Motion along a curve: finding velocity magnitude | AP Calculus BC | Khan Academy

A particle moves along the curve (xy = 16) so that the y-coordinate is increasing. We underline this: the y-coordinate is increasing at a constant rate of two units per minute. That means that the rate of change of y with respect to t is equal to two.

What is the magnitude in units per minute of the particle's velocity vector when the particle is at the point (4, 4)? So when (x) is 4 and (y) is 4.

Let's see what's going on. First, let's remind ourselves what a velocity vector looks like. Our velocity is going to be a function of time, and it’s going to have two components: the rate of change in the x-direction and the rate of change in the y-direction. So, the rate of change in the x-direction is going to be (\frac{dx}{dt}), and the rate of change in the y-direction is going to be (\frac{dy}{dt}). They tell us that this is that (\frac{dy}{dt}) is a constant two units per minute.

But they’re not just asking for the velocity vector or its components; they're asking for the magnitude. They’re asking for the magnitude of the particle’s velocity vector.

Well, if I have some vector, let me do a little bit of an aside here. If I have some vector, let’s say (\mathbf{a}), that has components (b) and (c), then the magnitude of my vector (sometimes you’ll see it written like that; sometimes you’ll see it written with double bars like that) is the magnitude of my vector. This comes straight out of the Pythagorean theorem; this is going to be the square root of (b^2 + c^2), the square root of the x-component squared plus the y-component squared.

So if we want to know the magnitude of our velocity vector—the magnitude of the particle's velocity vector—I could write that as the magnitude of (\mathbf{v}), and I could even write it as a function of (t). It’s going to be equal to the square root of the x-component squared, so that’s the rate of change of (x) with respect to time squared plus the y-component squared, which in this case is the rate of change of (y) with respect to (t) squared.

How do we figure out these two things? Well, we already know the rate of change of (y) with respect to (t). They say that's a constant rate of two units per minute, so we already know that this is going to be (2), or that this whole thing right over here is going to be (4).

But how do we figure out the rate of change of (x) with respect to (t)? We could take our original equation that describes the curve, and we could take the derivative of both sides with respect to (t). That’s going to give us an equation that involves (x, y), and (\frac{dx}{dt}, \frac{dy}{dt}).

Let’s do that. So we have (xy = 16). I’m going to take the derivative with respect to (t) of both sides. Let me do that in a different color, just for a little bit of variety.

So, the derivative with respect to (t) of the left-hand side, derivative with respect to (t) of the right-hand side. On the left-hand side, we view this as a product of two functions. If we say (x) is a function of (t) and (y) is also a function of (t), we’re going to do a little bit of the product rule and a little bit of the chain rule here.

This is going to be equal to the derivative of the first function, which is (1), times the derivative of (x) with respect to (t) (remember we’re taking the derivative with respect to (t), not with respect to (x)), times the second function (which is (y)), plus the first function (which is (x)) times the derivative of the second function with respect to (t).

So first, what’s the derivative of (y) with respect to (y)? Well, that’s just (1). Then what’s the derivative of (y) with respect to (t)? Well, that’s (\frac{dy}{dt}). That is going to be equal to (0) because the derivative of a constant is just (0).

So let’s see what this simplifies to. In fact, we don’t even have to simplify it more; we can actually plug in the values to solve for (\frac{dx}{dt}). We know that (\frac{dy}{dt}) is a constant (2), and we want the magnitude of the particle's velocity vector when the particle is at the point ((4, 4)). So when (x = 4) and (y = 4).

So now it’s a little messy right now, but this right here is an equation that we can solve for. There’s only one unknown here: the rate of change of (x) with respect to (t) when we are at the point (4, 4). If we could figure that out, we could substitute that in and figure out the magnitude of our velocity vector.

So let us write it out. This gives us (4 \cdot \frac{dx}{dt} + 4 \cdot 2 + 8 = 0).

So we have (4 \cdot \frac{dx}{dt} = -8). Subtract (8) from both sides, divide both sides by (4), and you get (\frac{dx}{dt}). Let me scroll down.

(\frac{dx}{dt} = -2). So when all the stuff is going on, the rate of change of (x) with respect to (t) is (-2). Then you square it, you get (4) right over here.

So the magnitude of our velocity vector is going to be equal to the square root of (4 + 4), which is equal to (8), which is the same thing as (4 \times 2).

So this is going to be (2\sqrt{2}) units per minute. So that's the magnitude of the velocity vector.