hey there everybody welcome back today we're going to take a look at 11.5 in calculus uh in the anton book this is on parametric equations of lines so let's just jump right into it let's we've always visualized lines as just being online and points on it but maybe for the sake of this argument let's picture it as a road and we or a yeah let's say a road and i've got some vehicle starting at p0 and it drives to pt and it takes time t to get there so it's heading along this road right here now it could keep driving in a straight line it could take as long as it needs but it's driving at a specific speed let's say its speed is locked uh speed is another way of saying speed in physics is velocity velocity is the same thing as speed but it has direction uh and so speed like velocity is a measure of distance over time like miles per hour miles is distance hour is a time frame so velocity is distance over time so what's the distance it's doing here well it starts at x0 and it goes to xt and if we don't know what the initial value is one way we can figure out what the distance for each component x y and z is to subtract the endpoint or subtract the starting point from the end point i could say this is uh x of t minus x naught and we've got y of t minus y naught and z of t minus z naught those are all components that are in the distance part right there uh and we're not actually going to calculate the distance i'm just doing each component separately so i don't need the square root thing i'm not going to do the overall distance we're just going to do x by itself and y by itself and z by itself uh time we usually use t so let's use t for time and the the rate of speed it might be going it's got a fixed speed going along that path but if i want to break it up into x y and z components maybe we could give it a speed stupid phone i could give it a speed of a b and c for the x y and z components so uh we can kind of solve this for x and t if i multiply t over i've got t times i'll put it back as v for a second t times v equals this is effectively point t minus point zero all right this up here and if i add p zero to the other side i get and if i write in terms of pt let's put pt on the left i got pt equals p0 plus t times this v so i could break this up and do all of its x y and z components so x of t equals x naught plus t a y of t equals y naught plus t b and z of t equals z naught plus t c so it's how far we've gone and i don't have a z component drawn on the page i i don't think let's just say z is not really changing much here for the intakes of this picture but every line is different so we could describe this as motion along a line well this line doesn't it could be anywhere i could have that it's the same thing if i use it the vector going from here to here it's still doing that same thing so it's it could be any vector it's not just related to p0 and pt any vector that's parallel or any line that's parallel to a vector like this would follow the same mechanics we just have a different p naught and pt so this this will work every time uh and so this is another way of looking at how to do lines you look at lines in terms of motion over time and this is called parameterizing [Music] the lines with time so these are called the parametric equations for the lines and if i show this formula initially up here pt minus p0 another way of writing that is the vector going from p naught to p t equals t v which equals t a b and c okay we're going to do a few examples of it uh several examples over it i thought i'd do some in this video just so it's not just boring equation generation so let's take a look real quick at that i want to bring this back down here and we'll flip that over so find the line parallel to 5 negative four eight and it's passing through the point negative two four and zero and i'm going to put those equations back up over here just as a reminder x of t is x naught plus t a y of t equals y naught plus t b and z of t equals z naught plus t c so the line to find the line parallel this vector we could treat this as our velocity the vector is going to be our velocity and so this will be our p0 right here so we're going to do let's do each component one at a time i've got x of t i don't know where that's going to end up but x naught will be its starting position negative two and then we're going to add t times this vector the x component of the vector so five t and we'll do the same thing with the other ones y of t and z of t y of t is the y naught component it's four and then z of t the z naught component is zero i don't even need to write it uh now we do tb tb this is b b is negative four so we're gonna go negative four t and z is just going to be that's c so ct 80 so you can check it if you like you plug in t equals zero what happens t equals zero x of t becomes negative two because the five t goes away y of t equals four and z of t equals zero because all the t components are zeroed out and we're left with p naught we could generate linear points on the line by doing t equals one and t equals two we'll do that in another example uh shortly next thing i i'll pause for a second let you finish writing that or whatever let's take a look at doing another problem find the parametric equations for the line that passes through two points p one is three negative seven and p two is negative two negative one and then write them in slope intercept form well we we already know how to do this the regular way so maybe at the end let's let's do the regular wave let's do y equals nx plus b the traditional algebraic way and that's what we're getting so maybe let's do the parametric equations and then see what we can get to get there for the parametric equations we need a starting point and a vector so right now we don't have a vector and i could use either point as a starting point but right now we don't have a vector so let's make a vector let's just do p1 to p2 and remember that is subtracting components so p2 minus p1 the x component negative 2 minus 3 and the y component is negative one minus negative seven and i'm getting negative five six is my v so i could do it two ways like let's do each point i could say x of t equals let's use p1 for the first set three because my x naught let me bring my equations back down so you can see them three is my x naught and a is going to be this component right here the first component of my velocity vector minus five t i could say y of t equals what's my first component there negative seven and my vector is six so that's one way to do it or we could use the other points negative 2 and negative 1 with the same velocity vector can i get the original points back from my parametric equations either set so we'll call this set a and we'll call this set b uh let's see what we need to do for set a first set a if i set t equal to 0 i'm going to get that 3 negative 7 right gives me the 3 negative seven and [Music] well by doing it subtracting it one from the other we're saying it's one time step so let's try t equals one that would be one time step t equals one i've got three minus five would be negative two and negative seven plus six would be negative one so that gives me p2 so that's p1 and p2 let's try it with the other set if i plug in t equals zero here i get negative 2 negative 1 which is p2 now i want to end up with negative 3 or positive 3. if you're not sure what time step it is i'm sure it's kind of obvious here but let's just work out what it would how to figure it out in case you don't know plug in the x component that you're looking for three equals negative two minus five t so i add the negative two over plus two plus two my green pen is dying five it should be it's completely full five equals negative five t and so t equals negative one if i do t equals negative one what do i get for the y component negative one plus negative six is negative seven and if i plug in t equals negative one for x negative two minus five times negative one is a plus five negative two plus five is three so we get back p1 so it's just going backwards in time if we go start from there now how can we get this set this up so we get our original equations uh let's pick either pair of either pair of these uh let's i don't know uh the numbers are smaller for this set negative two and every negative one i like that a little bit better so i'm going to use those two put a little line here showing that we're doing another step here what i'm going to do is what this y of t has a t in it i don't want t i want x so i'm going to solve this equation for t so if i solve for t i add 2 to both sides and then i can multiply by negative 1 5th and that should look like a subscript t right there and now that i have t in terms of x rather than x in terms of t i can take this and we can plug that in and right there for that t right there and what happens if we do that i've got y of t equals negative one plus six times negative one fifth times x sub t plus two so let's just clean it up some uh y of t equals negative one six times negative one fifth is negative six fifths so negative six fifths i'll distribute that minus six fifths x of t and then minus six fifths times two is minus twelve bits and if i rewrite this negative 1 as a 5 over 5 i'll do it at the end i'll put negative 5 over 5 here i can see that i get y equals negative 6 5 x minus 17 bits and we can optionally have the t component here but we don't need it anymore we've eliminated t so don't really need it anymore would that be what we get doing at the traditional method so doing y equals mx plus b we need the slope m is y2 minus y1 over x2 minus x1 it's probably been a while since you calculate students did a little bit of algebra so uh y2 let's just figure out what this is negative one minus seven and x2 minus x1 is negative two minus three and this does give me a slope of negative six over five so then we have uh the good old y minus y naught equals m times x minus x naught also probably been a day or a day and a half for you it's been a little while uh we can plug in either point i kind of like the negative two negative one again just because it's got smaller values let's use that so y minus y naught would be negative one equals m was calculated at negative six this x minus x naught is negative two so i got y plus one equals negative six fifths x these become a plus i got the two again two times six fifths is negative twelve this and if instead of doing one over one i do five over five if i subtract negative five over five from both sides what do we get y equals negative six fifths x minus seventeen fifths so it matches so the parametric equation of lines is it might seem like it's more work now but in 3d working with parameterized lines actually can make stuff a lot easier you've got things narrowed down to one component a component in terms of time rather than x y and z that'll be it for this section peace