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Parametric equations intro

AP.CALC:
CHA‑3 (EU)
,
CHA‑3.G (LO)
,
CHA‑3.G.1 (EK)
Sal gives an example of a situation where parametric equations are very useful: driving off a cliff! Created by Sal Khan.

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  • blobby green style avatar for user Bryan T. Robinson
    Other than a moving object in space, what are some real-life applications for parametric equations?
    (95 votes)
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  • blobby green style avatar for user Kendall Keener
    where did sal get (t^2)/2?
    (71 votes)
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  • mr pants teal style avatar for user bosa weluche-ume
    what is a parameter
    (34 votes)
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  • mr pants teal style avatar for user Ahmed
    What is a parameter? What makes it useful?
    (17 votes)
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    • piceratops ultimate style avatar for user jimstanley49
      A parameter is some constant that relates two or more functions. In the example, the x-position and the y-position are not related to each other directly, but they are both defined in terms of time. Time is the parameter that allows us to see what the x and y functions are doing together. Try graphing the x and y functions separately. (In a calculator, you may have to call them y1 and y2, and change the t's to x's.) Can you look at them and know intuitively what the car is doing? Parametric equations allow us to break up a complicated problem, like motion in two or more dimensions, into simpler problems that can be solved separately and then recombined (if we want) through their shared parameter.
      (63 votes)
  • blobby green style avatar for user dorytg
    at -- what makes Sal determine it's t2/2? Is this a formula?
    (21 votes)
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    • Well, I think the deduction of this equation comes out here:
      d=Va*t, where d is the distance,and Va means the average velocity.
      while Va=(Vf+Vi)/2, where Vf is the final velocity and Vi is the initial velocity (in this case Vi=0).
      In addition,we know that the difference of velocity Vdelta=Vf-Vi=g*t. So,Vf=g*t+Vi,since Vi=0, so Vf=g*t+Vi=g*t+0=g*t.
      Now replace Vf by g*t: d=Va*t=(Vf+Vi)/2*t=(g*t+Vi)/2*t=(g*t+0)/2*t=g*t/2*t=g*t^2/2.
      (35 votes)
  • aqualine ultimate style avatar for user Jorge R. Martinez Perez-Tejada
    when will there be exercises for parametric equations? are you considering making them at all? It'd be really nice, thanks!
    (24 votes)
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  • leaf green style avatar for user diviakallattil
    In "normal" functions (for the lack of a better word), like f(x)=x+2 for example, is "x" a parameter?
    (5 votes)
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    • piceratops ultimate style avatar for user jimstanley49
      Sort of. It's tempting to say so, but parameter has a special meaning in this context. Each of the functions in the example are 'normal,' separate functions. What makes them parametric is that they share a parameter. Parametric equations are used when x and y are not directly related to each other, but are both related through a third term. In the example, the car's position in the x-direction is changing linearly with time, i.e. the graph of its function is a straight line. In the y-direction, however, its position is changing exponentially with time. The unifying 'parameter' is time. The car is moving through time equally "in both directions." This allows us to graph (x, y) coordinates to show the position of the car, as Sal showed. This is much more useful and intuitive than looking at the graphs of y(t) and x(t) separately. You can also use the parameter to find a unifying function that does directly relate x to y, as Sal hinted at.
      Wikipedia has a pretty good blurb about the math uses of "parameter." http://en.wikipedia.org/wiki/Parameter#Mathematical_models
      (20 votes)
  • male robot hal style avatar for user Alejandro
    There is a topic on the "Precalculus" mission called "Parametric equations and polar coordinates" but it has no skill excercises. Did they took them awary?? Or are there non?? Anyone knows?? Thank you very much.
    (12 votes)
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  • mr pink red style avatar for user Donwacenske
    I'm pretty tired so I may be looking too far into this... Is weight a factor? For example we figured out if this car is falling off a cliff at 5 m/sec sqrd it will land in the area sal figured out. Let's say that we had a tank or a marble traveling at the same speed with the same gravity acting on it, and our object is falling from the same height. I can't imagine the results will be the same.
    (5 votes)
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    • piceratops ultimate style avatar for user Just Keith
      The actual, real-world results are far more complicated because of the friction between the falling object and air as well as some minor issues with buoyancy. But, those kinds of computations are very advanced physics and engineering questions. They are just too complicated for this level of study.

      I would anticipate, given its shape, that a marble would actually fall more quickly than a car. This would be because the car has an irregular shape and has lots of friction from the air slowing down its acceleration. A marble has a smooth, spherical shape, so it would have considerably less air friction.
      (5 votes)
  • male robot hal style avatar for user Wayming
    Did anyone else see that on the y-coordinate, as t went up, the amount that y decreased every time t went up increased by ten? (When t went from zero to 1, the difference in y was 5, when t went from 1 to 2, the difference in y was 15, when t went from two to three the difference in y was 25, and if there was no ground on the x-axis from t=3 to t=4 y would have decreased by 35.) Anyone know exactly why this happens the way it does?
    Thanks!
    (3 votes)
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    • aqualine tree style avatar for user Ted Fischer
      Yes, this is typical of quadratic functions. If you look at the successive differences between values, those differences form a linear pattern. This fact is sometimes used to fit a polynomial function to data, known as the "method of successive differences". In the quadratic case, you find that the "second difference" (the successive difference between the differences) is twice the quadratic coefficient. You observed that the second difference is -10, thus the quadratic coefficient is -5.
      (5 votes)

Video transcript

Let's say I have a cliff. Let's make this cliff, I don't know, let's say it's 50 meters high. And on this cliff, I have a car. And this car is not just sitting on the cliff, it's driving off of it. A very dramatic problem. So let's see, I have this car here. And it's driving off of this cliff at 5 meters per second. And I want to know, what is the path of this car as it falls off of the cliff? So let's set up a little coordinate axis here. So let's say that this is the y-axis right there. And then this will be my x-axis. So this is y, this is x. Let's say that this is the point-- well, we know that this is a 50 meter high cliff. Maybe y equals 0 is sea level. So this would be 50 right here. And let's say that this point right here on the cliff, that's at x is equal to 10. So this point right here's the point 10, 50. And let's say the car's right at this point, right about to drive off the cliff. And time is equal to 0. So this is that time equal to 0. So t for time. Time is equal to zero. So my question is, what happens to this car as it drives off the cliff? So this is a bit of a physics problem, and I won't go deep into the physics. And I won't prove some of the equations. And I encourage you to watch the [UNINTELLIGIBLE] videos, the projectile motion videos, if you want to know where the equations come from. But the point here is just to get the equations and see what the graph looks like. So if I want to know x as a function of time-- a suitably vibrant color-- so x as a function of time is going to be what? Well, we're going to assume that we're on a planet that has no air. We're in a vacuum. So if we start off in the x direction at 5 meters per second to the right, we won't be decelerated by air or friction or anything else. Newton's laws of motion: an object in motion stays in motion, unless it's affected by a net force. And there won't be any net force in the x direction. It's just going to keep moving to the right at 5 meters per second. And position, or distance, is just equal to velocity times time. Our velocity is 5 times time. And, of course, it didn't start at x is equal to 0. This is at time equals 0. So it started at x is equal to 10. So you want to know-- it's kind of x of 0, where it started off, so plus 10. And this should be a little intuitive for you, right? At time is equal to 0, this term cancels out, where x is equal to 10. That makes sense. At time is equal to 1, we should be a little bit-- we'll be 5 meters further out, so on and so forth. Fair enough. That's x as a function of the parameter time. As you probably realize, that this is a video on parametric equations, not physics. So it's nice to early on say the word parameter. Parameter. And time tends to be the parameter when people talk about parametric equations. Although it could be anything. It could be radius or angle or who knows what else? So let's figure out what y is as a function of time. So y as a function of time, it's going to be equal to the initial y position, or y of 0, which is 50. We're 50 meters up in the air. Plus our initial velocity in the y direction. And we don't actually have any initial velocity in the y direction. The car isn't jumping, or isn't diving. It's just moving horizontally to the right. And the cliff is supporting it. So it has no y velocity. But if you were curious, it would be the y velocity times time. But since there's no y velocity times time, at least initially, I'll put nothing there. Plus the acceleration of gravity times time squared over 2. We want to figure out the sine. And just so you know, I mean, it's nice to touch on the physics a little bit, just so you know where these formulas come from and you know the motivation behind why you would even use a parametric equation. Gravity goes downwards in this example, and downwards in this example is in the minus y direction. y is decreasing. And the real-- and you know, it's not exact-- but gravity is normally 9.8 meters per second squared in most textbooks. But for the sake of simplifying this, we'll say that it's approximately 10 meters per second squared. That's how fast everything will be accelerated downward on this planet. Since it has no air, let's assume it's a planet with a little bit more mass than earth. And since it's going downwards, its direction is negative. So in our formula up here, it's our initial position-- we had no velocity times time, so I won't put that there-- minus 10 meters per second squared times t squared over 2. And you can watch the projectile motion videos to figure out how I got these formulas right there. But that's not the point of this. The point of this is to graph what happens to the cars and learn a little bit about parametric equations. So what is the path of this car as it falls off the cliff? Let's make a table here. So x and y are a function of this third parameter, t. So we're going to set t at different values and we're going to figure out what x and y are equal to. And I'm just going to arbitrarily pick some t's. T is equal to 0. T is equal to 1, 2, and 3. At time is equal to 0, what is x? x of 0-- this is 0-- x is equal to 10 meters. At time is equal to 1, what is x? And this is x of 1, right? If I wanted to write that notation. So 5 times 1 is 5, plus 10 is 15. x of 2? 5 times 2 plus 10, that's 20. And it makes sense. Every second we're getting 5 meters more to the right. Or x is increasing by 5 meters. So when t is equal to 3, 15 plus 10 is 25. Easy enough. The y is a little bit more complicated. And just to simplify this, it's the same thing as 5, right? 10 divided by 2. So 50 minus 5t squared. So time is equal to 0, this term cancels out. We just have 50 meters up in the air. At time is equal to 1, 1 squared is 1 times 5, is five. 50 minus 5 is 45 five meters in the air. Is that right? Right, yeah, time 1, 50. Right. And then at time is equal to 2, 2 squared is 4. 4 times 5 is 20. 50 minus 20 is 30. And then finally, at time is equal to 3-- and I just say finally because that's the last number we picked-- time is equal to 3. 3 squared is 9. 9 times 5 is 45. 50 minus 45 is 5. So let's plot these points. So the time is equal to 0. That's what we got right there. At time is equal to 1, we're at x is equal to 15. That's roughly, you see this is 5, 10, 15-- let me do all of them-- 15, 20, 25. And then the y-axis-- let me label that while we're at it-- this is roughly 10, 20, 30, 40, 50. So at time is equal to 0, we're at 10, 50. That's that point right there. At time is equal to 1, we're at 15, 45. So x is 15, y is 45, which is right about there. So this is t is equal to 1. At time is equal to 2, or at the coordinate 20, 30, it's right about there. So this is at time is equal to 2. And then at time is equal to 3, we're at 25, 5. So we're right there. And if we kept going on, at some point we're going to hit the ground. And you can figure out, actually-- set this equal to 0 and you figure out the exact time you hit the ground. Actually, let's do that. If this is equal to 0, 50, you get t is equal to the square root of 10. Which is a little over 3 seconds. Which makes sense, right? A little over 3 seconds, we're going to be hitting the ground. But anyway, what's the path of this car? Well, it's going to look something like this. Ooh, it starts getting accelerated downwards, and then plunk! It hits the ground at 3 point something seconds. Now what was interesting here is that by setting the parameter, not only did we get the curve-- right? We got this curve, which is kind of half of a parabola, half of a downward shaping parabola-- and we could actually eliminate the t, and just get the equation for that parabola. And we'll do that in future videos. But what was interesting, by making it a parametric equation, we know the direction of the car. If you just saw this graph without the car and everything else I drew, you wouldn't know which way the car was falling. But now we know that as t is increasing, we're going in that direction. So we can draw some arrows here. So because it's a parametric equation, we can draw some arrows. And then the most important thing is we know exactly where the car is at any time t. You can substitute t is equal to 1.25 seconds and you'll know exactly where the car is. So you can plot these points and you can kind of get a sense that as time goes on, we're getting accelerated downwards. And that's why, for every second further, especially the y distance gets further and further apart. Anyway, I just wanted to give you this example. Although this was a good physics problem, the intention wasn't to teach you physics. The intention is to give you the motivation behind why parametric equations even exist. These two things are parametric equations. We defined x and y as a function of a third parameter, t, instead of defining y in terms of x or x in terms of y like we've done every other time since then. And this is super useful. I mean, you could imagine when you have really hard physics problems where you want to figure out the three dimensional position of something, then you'll have x as a function of t, y as a function of t, z as a function of t. All sorts of interesting problems come out of using parametric equations, not just in physics. But anyway, I thought a good place to start is the motivation. Because the first time I learned parametric equations I was like, why mess up my nice and simple world of x's and y's by introducing a third parameter, t? This is why. Because you can figure out the path of things. You can figure out the direction of something as it moves along a curve, and you can figure out its exact position at any, in this case, time.