two-dimensional plane that we're used to here the z-y-plane is this parabola and that parabola is going to go all the way down the x-axis and all the way down the x-axis the other direction and so you can see we kind of end up with this three-dimensional curve it's a parabola going all the way down the x-axis and this then becomes our three-dimensional graph of z equals y squared now this idea of graphing a two-dimensional version of the graph and then seeing how it stretches along the graph there's actually a name for that two-dimensional version for the graph so let's go ahead and formally define that here we call it a trace or a cross section that is created when a surface intersects a plane parallel to a coordinate plane we've been focusing thus far when one of the variables is zero so we've got the either the x y plane the y z plane or maybe the x z plane were on that center point but it doesn't have to be on that center point it could be anywhere so let's say we've got you know your typical cosine graph and let's see if i can make it 3d by drawing another cosine graph back here there's your three dimensional graph it's a cylinder because it's that same graph stretched all the way through and the idea is if we were to cut it in the middle what you end up with is a nice little cosine graph cut in the middle it's that two-dimensional representation when it's sliced down and then that two-dimensional is stretched throughout the rest of the graph to give you the 3d version of it these traces become very helpful when we try to graph other shapes that aren't just this cylinder shape that stretches throughout the whole graph but it changes as you go through the graph we talked about lines and planes the next level up lines and planes are made with x y and z the next level up are made with x squared y squared and z squared these graphs are called quartic surfaces and now in general a quartic surface is of the form ax squared plus b y squared plus z i'm sorry c z squared and then we've also got combinations of the x's and y's so we've got d x y plus e y z plus f x z and then we've got x's and y's that could appear by themselves g x plus h y plus we have to skip i because that's the square root of negative one so we'll jump to j z or there's also possibly having a constant k equal to zero some combination of x squared y squared z squared x y's and z's combinations thereof we call those the cortex surfaces this is the three dimensional version of our parabolas ellipses hyperbolas circles the conic sections now in 3d let's look at trying to graph one of these quartic surfaces let's try and sketch x squared over 4 plus y squared over 9 plus z squared over 25 equals one one thing you might notice is this kind of has the feel of an ellipse but we're going to see if the traces can confirm that suspicion what we can do with traces is we're going to take a trace in each direction first we're going to do a trace on the x y plane which means we don't want any z's we can make z equal to any convenient number we want the easiest one's probably going to be zero if that didn't work we'd have to pick another number if sometimes if z is zero everything goes to zero and it's useless to us but in this case it doesn't we end up with x squared over four plus y squared over nine plus zero equals one and we recognize this graph as an ellipse on the x y plane in the x direction we're going to go two each direction in the y direction it goes three each direction and so we end up with this nice little ellipse in the x y direction we need to know what's happening in all the directions so we're also going to draw a trace in the x z direction when we do this y is equal to zero so we end up with x squared over 4 plus z squared over 25 equals 1. all right in the xz direction the x's because the denominator is 4 the square root is 2 each direction the z's go this each direction 5 units and so on this xz graph we end up with this tall skinny ewips and i'm not going to quite fit it all on one screen but we do need to also do a trace in the y z direction which means our x is equal to zero when x is equal to zero the equation becomes y squared over nine plus z squared over 25 equals one and so if we were to try and draw that we end up with the y's we've got our y z axis the y's are going three each direction so it's a little wider than the green one the z's still going five each direction and we end up with this slightly fatter ellipse and now what we're going to try and do is put these three graphs together in one what we see is in the y z direction we get a fat tall ellipse in the x z direction it's a skinnier ellipse in the x y direction it's really a small ellipse and so when we put it all together into a 3d graph what we end up with is the short skinny one one direction tall and skinny another direction and fatter and skinnier pattern a little wider the other direction we end up with this three-dimensional what we call an ellipsoid so that's how we can use the surfaces to kind of combine them each together to end up with our three-dimensional ellipsoid what i've done here is i've drawn on a online calculator the 3d ellipsoid and what you can see is if we look at the top we see that ellipse that goes from two to three if i look from one side we see the ellipse that goes from two to negative two one direction and five to negative five another direction and if we look at the other side we see the slightly fatter 3 to negative 3 and 5 to negative 5 and this 3d shape is our ellipsoid we've been trying to draw now there are some common quartic surfaces that we're going to work to get familiar with in this section and your book has a really nice introduction to them on page 221 through 222 and so i want to hop over to a picture from the book of those shapes just that we can talk about some characteristics really quick about each of these we've already talked about the ellipsoid the ellipsoid is similar to the ellipse we've got an x squared over something plus y squared over something plus a z squared over something what really makes an ellipsoid defined is you notice that x squared y squared and z squared are all positive and they're not going to equal zero they're going to equal a number that's going to create the ellipsoid and if a b and c all are equal we actually end up with a perfect sphere because all of those major and minor axes are going to be equivalent and just like we saw before the traces were all ellipses in the various directions now if we make a minor change and we make one of them negative so we've got a positive a positive and you see that we've got that negative z squared if one of them is negative we end up with what's called a hyperboloid of one sheet so with two of the variables positive coefficient one with a negative coefficient and then the way we know which direction this hyperboloid is pointing is that the axis corresponds with the negative coefficient so in this case the z squared was negative and so you see the graph here goes around the z axis if the y squared was the negative one it would turn sideways and go around the y axis now if two of the terms are negative here we've got two variables with negative coefficients and one with a positive coefficient a positive a negative and a negative we end up with what's called a hyperboloid of two sheets the hyperbola with two sheets doesn't connect in the middle and what we also notice is that the axis of the surface corresponds to the variable with a positive coefficient so here the z squared was positive and that's why it goes around the z axis if the y squared was positive it would go around the y axis if the x squared was positive it would go around the x-axis now each of these have been equal to a number but sometimes our equations are equal to zero which changes things slightly one example here is the elliptical cone with the elliptical cone we've got two positives and one negative but this time it equals zero if it equals zero then the axis of the surface will correspond to the variable with the negative coefficient so again here the z is negative that's why the cone goes around the z axis we end up with an elliptical paraboloid if we have a constant term z and the x and y terms are each squared that constant term is called the linear term and the axis will correspond with the linear variable which is why this graph goes around the z-axis finally the last one we're going to look at for our purposes is the hyperbolic paraboloid the hyperbolic paraboloids really interesting because in one plane you have a hyperbola in the other planes you have parabolas and the way we get that very similar to our last example except we've got the middle term is negative or there's a negative in between the x squared and y squared still equal to a linear term and that linear term that it's equal to is going to be the axis of the surface that it's all built around also interesting is it's going to wrap around the positive variable so these six shapes you can see them in your book it also lists all the traces that you're going to get from each of these x y z directions there's going to be some problems on the assignment where it's going to ask you to identify what shape you're dealing with and the way we can do that is by paying attention to what's positive what's negative and what does it equal so let's see if those are our common quadratic surfaces let's see if we can identify some quartic surfaces let's say we're given the surface 16 x squared plus 9 y squared plus 16 z squared equals 144. what i would notice on here is looking at the squared terms x squared y squared and z squared all the squares are positive we've got a positive 16x squared a positive 9y squared a positive 16y squared it also does not equal zero so if i were to go through and divide the whole thing by 144 it would be equal to 1 which is exactly the form of the ellipsoid let's take a look and see what this graph looks like in the 3d graphing calculator so here's our 3d graph just as we expected it's an ellipsoid all the traces are ellipses just like we predicted so we're doing good on that one let's try another example let's see if we can identify the shape of 9x squared minus 18x plus 4y squared plus 16y minus 36 z plus 25 equals zero well what are we going to note about this shape i do see that there's an x and a y term but those could be paired with the x squared and the y squared and through a process of completing the square we could make that x or y plus or minus some number the whole thing squared so those kind of come in groups but i do notice that the z is a linear term so we have x squared and y squared are positive and the z is linear going back to our graph we saw x squared and y squared positive and the z linear with the elliptic paraboloid and it could be either variable linear what we have is two squared variables in one linear variable and the linear variable is going to tell us which axis this graph is going to wrap around so we end up with an elliptic paraboloid here and if we go to the graphing calculator let's see if our guess was correct and here is our shape graft what you see is it does have that bowl shape that we were expecting from the elliptic paraboloid it's an ellipse from one angle and a parabola from the two other angles we have an elliptic paraboloid let's do one last example of identifying our shape we're going to do 9x squared plus y squared minus z squared plus 2z minus 10 equals 0. and let's see what we can note about this shape again we've got a z linear term but it can be paired with the z squared through a process of completing the squared so we have x squared and y squared are positive but you notice that z squared is negative so we have two positive squares and one negative square and that plus 10 can bump to the other side so it's not going to actually be equal to zero which one did that two positive squares and one negative squared that's going to be a hyperboloid of one sheet it is a hyperboloid of one sheet let's look at this example on the graphing calculator here's our hyperboloid of one sheet on the graphing calculator you notice one direction it's a hyperbola other direction it's an ellipse and the other direction it's a skinnier hyperbola we've got our hyperboloid of one sheet because it's going to go through the origin the hyperbola hyperbolas connect in the middle if we had two negatives then they wouldn't connect in the middle but it is hollow in the center so today that's what we're taking a look at three dimensional shapes whether they're cylinders or aquatic surfaces hopefully you can get some familiarity with the different shapes as you work through the homework assignment and in class we'll discuss them further and answer any questions and that you may have as we wrap up chapter 2 on our discussion about graphing in 3d and how vectors can help us with that i want to address another question that asks what other coordinate systems do we have for graphing in space just like with the rectangular coordinate system in 2d we had the x y coordinates but we also had polar coordinates that used a radius and an angle theta with the x-axis sometimes the polar system was easier for graphing equations and interpreting situations and sometimes the rectangular was better do we have the same situation with three-dimensional graphing and the answer is yes we've got two other coordinate systems that we can use the first of the two are called cylindrical coordinates and cylindrical coordinates are kind of like polar coordinates in 3d with polar coordinates we had a radius and an angle theta to make it in 3d we're just going to add an extra coordinate z that tells us how far vertically the graph point is in that z direction so let's formalize that here let's say it is similar to polar coordinates which have r theta a radius at an angle theta and angle with the x-axis we're just adding this z which is the vertical distance in the plane or in space so visually what we're talking about here is we've got our x y z axis from rectangular and instead what we're going to do is we're going to have some radius that makes some angle theta to a point and then we go vertically to some point some distance z and that vertical z makes a right angle with the radius r so if we've got this cylindrical coordinate system let's talk about how we can make conversions between the rectangular system and the cylindrical system very similar to polar coordinates x is equal to r cosine theta and y is equal to r sine theta we just have this new coordinate z which is just the vertical distance which works the exact same way in both systems and similarly if we're going the other way we know that r squared is equal to x squared plus y squared from the pythagorean theorem and that the tangent of theta is equal to the y distance over the x distance and again the z coordinate is just the z coordinate now there is one thing we got to be careful of in three space with the tangent of theta we want to make sure that we check the domain as we take the tangent inverse because on our calculators tangent inverse is always between pi over 2 and negative pi over 2 and sometimes we want to be in the opposite half of the grid so we have to really think about where our point is and we'll look at an example of that right now let's do some examples where we can use these conversions let's say we have the cylindrical point 5 pi over 6 4 and we're going to convert it to a rectangular point well very similar to what we did with polar coordinates x is equal to the radius of 5. let's go ahead and label 5 so the radius pi over 6 is the theta and 4 is the z so x is equal to the radius of 5 times the cosine of the angle pi over six y then is the radius of five times the sine of the angle pi over six and the z coordinate is just what the z coordinate is which is four simplifying our x and y then cosine of pi over six is root three over two so we end up with five root three over two for the x coordinate the sine of pi over 6 is one half so we end up with five halves for the y coordinate and so our cylindrical point change to rectangular x y z becomes 5 root 3 over 2 comma 5 halves comma 4 and we have our rectangular point that is equivalent to the cylindrical point 5 pi over 6 4. we can also go the opposite direction we can take a rectangular point and convert it into a cylindrical let's take the rectangular point negative eight five negative seven and convert it to a cylindrical point for the cylindrical point we now have x y and z that we're converting to an r a theta and a z our formula says that r squared is equal to x squared 64 plus y squared 25 which is equal to 89 taking the square root of both sides the radius must be equal to the square root of 89 theta comes from the tangent the tangent of theta is y over x so we're going to take the tangent inverse of y over x 5 over negative 8 which is equal to negative doing it on our calculator it'll give you a negative point five five eight six but here's where we have to be careful with the tangent inverse if i think about just the x y plane z it's going to be vertically or below this looking at the x y x is negative eight y is five our point is going to be above some point in the second quadrant however theta of negative 0.55 is a negative angle negative 0.5586 goes the wrong direction we need to move it so it goes to the correct direction well the distance that we need to move it is halfway around the unit circle we need to add pi to get to the correct spot so when we do negative 0.5586 plus pi on our calculator we'll get approximately 2.583 and that's going to be the number of radians that we want it to open up to get us in the correct quadrant so again we want to make sure we're careful tangent inverse on our calculator will always stick us on the right half of our xy graph which means if your point ends up in the left half you have to add a pi to it to get to the correct place and a quick sketch can help you see what's happening there now with z z straightforward with these cylindrical points z is just the z so negative seven so we end up with our cylindrical point of the square root of 89 for the radius 2.583 radians for the angle and negative 7 for the z value we've got our rectangular point we can even switch between equations equations in rectangular and equations in cylindrical let's say we've got the equation that r is equal to 3 sine of theta and we want to convert that equation to a rectangular equation well we know that r sine theta is equal to something so if we multiply both sides by r we end up with r squared equals three r sine theta and we know from our formulas that r squared is x squared plus y squared three is just a constant we know that r sine theta is equal to the y coordinate and so we've converted this to a rectangular equation x squared plus y squared equals 3y represents the same equation in cylindrical as r equals 3 sine theta so that's cylindrical coordinates very very similar to polar coordinates which we're already familiar with but there is a second three-dimensional graphing system that i want to talk about called spherical coordinates and with spherical coordinates we end up with three greek letters for our coordinates rho theta and phi rho theta and phi rho is the distance from the origin to the point and i want to be careful not to confuse that with the r from the cylindrical coordinates r was the radius in the x y plane and then we moved vertically up rho is going to be the diagonal distance from the origin to the point almost like a vector that goes diagonally through space it's the actual distance to the point not just in the x y plane but through space theta is the angle with the x-axis just like we did before it's the same theta that we saw in the cylindrical coordinates also from polar coordinates theta is the angle with the x-axis and phi our new greek letter v is going to be the angle with the z-axis so visually what we're talking about here is there's some distance diagonally through space rho that comes off of a projection if it makes a right angle it comes off of an angle theta that's made with the x-axis and an angle phi made with the z-axis so we end up with two angles and a distance that give us the coordinates to the point so then the next logical question is how do we do conversions between the rectangular system that we're more familiar with and this new system the spherical coordinates well our x-coordinate is equal to rho times the cosine of theta times the sine of phi notice the first part looks very similar to what we had with polar coordinates and cylindrical coordinates the distance times the cosine of the angle we're just also multiplying by the sine of the last angle y is going to be rho sine theta sine phi and z is rho times the cosine of phi if we're going the other direction from the pythagorean theorem rho squared is equal to x squared plus y squared plus z squared the pythagorean theorem in three dimensions because this time we're going straight to the point instead of just the x y plane because theta is the same theta we saw before it's still tangent of theta is y over x just make sure that you check the domain again and finally to find phi we solve the last equation here for phi z equals rho cosine of phi dividing both sides by rho we get that the cosine of phi is equal to z divided by rho that's usually going to be easier to solve but it does require us to know rho first so if we don't know rho we basically just calculate it it's the square root of x squared plus y squared plus z squared but it'll be usually quicker just to do z divided by rho all right let's see if we can do some examples where we make these exact conversions let's start with taking a spherical point let's take 2 negative 5 pi over 6 comma pi over 6 and convert that to a rectangular point well we need to know what x y and z are in rectangular points and our conversion formula say it's going to be rho which is 2. let's go ahead and label these as rho theta and phi times the cosine of the first angle negative 5 pi over 6 times the sine of the second angle pi over 6. y is 2 sine of negative 5 pi over 6 times the sine of pi over six and z is equal to rho two times the cosine of the last angle just pi over six simplifying these will give us our rectangular point that's equivalent to the spherical point so we've got 2 times the cosine of negative 5 pi over 6 is negative root 3 over 2 and the sine of pi over six is one half and that gives us negative root three over two for the y coordinate we have two times the sine of negative five pi over six that's in the third quadrant so it's negative one-half sine of pi over six is still one-half and so we end up with negative one-half for our y-coordinate and finally with z we have 2 times the cosine of pi over 6 which is root 3 over 2 which is just the square root of 3. so our spherical point of 2 negative 5 pi over 6 pi over 6 is the same as the rectangular point negative root 3 over 2 negative one-half square root of 3. let's try going the other direction let's look at the rectangular point of negative one one square root of six and we're going to convert that to a spherical point well to get spherical we need to know what rho theta and phi are going to equal from our x y and z well our row formula actually says rho squared is equal to the sum of all the squares so negative 1 squared is 1 plus 1 squared is 1 plus the square root of 6 squared is 6. and so that gives us rho squared is equal to 8 taking the square root of both sides rho is equal to the square root of 8 which simplifies to 2 root 2. with theta theta is equal to the tangent of y over x so if we do a tangent inverse of y over x 1 over negative 1 or the tangent inverse of negative 1 negative one tangent inverse is going to be pi over negative pi over four but again we need to take a quick look at where this point is gonna be it's a rectangular point of negative one comma positive one and negative pi over four sticks us in the wrong quadrant so we need to add pi to it and when we add pi to a negative pi over 4 we're adding 4 pi over 4 we end up with 3 pi over 4 is the actual point for phi the cosine of phi is equal to z over rho z is the square root of six divided by rho which we just found out was two root two and when we simplify that we're actually doing the cosine inverse dividing by the square root of two we get root three over two and we know the cosine inverse of root three over two is pi over six and so we've ended up with a spherical point that is equal to the rectangular point of negative 1 1 root 6 that point is 2 root 2 3 pi over 4 pi over 6. similar to what we did with the cylindrical coordinates spherical coordinates can also be expressed as equations or surfaces can be expressed as equations let's do rho equals 5 cosine of phi we can manipulate these equations to try and convert them into the other form into rectangular form and you can see on this one similar to the other example if we multiply both sides by row we end up with rho squared equals 5 rho cosine of phi and rho squared we know from our formulas is x squared plus y squared plus z squared equals the constant of five and then rho cosine of phi that is directly equal to z and so we end up with the equation x squared plus y squared plus z squared equals 5z as being the equivalent equation in spherical coordinates as rho equals 5 cosine of phi so those are our two new systems we're looking at spherical coordinates and cylindrical coordinates sometimes we might want to make a conversion though between these two new systems and it is a lot of work to have to go to rectangular in the middle so let's take a look at how we can make a direct translation between the two how we can convert between our new systems between spherical and cylindrical coordinates well just as before we have some conversions remember that spherical we just did is expressed as rho theta phi and cylindrical is what we started with is expressed as a radius theta times the z distance so if we want to change spherical into cylindrical our conversion for r is that r is equal to rho times the sine of phi theta i mentioned earlier is the same theta between both systems so there's really no conversion needed for that center coordinate and z is going to be rho times the cosine of phi if we're going the other direction we have cylindrical and we want the spherical rho squared from the pythagorean theorem is the radius squared plus z squared again theta is the same theta so there's no work to do there we just need to get after the phi which if we take the z equals equation that we just had and divide both sides by rho we end up with the cosine of phi is equal to z over rho which usually by this point we will have already calculated rows so that's easier but if we haven't calculated rho we just use our row equation and rho is the square root of r squared plus z squared so let's see if we can do some examples where we convert directly between these new systems of graphing in space between spherical and cylindrical let's do a few examples let's take the spherical point first the spherical point two negative five pi over six and pi over six we're going to convert that to cylindrical well for cylindrical we need to know the radius the angle theta and the vertical distance z the radius is equal to rho 2 times the sine of the angle pi over 6 the phi the last angle theta is just theta of negative 5 pi over 6 and z is rho 2 times the cosine of phi times the cosine of pi over 6. simplifying the sine of pi over six is one-half so we end up with a radius of one the cosine of pi over six is the square root of three over two which gives us square root of three and so we now have our cylindrical point of one negative five pi over 6 square root of 3 which is equivalent to the spherical point of 2 negative 5 pi over 6 comma pi over 6. let's do one more example where we go the other direction let's take the cylindrical point of the square root of 2 3 pi over 4 square root of 6 and we're going to convert that to spherical so because it's a cylindrical point we've got a radius a theta and a z i didn't label on the spherical back then we had the rho the theta and the fee so for finding the spherical point we need to know what rho is what theta is and what phi is our row equation says that rho squared is equal to the r squared plus the z squared r squared is two plus the z squared is six which means rho squared is equal to eight taking the square root rho is the square root of eight or two root two theta is the same in both graphing systems so we have three pi over four for phi we know the cosine of phi is equal to z over the rho so we'll do the cosine inverse of z the square root of 6 over the rho to root two and i think we simplified this one before we have the cosine inverse dividing out the square root of two of root three over two cosine inverse of root three over two is pi over six and so we've taken our cylindrical point of square root of two three pi over four square root of six and we've ended up with the spherical point two root two three pi over four and pi over six so that's what we're focusing on today with these new coordinate systems spherical graphing and cylindrical graphing can we convert between all three systems spherical cylindrical and rectangular i want to wrap up really quick by summarizing the three graphing systems visually in three dimensions so we'll make our little coordinate plane here and normally we label them x y and z but just to make it really clear i'm going to call this i j and k as vectors in the x y z direction when we're graphing with the rectangular system with the rectangular system we go a distance x and then we go a distance y and then we can vertically go a distance z to our point that is x y z and each turn makes a right angle that's the rectangular coordinate system when we do cylindrical we say instead what we're going to do is we're going to take a radius that stretches out to where we go vertically at z and that radius is going to make an angle theta with the x-axis and when we do we end up with a point that is the radius theta z with the spherical coordinates instead what we're going to do with spherical coordinates is we're going to go directly from the origin to the point and we're going to call that distance row and that row is going to be formed with an angle of theta with the x-axis it's the same theta as we had with the cylindrical coordinates but also with an angle phi from the z axis and in that way we end up with our point rho theta phi so we now have three ways to represent that same point in space and depending on the context sometimes one context is much easier to work with than the other context but for now we're focusing our energy on learning about the three systems converting between the three systems and interpreting where those points will be so take a look at the homework assignment practice some of those and we will discuss them more in class our next chapter is going to focus exclusively on these things called vectored valued functions so this first lesson we're going to answer the question what are vector valued functions and put quite simply a vector valued function can be defined and 2d as r of t the vector valued function is going to be some function of t times the i vector plus some function g of t times the j vector or in component form f of t comma g of t and similarly it can be defined in 3d as the vector value function r of t is equal to some function of t times the i plus some function of g times the j and some function of t times the k or in component form f of t comma g of t comma h of t basically it's a vector made up of functions in fact we've seen this type of a function before noting that this is really just a parametric equation or a parametric function in vector form when we had parametric equations or parametric functions from a previous chapter we said x was equal to some function f of t y is equal to a g of t and z is equal to some h of t and as t changed it gave us our entire graph our function now we're just expressing those three functions as one function as a vector so now if i was asked to evaluate some vector valued functions for example let's say we have the vector r of t is equal to t squared minus three t i plus four t plus one j and we were asked to find the vectored value function when time is negative four all that really means is we're going to plug negative four into our vector valued function and so we have negative four squared minus three times negative four times the i component plus four times negative four plus one times the j component and simplifying 16 plus 12 is 28 i plus actually we're going to make it minus because we have negative 16 plus 1 is negative 15 j and we've evaluated the vector valued function at time of negative 4. let's try one more let's find the vector valued function r of t is equal to the cosine of t to sine of t and the natural log of one minus t and we're going to find the vector function at pi over 6. well that means we're just going to plug pi over 6 into each component so we have the cosine of pi over six comma two sine pi over six comma the natural log of one minus pi over six and when we simplify we know the cosine of pi over six is root three over two the sine of pi over six is one half times two is one and then we're left with the natural log of getting a common denominator that's going to be 6 minus pi over 6 and we've evaluated the vectored valued function at pi over 6. now as with any case since a function is just a mapping from the domain to the range we can talk about what exactly is the domain of our vectored valued function and similar to functions that you've seen in the past for example we can't have 0 in the denominator we can't have negatives under the square roots or in a natural log similar to those we can apply those concepts of a domain to our vectored value function the only rule that we have to add is our domain must work for all components of the vector valued function so if r of t is the function 3 over the square root of t plus 1 comma the natural log of t minus 5 comma secant of 3t we can go piece by piece to start building a domain that works for the entire vectored valued function one thing we see is t plus 1 is under a square root so we know t plus one must be greater than zero or equal to zero subtracting one we find out t is greater than or equal to negative one another thing we see is we've got a denominator and we know denominators cannot be equal to zero squaring both sides and subtracting one t cannot be equal to negative one which then if we combine that with the first conclusion now we're saying t is just strictly greater than negative one it cannot be equal to negative one because of our work there in the green continuing on we see a t minus five inside a natural log well natural logs must always be greater than zero so we can say t is greater than five which already is greater than negative 1 so we kind of have built this new domain t greater than 5 is the more restrictive domain so we're going to stick with that for now but we still have to deal with the secant because the secant we can take the secant of anything but secant is the reciprocal of the cosine and we can't have zero in the denominator so cosine of three t cannot be equal to zero well if we do the cosine inverse then we get that 3t cannot be equal to cosine is 0 at pi over 2 and then again it's 0 at 3 pi over 2 5 pi over 2 7 pi over 2 basically every plus n pi where n is the natural numbers to get the t alone then we just have to divide both sides by 3 and we get t is equal to pi over 6 plus n over or n pi over 3. and let's get a common denominator by multiplying the second one by two so we have pi plus two n pi over six and that's actually not equals two because that'll make the secant undefined at all of those points and so we can combine all of this together and say our domain is all t greater than 5 where t is not equal to pi plus two n pi over six and that covers all of the components of this vectored valued function now since ultimately we're interested in doing calculus with these vector valued let's talk about the first step of calculus and that's finding the limit of a vectored valued function and we'll start with a definition which might be exactly what you expect that the limit as time goes to any value we'll call it a of the vector valued function r of t is really asking us to just take the limit of each component the limit as t goes to a of the first function comma the limit as t goes to a of the second function and if it's in three space the limit as t goes to a of the last function so doing a couple quick examples if we're trying to find the limit as t goes to negative 2 of the vectored valued function the square root of t squared minus 3t minus 1 times i plus four t plus three times j plus the sine of t plus one times pi over two okay well basically we're just going to take the limit of this vector valued function component wise so we're going to plug negative 2 in for all of the t's as long as it doesn't make anything undefined so we get the square root of negative 2 squared minus 3 times negative 2 minus 1 i plus 4 times negative 2 plus 3 j plus the sine of negative two plus one pi over two okay simplifying negative two squared is four four plus six is ten minus one is nine square root of nine gives us three i four times negative two is negative eight plus three negative eight plus three is a negative five j and then when we do negative two plus one we get negative one half or negative pi over two the sine of negative pi over 2 drawing a quick unit circle negative pi over 2 is straight down at the point zero comma negative one sine being the y coordinate so we get negative 1 or negative k and so this vector valued function is approaching 3i minus 5j minus k as time is approaching negative 2. let's do one that's a little more involved than just plugging in values for the limit we're going to find the limit as t goes to zero three t squared minus five t all over six t squared plus t comma the cosine of t minus 1 over t squared and to make this a little more straightforward i'm going to look at one component at a time first looking at the i component or the x component depending on how you want to look at it if we were to plug 0 in we'd end up with 0 in the denominator of this fraction and that is not acceptable so instead what i'm going to do is i'm going to factor a t out leaving behind three t minus five and a t out of the denominator leaving behind six t plus one and that allows the t's to divide out so now we can take the limit as t goes to zero by plugging zero in for the t's which will leave us with negative five over one which equals negative 5 because we've removed that discontinuity if we look at the j component what we find is plugging zero in for cosine we end up with one minus one or zero over zero squared which is zero and so this is an indeterminate form of zero over zero so what we're going to do is use our old friend l'hopital and take the limit as t goes to 0 of the derivative of cosine is negative sine of t the derivative of t squared is two t but now if we plug zero in the sine of zero is zero and this two times zero is zero we're still going to zero over zero so we'll use l'hopital one more time to take the limit as t goes to zero of negative cosine of t over two and now when we plug the zero in cosine of zero gives us one so we are left with negative one over two which means the limit as t goes to zero of our vectored valued function is going to be negative five comma negative one-half taking the limit component-wise of each part speaking of limits and i've kind of alluded to it with this idea of continuity to remove the discontinuity in order to find the limit let's talk about when a vector valued function is continuous a continuous vectored valued function a vector-valued function is continuous if and only if the limit as time approaches a of the vector valued function r of t is exactly the same as the vectored valued function evaluated at that value a the limit is the same as the value which means the limit has to exist and the value has to exist they have to both be defined and quite often instead of determining where something's continuous as we found in calc 1 it's easier to describe when it's discontinuous when the limit is not equal to the value at the function when we have a gap a jump or an infinite discontinuity so let's say we have the vector valued function r of t is equal to e to the t i [Music] plus the sine of t j what we know about the part e to the t is e to the t is continuous for all values we never have to lift our pencil as we graph it similar with sine of t sine of t is continuous for all values there's no gaps no jumps no asymptotes vertical asymptotes it's actually defined for all real numbers and the limit is also defined for all real numbers so this is continuous for all real numbers so let's do an example that might be a little more exciting let's take the vector valued function r of t is equal to the vector 1 over x comma tangent inverse of x well looking at the first component 1 over x is undefined at 0 which means while the limit exists at 0 the function does not and so we're going to say it's not continuous at zero because we have an asymptote we're also not continuous when tangent inverse is undefined we know tangent inverse is undefined tangent is y over x so it's undefined whenever the x coordinate of the unit circle is zero that happens at pi over two and every n pi after that so we have a two points of discontinuity from one from each component so this video is a real brief introduction to these things called vectored valued function really it's just a fancy way to take a parametric equation and put it into a vector form so it's one single function we'll be using these vector valued functions to do a lot of modeling of different situations do a lot of different calculus tricks but for now we're going to do some homework problems get comfortable with vector valued functions in general and we'll see you in class to answer any questions you may have now that we've taken a look at what vector valued functions are we're ready to answer the question that we really care about because this is a calculus class how do we do calculus of vectored valued functions and let's start with the first part of calculus calc 1 where we took derivatives and quite simply to take the derivative of a vector valued function we're going to take the derivative of each component so let's do some examples where we do just that first let's take the vectored valued function r of t is 6 t plus 8 i plus 4 t squared plus 2 t minus 3 j the derivative of the vector valued function which will denote r prime of t is simply the derivative of each component the derivative of 6 t plus 8 is just six i's plus the derivative of the polynomial is eight t plus two times the j vector and there is our derivative let's try and make them a little more interesting let's say the vector valued function r of t is e to the t sine t times i plus e to the t cosine t j minus e to the 2t k and we're going to find the derivative of our vector valued function r prime of t well our first pr derivative is a product so we remember we have to use the product rule which is the derivative of the first part e to the t times the second sine of t plus the derivative of the second part which is cosine of t times the first e to the t times the i component plus again using our product rule the derivative of the first times the second plus the derivative of the second which is going to give us negative sign times the first which is e to the t times the j component minus the derivative of e to the 2t is e to the 2t using the chain rule times the derivative of the inside is 2 times the k component and now we've got our derivative let's try r of t is equal to t natural log of t i plus 5 e to the t j plus the cosine of t minus the sine of t k and so our derivative r prime of t again we've got a product rule so we take the derivative of the first which is one times the natural log of t plus the derivative of the second which is one over t times the first which is t all times i plus 5e to the tj that's a pretty straightforward derivative plus derivative of cosine is negative sine the derivative of sine is cosine times the k a little bit of simplifying just simplifying that t over t is one so we find out our derivative of the vector vector-valued function is the natural log of t plus 1 times i plus 5 e to the t times j plus negative sine of t minus the cosine of t k and we have our derivative of the vector valued function one of the most common and useful purposes for the derivative of vector valued functions is to find what are called tangent vectors and similar to a tangent line being tangent to the curve at a point being the first derivative it works exactly the same with vectors we're going to find the tangent at a point let's say the vector r of t is equal to the cosine of t times i plus the sine of t times j and we're going to find the tangent vector at the point t equals pi over 6. well the derivative is just component wise gives us the negative sine of t times i plus the cosine of t times j and we're being asked to find the tangent vector specifically at the point pi over six negative sine of pi over six i plus the cosine of pi over six j which if we simplify the sine of pi over 6 pi over 6 is at root 3 over 2 comma one half sine being the y coordinate gives us the one half so we have negative one half i plus the cosine of pi over six is root three over two j this vector then can be positioned so it's tangent to the point at time equals pi over six visually what we're looking at if you remember from parametric equations graphing x equals cosine y equals sine that's actually just the unit circle going around counterclockwise and we're specifically interested at when time is pi over six and if we were to draw a tangent vector right there it could be written in component form as negative one half i plus root three over two j and that would be the same as the component vector coming out of the origin it's going in the same direction because remember vectors don't have a location they just have a direction in magnitude speaking of magnitude there's no rule by just taking the derivative that really dictates the magnitude of that tangent vector and so usually we like to standardize how we think about tangent vectors by saying we're going to be particularly interested in the unit tangent vector that way no matter how you get your tangent vector it's always going to be the exact same length the unit tangent vector is a very special vector that we're going to use throughout this chapter we're going to call it capital t of t and it is simply just the derivative of the basic vector r of t divided by the magnitude of that derivative so first we're going to find the tangent vector and then we'll make it a unit vector by dividing by its magnitude so let's try an example where we do just that let's say the vector r of t is equal to t squared minus 3 i plus 2t plus 1 j plus t minus 2 k and we want the unit tangent vector to this vector we're going to do this generally not at any specific point we're going to do it generally for any point t we could find it at a specific point if we had a t value we were particularly interested in but for this example let's just do it generally so first we need to know the derivative of our vector valued function r which is two t i plus two j plus 1k but that is not a unit vector so to get our special vector our unit tangent vector we're going to find the magnitude of the derivative and we remember the magnitude is the square root of the square of the components two t squared plus two squared plus one squared which gives us the square root of four t squared plus four plus one which is five that tells us then that the unit tangent vector that special vector t of t is 1 over the square root of 4 t squared plus 5. times the derivative which is 2t comma 2 comma 1 in component form if we wanted to we could distribute that divide by the square root giving us just the vector two t over the square root of four t squared plus five comma two over the square root of four t squared plus five comma 1 over the square root of 4 t squared plus 5. and now we have a vector of unit length it's always going to have a length of 1 that will always be tangent to the vector r of t we're going to come back to unit tangent vectors in our future sections and explain why they're so important and helpful to us as we're working with vector valued functions but for now we're just going to find them generally like we did in this example or we could find them specifically at a point if i knew t was equal to 7 seconds we could evaluate that but for now let's step away from derivatives we still have to look at the second half of calculus which is integration and just like derivatives we can take the derivative component at a time we can take the antiderivative of each component but just like before in calc 2 when we took an integral we would always have plus a constant we need to reflect that here also but we need to express that constant is a vector so we're going to say it's plus a constant vector which we will represent with a c with the vector symbol above it representing some vector made up of three constants constant one constant two constant three so for example if we're being asked to find the integral of two t plus four i plus three t squared minus four t j dt we can integrate this one component at a time integrating the i component we end up with t squared plus 4t times the i plus the j component becomes t cubed minus 2 t squared times the j component but because this is not a definite integral we need to also add a constant vector to our final solution and so we're integrating a component one component at a time to get our final solution let's do one last example let's do a definite integral this time let's go from zero to pi over three of the vector valued function sine of two t i plus the tangent of t j plus e to the negative 2 t k dt evaluating one component at a time the antiderivative of sine is the cosine of two t actually the negative cosine to account for the times two we need a one half out front times the i component plus the antiderivative of tangent might remember is negative natural log of the cosine of t j remember that comes from writing tangent to sine over cosine and then using substitution to integrate plus antiderivative of e to the negative 2t is e to the negative 2t but we have to multiply by a negative one-half to account for that 2 times the k component and we're just going to integrate this entire thing from 0 to pi over 3. and we'll do that substitution one component again at a time so first with the i we have negative one-half cosine of two t or two pi over three plus one half cosine of two times zero is zero times i plus plugging pi over 3 into the natural log we get negative natural log of the cosine of pi over 3. plus the natural log of the cosine of zero all of that's your j component plus negative one half e to the negative two pi over three plus one half e to the zero all that times the k component simplifying let's see what we have left cosine of 2 pi over 3 is going to be 2 pi over three is over here on the unit circle so that's going to be negative one half comma root three over two so the x coordinate is the negative one half so negative one half times negative one-half plus one-half times the cosine of zero is one i plus the negative natural log of the cosine of pi over three one pi over three is a positive one half root three over two so the natural log of one half plus the natural log of cosine of zero is one that's going to be nice because the natural log of one is zero times the j component plus not much simplifying that we're left with on this last part i'm going to put the positive one-half in front just because it makes it pretty minus one-half e to the negative two pi over three times the k component and so one time through left to simplify we end up with one fourth plus one half is three fourths high minus the natural log of one half j plus one half minus one half e to the negative two pi over three k and now we found the antiderivative of our vector valued function from zero to pi over three so calculus with vector valued functions not too difficult it's just basically extending what we saw in calc 1 and calc 2 with derivatives and integrals just doing it with every single component all the way through we also introduced this idea of the unit tangent vector that's going to be very important to us as this chapter progresses so take a look at a few of these on the practice assignment we will see you in class to answer any questions you may have today's video is going to look at answering the question of how do we measure arc length and curvature and we'll talk about curvature a little more when we get to it but first we're going to address the question of arc length for our vectored valued functions and we're going to build this based on what we know from parametric curves when we did parametric curves we said arc length s is equal to the integral from a to b where a and b were the edges of the curve of the square root of x of t prime squared plus y prime of t squared dt well similarly if the vector valued function r of t is equal to the component form f of t g of t h of t then we have arc length is equal to the integral from a to b of the square root of the first component's derivative squared plus the second component's derivative squared plus the third components derivative squared dt and really actually finding the square root of all the components squared we've seen that formula before this is the formula for the magnitude of the vector so instead we can simplify our arc length formula to the integral from a to b of the magnitude of the derivative of the vector and that becomes our formula for arc length we take the derivative of the vector we find its magnitude and then we integrate it over its range so let's do an example using this new formula let's say r of t the vector valued function is equal to two t squared plus one comma two t squared minus one comma t cubed and we're gonna find the arc length as time goes from 0 to 3. well first we need to find the derivative of our vector valued function which is going to be 4t comma 40 comma 3t squared next the formula says we need to know the magnitude of the derivative of the vector-valued function and magnitude we know is the square root of the square of its components 16 t squared plus 16 t squared plus 9 t to the fourth which simplifies to the square root of 32 t squared plus nine t to the fourth and actually if we factor out a t squared we get the square root of t squared times 32 plus nine t squared and we can take the square root of the t squared to get our total magnitude is t times the square root of 32 plus nine t squared now we're ready to go to our formula which says that the arc length is equal to the integral we're going from zero to three of the magnitude of the derivative of our vector valued function which we just found out was t times the square root of 32 plus 9t squared dt we can take this integral quite quickly using u substitution where u is 32 plus nine t squared that makes d u equal to 18 t dt so we'll multiply by 18 inside and one over 18 outside to get one over 18 times the integral plugging 0 and 3 into the u equals equation we get 32 plus 0 and plugging the 3 in 3 squared is 9. 9 times 9 is 81 81 plus 32 is 113 the 18t dt becomes our du and the square root becomes u to the one half which we know is u to the three halves times two-thirds with the one over eighteen on the outside integrated from thirty-two to one thirteen two over eighteen is going to reduce to 9 so we're going to have 1 over 27 times and i don't think this is going to simplify so we'll leave it as 113 to the three halves minus 32 to the three halves and that's an ugly decimal so we'll call that good for the arc length of our vectored valued function so using our formula arc length is simply the magnet the integral of the magnitude of the derivative of the original vector valued function really we've done arc length before when we worked with parametric curves so we're going to move to the next topic which is curvature and curvature is an important concept to us as we look at a curve curvature measures how sharply a smooth curve turns if it's got a real sharp tight curve then you'll get a larger curvature if it's a nearly flat wide curve you're going to get a much smaller curvature curvature measures how sharply the curve turns similar to slope or the derivative measuring how fast the curve changes curvature measures how sharply the curve turns and there's a couple formulas for curvature depending on the context that we're working in we use looks like a k it's kind of a cursive k for our curvature symbol curvature is equal to the magnitude of the unit tangent vector derivative divided by the magnitude of the regular vector's derivative this function works best in 2d so if we have a two-dimensional vector we'll find the unit tangent vector and the magnitude of its derivative divide by the magnitude of the derivative of the valued function however if we're in three dimensions there's a quicker formula that can help us because we can use the cross product in three dimensions in three dimensions we'll take the derivative of the vector valued function and we'll cross it with the second derivative of the vector valued function find its magnitude and divide by the magnitude of the derivative raised to the third power that formula is going to be best when we're working in three dimensional vector space now this can be simplified using functions to show that this is the absolute value of the second derivative of y divided by one plus the first derivative of y squared raised to the three halves power and this function this equation is the one we'll use best with y as a function of x so if we just have a function y equals we'll use the last equation if we have a two-dimensional vector valued function we'll use the first equation if we have a three-dimensional vector valued function we'll use that center equation so let's take an example where we use each of these equations to measure the curvature or how sharply the graph curves these three equations are going to be really key to this section so let's highlight them as we go into our first example let's say we have the function r of t is equal to 5 cosine of t i plus 5 sine of t j since we're just working with i and j we have a vector valued function in the plane in two dimensions so we're going to use that first equation the magnitude of the derivative of the unit tangent vector divided by the magnitude of the derivative of the vector valued function probably easier just to start finding all the pieces and then dividing at the end so first the derivative of the vector valued function r of t is negative five sine of t i plus five cosine of t j well for the denominator we need to know the magnitude of that so the magnitude of the derivative of the vector valued function is the square root of the component squared which is 25 sine squared of t plus 25 cosine squared of t which is really nice because if we factor out the 25 we get the square root of 25 times sine squared of t plus cosine squared of t and we know sine squared plus cosine squared is one so this is just going to be the square root of 25 or five which means it should be really easy to find the unit tangent vector we know the unit tangent vector is the derivative vector divided by its magnitude so one-fifth dividing by the magnitude times five sine of t i plus i'm sorry that's negative five sine of ti plus 5 cosine of t j distributing the 1 5 through making it a unit tangent vector it becomes negative sine of t i plus the cosine of t j the numerator is the derivative of the unit tangent vector so we'll take this unit tangent vector that we just found negative sine of t i plus the cosine of t j and find its derivative which gives us negative cosine of t i minus the sine of t j well the numerator is the magnitude of this vector the magnitude of the unit tangent vector which in this case is really nice because it's the square root of the components which is cosine squared of t plus the sine squared of t and we know sine squared plus cosine squared is one and so now we're ready to go to our formula for curvature in 3d in 3d the curvature is the magnitude of the unit tangent vector's derivative divided by the magnitude of the vector-valued function's derivative well the tangents derivatives magnitude we found out was one the magnitude of the derivative of the vector we found out was five and so we have a curvature of one-fifth it turns out in this case that the curvature is constant because this entire shape is a circle it's a circle of radius five and so it curves at a rate of one-fifth all the way around the entire graph but that's just a 2d example let's take a look at an example in 3d let's take the vector valued function r of t is equal to 4 cosine of t for sine of t and three t now it's off my screen so i'm just going to copy again that curvature in 3d is going to be the magnitude of r prime of t crossed with r prime prime of t divided by the magnitude of r prime of t cubed just to remember the formula which is off my screen so we just need to start finding these pieces so let's start with finding r prime of t r prime of t is negative four sine of t comma four cosine of t comma three we also need to know r prime prime of t which is negative four cosine of t negative four sine of t and zero to make the numerator we need to cross those two vectors so r prime of t crossed with r prime prime of t and when we cross these we get 0 minus a negative 12 so positive 12 sine of t comma 3 times negative 4 is negative 12 cosine of t minus 0 comma 16 sine squared of t minus a negative so plus 16 cosine squared of t which if you factor on that k component factor out a 16 we have sine squared plus cosine squared so that's actually going to simplify just to 16. so let's rewrite this vector simplified with 16 in the k component the numerator of the curvature is the magnitude of r prime crossed with r prime prime of t so let's find the magnitude of this vector which is the square root of 144 sine squared of t plus 144 cosine squared of t plus 16 squared which is 256. factoring out the 144 we're left with cosine squared plus sine squared which is one so we're left with 144 plus 256 which is the square root of 400 which is 20. so our numerator for the curvature is 20. the denominator is the magnitude of our prime of t cubed well we found r prime of t already in our first step up above in blue so let's just run the magnitude formula which is the square root of the components 16 sine squared of t plus 16 cosine squared of t plus nine again if you pull the 16 out of the first two terms you're left with 16 times sine squared plus cosine squared which is one plus nine we get the square root of 25 which is just five and so finally we're ready to calculate the curvature of our vector valued function curvature we said using the formula in the top right here is the numerator is the magnitude of the cross product we found out was 20 divided by the magnitude of the derivative which is 5 cubed which is 20 over 125 which dividing by 5 will give us 4 over 25 as our final curvature curvature in three dimensions measuring how sharply a corner is turning we had a third formula though for curvature so let's take a look at it the third formula is used best when we have a function written out like f of x equals x squared how sharp is this function turning how is the curvature measured well again it's off my screen but you have it on your notes up above that the curvature when we have a function is the absolute value of the second derivative divided by one plus the first derivative squared raised to the three halves power so let's find all those pieces first derivative is two x the second derivative is just 2. and so if we plug into our curvature function we get the absolute value of the second derivative which is 2 divided by one plus the first derivative which is two x squared all raised to the three halves power cleaning up a little bit the absolute value of two is just two over one plus four x squared raised to the three halves power and this time our curvature is not a constant but it depends on how big x is and that makes sense because we know x squared starts out with a real sharp curve in the center but on the outsides it's becoming more and more flat in fact let's take a look at what the graphs of x squared looks like compared to the graph of its curvature so here i've graphed the graph of x squared and now i'm going to graph its curvature which we said was two over one plus four x squared raised to the three halves power and what we see is the curvature is at its largest value its steepest point at zero but as it gets out further and further and further the curvature gets less and less and less and so we see the curvature increases as the curve sharpness increases and that's what we're measuring with curvature how sharply how fast is this smooth curve turning so we've got three formulas for curvature one formula that helps in two dimensions one in three dimensions and one when we have y as a function of x instead of a vector valued function take a look at the homework assignment also doing some arc length problems to see if you can get comfortable with them come to class with questions and we will see you then now that we've gotten comfortable working with these vector valued functions we're ready to address the question how do we model motion in space over time and the idea is very similar to when we were working with derivatives and integrals we can talk about the position of a particle we can talk about its velocity and we can talk about its acceleration in fact there's one more thing we can talk about too now that we have vectors and that's the idea of speed but we'll get to that in just a minute and what's nice with vector valued functions is now we can extend those ideas of position velocity and acceleration into three dimensions the vector valued function r of t can represent the position of a particle at time t in each direction so the i component can represent how far it's moving horizontally over time the j component represents vertically over time and the k component can represent the third dimension over time and so with each component we can represent motion in each of the directions then we can come up with a velocity vector which is equal to as you might expect from calculus 1 the derivative of the position vector and this represents the velocity at time t in each direction so it might be moving faster vertically than it is horizontally maybe if you shoot a canon vertically up it's moving vertically really fast and maybe horizontally just a little bit the vector v of t can show that velocity in each direction and then as you might expect the acceleration vector a of t is the derivative of the velocity vector or the second derivative of the position vector representing the acceleration at time t in each direction and we're going to introduce a fourth thing to this discussion and that's going to be what we call v of t and notice this is not marked as a vector this is a function that's not a vector but it's going to have a scalar solution and the reason for that is the v of t is going to represent the speed and the way we calculate that is we take the magnitude of the velocity vector so we want to be very careful that if it's a vector it represents the velocity if it's a scalar it represents the speed another way we could say that because the velocity is the derivative of the position we could say that's the magnitude of the position vector this represents speed not in any particular direction but the overall speed of the particle so let's do some examples where we look at these concepts let's say we've got the position vector r of t is equal to t squared i plus the square root of five minus t squared j if we wanted to find the velocity vector with respect to time then we just take the derivative of the position vector taking the derivative of each component we get 2t times i plus square roots of one-half power so we have one-half times five minus t squared to the negative one half times the derivative of the inside which is going to be negative two t times our j cleaning that up a little bit we've just got two t i dividing out the twos minus a t times five minus t squared to the negative one half j this then is our velocity vector at time t in the direction of i and j if i wanted to find the acceleration vector at time t we just take the derivative of the velocity so on the i component we just have 2i minus on the j component we've got a product rule the derivative of t is one times five minus t squared to the negative one half plus the derivative of the negative one half power is negative one-half times five minus t squared to the negative three-halves times the derivative of the inside which is negative two t then we have to also multiply by the first factor which is t all that times j well let's clean that up a little bit so our final acceleration vector is two i minus five minus t squared to the negative one half the twos divide out and the negative times a negative is a positive so we're gonna have plus t times t is t squared times five minus t squared to the negative three halves all that times our j a little bit more complex but that is our acceleration vector we can also find our speed speed is v of t it's the magnitude of the velocity vector so magnitude is the square root of each component squared so we take the 2t squared is 4t squared plus we have a t squared and then the 5 minus t squared to the negative one because when we square one-half the negative one-half the twos divide out and that does not look like it's going to simplify at all so that represents the speed of our particle represented by the vector r of t let's try one more example to make sure we can do velocity acceleration and speed let's say we've got a vector r of t which is equal to let's do component form this time t squared minus three t two t minus four and t plus two the velocity vector then is the derivative of that which is two t minus three 2 and 1. the acceleration vector is the derivative of that vector derivative of the velocity which is 2 0 0. so it's only accelerating it turns out in the x direction or towards the i vector every other vector the acceleration is not changing it's a constant velocity of two or one if we just want v of t remember that's the magnitude of the velocity vector so we take the square root of each component squared which becomes squaring two t minus three we get four t squared minus twelve t plus nine plus two squared is four plus one squared is one and so for our speed it's going to be the square root of 4 t squared minus 12 plus 14. and so we have our acceleration velocity and speed values one nice application of these vector valued functions with velocity acceleration position is this idea of projectile motion projectile motion is when an object is moving through space and the only force acting on it is gravity and in two dimensions projectile motion can be represented with the vector valued function s of t the initial velocity times time times the cosine of the angle that it begins moving at times the i vector plus the initial velocity times time times the sine of the angle but since the second component represents the vertical distance we also have to subtract the force of gravity which is one half times gravity times time squared times the j vector this formula can be used to model projectile motion into space so let's see if we can do an example where we use that formula let's say a cannonball is fired from a cannon on a cliff towards water the cannon is aimed at a 30 degree angle or pi over 6 above the horizon which just means above the flat line and the initial speed of the cannonball is 600 feet per second we also know that the cliff is 100 feet above the water so if we were to draw a picture we've got a cliff with a little cannon and it's all above water and the idea is we're going to shoot this cannonball and it's going to go up and ultimately land in the water we know the cannon makes a 30 degree angle with the horizon we know the initial speed coming out is 600 feet per second we know the cliff is a hundred feet above the water and the first thing we're going to do is model this situation with a vector valued function and as a hint since we're in feet we're going to do the english version of gravity which is 32 feet per second squared well using our model up above the vector s of t is equal to the initial velocity which is 600 times time times the cosine of my angle i plus 600 times time times the sine of my angle minus one half of gravity and since we're in the english system we're going to use 32 feet per second times time squared times the j component we can do a little bit of cleanup that our vector s of t is equal to the cosine of 30 a 30 degree angle is at the point root 3 over 2 comma one half so the cosine of 30 is going to be root 3 over 2 and then divide by 2 will reduce with the 600 leaving behind 300 root 3. actually let's put the t in there 300 t root 3 times the i vector plus the sine of 30 is going to be one half so half of 600 is 300 t minus half of 32 is 16 t squared times the j vector this vector valued function then is going to represent both the horizontal motion over time in the i component and the vertical motion over time with the j component so now that we have our equation we can answer questions such as finding the maximum height if i'm interested in the maximum height all i really care about is the component that describes the height and that height is going to be at its maximum point when the velocity is zero or as we know in terms versus put on velocity equals zero or in terms of what we talked about in calc one the maximum height is when the first derivative s prime of t is equal to 0. and we particularly only care about the y component because we just want to know how far vertically it is so we're going to look just at the 300 t minus 16 t squared and we're going to take its derivative which is 300 minus 32 t and set that equal to zero which is a quick solve 300 equals 32 t and dividing by 32 t is equal to about 9.375 so at 9.375 seconds this cannonball has reached its maximum height but the question wanted us to actually find the maximum height not just the time when it occurs so what we need to do is plug that time in to the position vector to see how high it actually is so we've got 300 times 9.375 minus 16 times 9.37 squared and if i work that out on my calculator we should get 14.06.25 feet so it shoots from 100 feet up another 1400 feet into the air at its maximum height after 9.375 seconds if i was also interested in how far horizontally this reaches over the water we could plug that 9.375 into the x component or the i component to see how far horizontally over the water we are but the question didn't ask about that so let's move on to question c question c i want to know how long until it hits the water i'm going back to my picture cannonball hits the water all the way out on the right side here what we know there is the y component the vertical distance is not zero because zero is when it hits the same level of as the original cannon we've got to drop another 100 feet down so we actually are asking when does the vertical component equal negative 100 100 feet below the cliff and we know that the vertical component is 300 t minus 16 t squared so we want to know when that equals negative 100. well putting things in order and moving the 100 to the other side actually let's make everything positive let's move everything to the right side when we do we get 0 equals 16 t squared minus 300 t minus 100 and we know we can solve this using the quadratic formula t is equal to the opposite of b plus or minus the square root of b squared minus 4 a c all over 2 times a although we know that we want time to be a positive number so we're probably not going to need to worry about the subtraction that's probably going to give us a negative number and so when i put this in my calculator we get 19.08 seconds which means putting our answers together after nine and nine seconds and change the cannonball reaches its maximum height and then after 19 seconds and change the cannonball has crash-landed into the water well if it crash lands into the water what i might be interested in as a military personnel is how far out to see will the cannonball travel in other words what is my range with my cannonball on shore here well we just found out in part c that it hits the wall hits the bottom or hits the water at 19.08 seconds and we're asking how far out does the cannonball travel horizontally that's going to be the x component this time we're going to be interested in the 300t divided by the square or times the square root of x so 19.08 seconds in the x component or the i component so the i component then is 600 times time and we just found out time was 19.08 seconds times the square root of 3. plugging that in the calculator we end up with 9914.26 feet the cannonball can travel just shy of 10 000 feet in those 19 seconds when it crashes into the water so today we're looking at an application of these vector valued functions motion in space we're basically taking our vectors which represent the position velocity and acceleration in each direction knowing that the velocity is the derivative of the position the accelerations the derivative of the velocity and if we're also interested in general speed that's not a vector it's a scalar it's the magnitude of the velocity vector take a look at the assignment practice a few of these and we will see you in class to discuss them further