hello and welcome to college physics 1 lecture 1 mathematics review before we begin this series of lectures i just wanted to point something out that this is a series of lectures intended for my students in my class but i have made these videos publicly available so for those of you that might be here from a public viewing on youtube first of all understand that you might see college physics and university physics the difference between the two is the math that they're based upon college physics is an algebra based physics while as university physics is mostly a calculus based course so being college physics we are using algebra in this case a second point of note is that because this is for my students personally you might not find exactly all the material you need or you're looking for if you've come publicly to see these videos so please understand that might be the case uh so with that said let's go ahead and jump right in a study of physics at this level in this course requires some basic math skills the relevant math topics are summarized in these first two lectures that we'll have in the first lecture here we're going to cover uh exponents fractions and solving equations it is strongly recommended that you review the material in these lectures and become comfortable with it as quickly as possible so that during your physics course you can focus primarily on the physics concepts and procedures that are introduced without having to become you know too distracted by the unfamiliarity with the math that is being used so i like to go through a math review before we begin so let's go ahead then and jump straight into the material it's going to start out pretty basic i think most of us will know what an exponent is but still it's good to cover so anytime you see a value with a superscript in other words a smaller number on the top right such as the standard form x raised to the m the superscript m is called an exponent and the base number uh x is said to be raised to the nth power so in the case of the example on the right we would say the base number 4 is raised to the third power because the exponent is a three the exponent itself defines how many times you're going to multiply the base number by itself so again using the example on the right an exponent of 3 indicates that we're going to multiply 4 by itself 3 times in other words 4 times 4 times 4. also note that you can have an exponent that is a fraction so the standard form here being x raised to the 1 over m the denominator in other words the m in this case is the root for example the exponent of one-half so if we had say x raised to the one-half that is the square root of x now you could have a fraction that's one-third or one-fourth or what have you for our purposes we're only going to end up dealing with square roots so we're not going to talk about cube roots or anything like that so just know that fractional roots are or fractional exponents are roots now exponents will obey several rules in mathematics so what i mean by this is you know we have to ask okay well what's going to happen to exponents as we multiply two values or add two values together so let's go through some of these rules we'll begin with the product rule which means multiplication in mathematics so let's say we have some base number x raised to the nth power times x raised to the nth power the product rule for exponents states that if you multiply two values with exponents you take the base number and raise it to the sum of those two exponents in other words x raised to the m plus n so as a general example let's say you have three raised to the third power times three raised to the second power you keep your base number three and then raise it to the sum of the exponents which is three plus two three plus two hopefully we can do that is five so that gives us three raised to the fifth power or two hundred and forty three next up is a very similar situation but this time instead of multiplying our two values we're going to divide hence the quotient rule so this is a situation a situation in which we have x to the m divided by x to the n in this case instead of adding exponents we subtract them so you'll have your base number x raised to the m minus n so for our example here let's just go ahead and use these same numbers 3 to the third power divided by three to the second power we keep our base number three raise it to the difference of the two exponents so three minus two which gives us one an exponent of one is basically meaningless you could just get rid of it and it wouldn't change anything that's just 3. next up is what we call the first power rule so this is when you take a number raised to an exponent and then raise that to another exponent so we could say x raised to the nth power raised to the nth power in this case you just multiply the exponent straight across so we have x to the m times n so let's say we have two raised to the second power then raised to the third power that gives us two raised to the sixth power or 64. okay there are a few more uh i will say that there's no like technical name or like detailed name for each of these so i kind of made up my own um so the next one we'll call the distributive rule so i call it this because this is when you have two different base numbers that are then raised to an exponent in such a case you distribute that exponent to everything inside so in this case you have x to the m times y to the m the m is distributed to both base numbers this also works if you have values being divided and then raised to the nth power again the m exponent gets distributed to both even if you have like a hundred different terms inside of this parentheses which you know hopefully that never is a thing you experience but if even if you have 100 things in here every single one of them will be raised to the nth power in this situation next up is the negative exponent rule oh sorry i forgot my example so if we have 3 times 2 raised to the 4th power that's the same as saying 3 to the 4th power times 2 to the fourth power then you can kind of just solve from here that gives you 81 times 16 or 1296. uh our fifth rule is the negative exponent rule this is more so just a statement than it is really a mathematical operation but exponents can also be negative anytime you have a negative exponent that is indicating that your base number is in the denominator of a fraction so this is the same as saying 1 over x to the m this one tends to stump students every once in a while when it comes up in the middle of math problems so be careful with this one so as an example 2 raised to the negative 1 power it looks kind of weird so let's rearrange it and say that's 1 over 2 to the first power well remember the first power is basically nothing so it's just one over two which is one half so it's 0.5 if you move it into decimal form the last one we'll look at is one we already covered a few minutes ago this is the roots rule so whenever you have a exponent that is a fraction remember that is a root and in our purpose for our course again we're only going to deal with square roots so that would be a number raised to the one-half power so for example 4 raised to the one half power that's the same as saying the square root of four which is two okay so that's the first of our three topics for this first lecture next up uh we're just gonna take a look at fractions we're not going to do any examples with these we're not going to do anything detailed all we're going to do is glance over the rules for dealing with fractions in mathematical operations so we'll start with the simplest one multiplication so we're going to ask what happens if you have two fractions and you multiply them well my simplest way of saying it is you multiply straight across the fractions so what does that mean well for example let's say we have a over b times the fraction c over d a longer version of what to do states that you multiply the numerator values together separately multiply the denominator values together and then combine them but for a shorter way of saying it i'd just say multiply straight across so a times c over b times d that's it that's the easiest of our rules for fractions division tends to stump students a little bit more the shortest way of saying this rule is to multiply by the reciprocal of the denominator that's a little technical so here's a straighter definition you can invert the fraction in the denominator then multiply it by the fraction in the numerator so let's take a look at this let's say we have the fraction a over b divided by the fraction c over d so what we're going to do let's follow this in parts invert the fraction in the denominator what that means is flip it over so let's say now it's d over c you then take that and multiply it by the fraction in the numerator okay so flip this it's d over c and now we're going to multiply it by the value in the numerator which was a over b so this is what we say as multiplying by the reciprocal of the denominator now it's just the multiplication rule multiply straight across you end up with a d over bc okay so for the students in my course you will encounter this every once in a while in some math problems down the road so make sure you're comfortable with this one moving forward the next one is arguably one of the more challenging ones or at least the more involved the good news is i don't think we use it much if at all in our course but i still want to go over it since we're already here this is when you add or subtract fractions in general the shortest way of saying this one is you need to find a common denominator stating this in full you multiply each fraction by a factor that equals 1 so that they have the same denominator so let's say we start with a over b plus or minus c over d and note i'm saying plus or minus because this rule works for both addition and subtraction so plus or minus just means whichever one you're dealing with so to do this you need to find a common denominator so on the left we have a fraction being divided by just b on the right we have one divided by just d these are not common denominators they're not the same so our goal is to set it up so that they do have the same denominator in other words we need to have a d on this left hand side and we need a b over here on the right hand side so to do that we're going to multiply each side by one now this might look a little sloppy i'll probably get rid of this in a second but we're going to try to multiply both sides by one to do that notice that we need a d value in the denominator on the left side so why don't i multiply by d over d d over d is just one so i'm not changing the value here at all but i am allowing a d to be in the denominator well do the same kind of thing on this side this is going to look sloppy again just kind of ignore that we need to get a b over here so why don't i multiply by b over b which is again just one so i'm not changing the value but now it gives me a b in the denominator so what i've done is just written as the next part of this equation so that's what i did to get to the second step okay so here we have that so d over d times a over b plus or minus c over d times b over b now again we can just use our multiplication rule and multiply straight across this will give us if you rearrange it alphabetically a d over bd plus or minus bc over bd notice they're both divided by bd so we have now found a common denominator so we can have it as a single denominator across everything so we have just bd on the bottom and then ad plus cb on the top that is in very quick terms how you find your common denominator to simplify the addition or subtraction of fractions to close out this lecture i mean it's still only about halfway over but our last of the three concepts we're going to cover in this one is solving equations so oftentimes one of the symbols or variables that you find in one of your equations will be considered unknown in other words you don't know what the value is you're going to have to solve the equation to get it and so one thing to keep in mind is anytime you perform a mathematical operation on one side of the equation you have to do it to the other side of the equation as well for the equation to remain valid in other words if you multiply one side by five the other side has to also be multiplied by five otherwise it's not equal anymore so what we're going to do is just go through a series of random examples so for example here let's solve a bunch of equations for x we're going to start with a really simple one as you can see and i think we're going to do six of them each one may be adding in an extra step perhaps so let's go ahead and try to solve some of these more simple equations so equation one is y is equal to x over five our goal here is to solve for x to do this there's really only one step we need to get x on its own which means we have to get rid of this five since the 5 is in the denominator to get rid of it we're going to multiply both sides by 5. right so multiply both sides by 5 to get rid of it and we should get our answer immediately so let me write this out here i'll use an arrow to indicate we're doing something new so x if we multiply both sides by 5 will be equal to 5y and that's it can't really have a simpler equation to solve note if you're my student i do always ask that you show work in this case you don't have to do what i did initially and write a 5 times y equals x over 5 times 5. if anything i actually think that clutters your work up and makes it a little messier so don't feel like you need to do that if you don't want to also recognize there's no numbers we're just rearranging the equations for x uh the next one we have ax minus y equals 2y okay so one of the things you're going to start to recognize is that there are multiple ways to solve an equation typically the first thing i would do here is i recognize that there's an x term here with this extra little negative y term well we want to get x on its own so typically what i'll do in this case is take that extra term in this case the minus y and get it to the other side so let's add y to both sides that will leave us with ax on the left-hand side equal to 2y plus y on the right-hand side so again all i did was add that negative y to the other side so this is just 3y right so 2y plus y is 3y okay so we have ax equal to 3y so if we want to solve we have one step remaining we have to get rid of the a to get rid of the a we divide by a so this leaves us with x is equal to 3y divided by a and there is our solution uh one thing to note for my students is i do ask that you box your answers especially when we get into longer and more complicated problems i want to make sure i know what you are telling me is your answer so make sure you box answers just get in the habit of doing that throughout the semester okay several more again trying to up the difficulty a little bit each time number three is a times the quantity x plus b equal to c now this is one where you definitely have students split on what to do first i'm going to just choose the shorter of the two options but i'll tell you what the longer one is you could take this a and distribute it to both the x and the b in other words you can end up with ax plus a b but we can save ourselves a step and just take that a and divide it straight over to the other side before we even worry about distributing so let's go with that route let's take that a and divide it to the other side that leaves behind the x plus b on the left and leaves us with c divided by a on the right hand side now there's only one step remaining subtract the b to the other side and so we have x is equal to c divided by a uh minus b and now we have x on its own so that's our solution okay number four y divided by x plus a is equal to b now this one's a little bit different because we now have our x value in the denominator of a fraction so before we worry about that i'm going to do what i did in the second problem and recognize that there's this additional term hanging out here that i want to get to the other side you don't have to do this first but i would take the a and subtract it to the other side so that leaves y divided by x equal to b minus a now this one this situation here does come up every once in a while in physics not with this exact equation but a similar situation and students again tend to get confused because the x is in the denominator we're gonna have to multiply it across to the other side so that it's out of the denominator excuse me so that leaves the y on the left hand side equal to b minus a times x okay so i multiplied x to the other side we have one step remaining x is being multiplied by this quantity b minus a so let's divide it to the other side that will give us x is equal to y divided by b minus a and there is our solution for number four okay uh we have two more number five this one's not particularly difficult but it does introduce something new again and that is a square root so we have the equation the square root of x plus 3 then minus y equal to 0. once again we have this little extra term hanging off on the side so i would first move that over so that gives us uh the square root of x plus 3 equal to y right so i just added y to the other side now we have to deal with this square root because the x is underneath it we can't subtract 3 over yet because it's underneath the root so we have to deal with that square root so how do we get rid of it well remember a square root is the exponent one half to get rid of one-half you multiply by two in other words we square this term so to get rid of a square root you square it which means we're gonna have to square the right-hand side as well so this will give us x plus 3 equal to y squared okay square both sides at this point the last step should be pretty apparent we just have to subtract the 3 to the other side so we have y squared minus 3 as our solution getting a little shaky there okay fantastic so last one this one will be the most steps we've seen because in this one we have x in two different places the equation says a times the quantity x minus y is equal to b times the quantity x plus y now again there are multiple rods you could take to solve the first thing that i would do is distribute a into the parentheses and distribute b into the parentheses so let's go ahead and do that we're distributing a so we have a x minus a y and now we distribute the b b x plus b y so that gets rid of the parentheses at least now notice we have two terms containing x and two containing y our goal here is to get the x terms on one side of the equation and our y terms on the other so in this step all we're going to do is some rearranging i'm going to take my bx and subtract it to the left hand side and take my a y and add it to the right hand side this will leave us with ax minus bx equal to a y plus b y to rearrange my hand here a little bit uh a y plus b y okay so i've just moved two terms now you can see that we have all x terms on one side all y terms on the other so we're getting there but we do still have x in two places we need to find a way to combine those x's well notice x is being multiplied to both a and b so why don't we factor that value out in other words factor out x so that you have x times the quantity a minus b and you don't have to do this but let's factor out the y as well equal to y times a plus b so now we only have one x and one y that kind of helps a little bit we only have one step step left um i'll write this down in the bottom here we need x on its own it's being multiplied by the quantity a minus b so let's divide that to the other side x is equal to y times a plus b and divided by a minus b so this would work as your solution we have got x on its own in terms of y now again you might have a different brain in terms of how you operate some of these equations so i might have done something differently than you might have seen when you were first looking at each example again that doesn't mean you're wrong it just means you're looking at it in a different way as long as you get a identical answer you're fine you just figured out a different way to solve the problem okay we're going to look at one more thing here oh and by the way notice that i didn't plug any numbers into these all i did was rearrange 4x so i didn't plug any numbers in for a or b this is very common in physics so it is widely believed that it's easier to work physics problems by substituting numbers in at the very beginning of the problem a lot of my students will do that they'll plug their numbers in right away in the beginning however expert physics problem solvers will typically solve the physics problem using the variables so for example a b y what have you and then substitute the numbers in only as a final step so once we have these boxed answers then you plug the numbers in because this approach is not only generally faster and easier but it also reduces the number of stakes you'll the number of mistakes you'll make and results in a better and more general understanding of the physics in the problem it might take a little bit of practice to get used to doing that but it has some very important benefits so i encourage my students or anyone watching to first rearrange your equation and then plug your numbers in it took me a while to get used to doing that but i will never go back it makes things way better so please get used to doing that okay so to conclude this lecture we're going to look at one more type of equation and it's a very specific one and that is quadratic equations we only encounter these i think twice in my particular physics class but it's good to get a primer on it now so a quadratic equation is any equation of the form ax squared plus bx plus c so the key here is it has a term that's squared one that's to the first power and then one that's to the zero power or there's no variable x in the third term the letters a b and c are just constants so there are a couple ways you can sometimes solve these you could try to use like a foil method but in our purpose we're going to use the quadratic formula to solve this is a pretty commonly remembered equation even though it's pretty hideous looking the quadratic formula says your value x is equal to negative b plus or minus the square root of b squared minus 4ac all divided by 2a now notice there's a plus or minus symbol that means that you're going to end up with two solutions one when you use a plus sign and one when you use a minus sign in the context of our physics problems that we deal with in the course you'll be able to determine if you understand the problem which of the two values is the real solution you'll we'll talk more about that when we get there so how do you solve one of these let's just go through the steps and show an example first step get your equation into the form of ax squared plus bx plus c that sounds weird because i mean if it's a quadratic equation that should be its form right well yes but it might not always look exactly like that as an example it could be say 2x squared equals 3x plus 5. it doesn't necessarily look exactly like the equation above so let's rearrange it so that it does take the two terms on the right-hand side and subtract them over we have 2x squared minus 3x minus 5 equal to 0. so now if you compare it to what's above it's very clearly in the same form this is indeed a quadratic equation our next goal is to identify the values of a b and c so in the context here a is whatever is out in front of your x squared term b is whatever is out in front of your x term and c is whatever number is out on its own so in the context of this example a is 2 b is negative 3 and c is negative 5. note the negatives do matter once you know your abcs you substitute them into the quadratic formula this is a tedious process but you just kind of have to pray to the calculator gods that you do it right so this is what the equation looks like with our numbers plugged in and hopefully if we solve it correctly using both a plus and a minus you can then solve to get negative 1 and 2.5 okay so that is it for our very first lecture of physics 1. this is just a primer so this really isn't even the physics material yet in our next lecture we're going to finish our math review by talking about direct and indirect relationships the equation of a line plane geometry areas and volumes and then conclude with the most important of all those uh trigonometric functions sine cosine and tangent until then thanks for watching and have a great day