Motion along a curve: finding rate of change | Advanced derivatives | AP Calculus BC | Khan Academy

We're told that a particle moves along the curve (x^2 y^2 = 16), so that the x-coordinate is changing at a constant rate of -2 units per minute. What is the rate of change, in units per minute, of the particle's y-coordinate when the particle is at the point (1, 4)?

So let's just repeat or rewrite what they told us: the curve is described by (x^2 y^2 = 16). They tell us that up there, they tell us that the x-coordinate is changing at a constant rate. Let me underline that: the x-coordinate is changing at a constant rate of -2 units per minute.

So we could say that (\frac{dx}{dt}) (the rate of change of the x-coordinate with respect to time) is equal to -2, and they're saying units; some unit of distance divided by minute, units per minute. And what they want us to figure out is: what is the rate of change of the particle's y-coordinate?

So let me underline that: what is the rate of change of the particle's y-coordinate? So what they want us to find is: what is (\frac{dy}{dt})? What is that equal to? And they say when the particle is at the point (1, 4).

So when (x = 1), (y = 4). Can we set up some equation that involves the rate of change of (x) with respect to (t), (y) with respect to (t), (x), and (y)?

Well, what if we were to take the derivative of this relation that describes the curve? What if we were to take the derivative with respect to (t) on both sides? So let me write that down.

So we're going to take the derivative, actually let me just erase this so I have a little bit more space. Alright, and so that way I can just add it.

So let's take the derivative with respect to (t) of both sides of that. And if at any point you get inspired, I encourage you to pause the video and try to work through it.

Well on the left-hand side, if we view this as a product of two functions right over here, we could take the derivative of the first function, which is going to be the derivative of (x^2) with respect to (x). So that is (2x).

And remember we're not just taking the derivative with respect to (x), we're taking the derivative with respect to (t). So we're going to have to apply the chain rule.

So it's going to be the derivative of (x^2) with respect to (x), which is (2x), times the derivative of (x) with respect to (t) (so times (\frac{dx}{dt})), and then we're going to multiply that times the second function (so times (y^2)).

And then that's going to be plus the first function, which is just (x^2), times the derivative of the second function with respect to (t). And so once again, we're going to apply the chain rule. The derivative of (y^2) with respect to (y) is (2y) (let me do that in that orange color).

It is equal to (2y) times the derivative of (y) with respect to (t) (times (\frac{dy}{dt})). And then that is going to be equal to the derivative with respect to (t) of 16. Well, that doesn't change over time, so that's just going to be equal to zero.

And so here we have it. We need to simplify this a little bit, but we have an equation that gives a relationship between (x), (\frac{dx}{dt}), (y), and (\frac{dy}{dt}).

So actually, let me just rewrite it one more time so it's a little bit simplified. So this is (2xy^2 \frac{dx}{dt} + x^2 (2y \frac{dy}{dt}) = 0).

And so, let's actually just substitute the values in. We know we want to figure out what's going on when (x = 1). So we know that the (x)'s here are equal to one; this (x^2) well that's just going to be (1^2), so that's going to be equal to (1).

We know that (y = 4), so this is going to be (2 \cdot 1 \cdot (4)^2) which simplifies to (2 \cdot 1 \cdot 16 = 32). And this is going to be (1^2) times (2 \cdot 4 \cdot \frac{dy}{dt}).

We know (\frac{dx}{dt} = -2) (they tell us that in the problem statement, (-2)). And so now this is a good time to simplify this thing.

So this will simplify to (2 \cdot 1 \cdot -2 \cdot 16 = -64). And then we have (let me do this in a color you can see) and then we have all of this.

Well, this is just going to be (1 \cdot 8 \cdot \frac{dy}{dt}), so this is going to be (8 \frac{dy}{dt}).

So, plus (8) times the derivative of (y) with respect to (t) is equal to (0). Add (64) to both sides and we get (I'll switch to a neutral color) (8 \frac{dy}{dt} = 64).

Divide both sides by (8) and you get (\frac{dy}{dt} = \frac{64}{8}), which is just (8).

And if you want to look at the units, it will also be in units per minute, some units of distance per minute. And we are done.