Second derivatives (vector-valued functions) | Advanced derivatives | AP Calculus BC | Khan Academy

So I have a vector valued function H here. When I say vector valued, it means you give me a T; it's a function of T. So you give me a T, I'm not just going to give you a number; I'm going to give you a vector. As we'll see, you're going to get a two-dimensional vector.

You could view this as the X component of the vector and the Y component of the vector. You are probably familiar by now that there's multiple notations for even a two-dimensional vector. For example, you could use what's often viewed as engineering notation here, where the X component is being multiplied by the horizontal comp unit vector.

So you might see something like that, where that's the unit vector plus the Y component, 4T^4 + 2T + 1, is multiplied by the vertical unit vector. These are both representing the same thing; it just has a different notation. Sometimes you'll see vector valued functions with an arrow on top to make it explicit that this is a vector valued function.

Sometimes you'll just hear people say, "Well, let H be a vector valued function," and they might not write that arrow on top. So now that we have that out of the way, what we are interested in is, well, let's find the first and second derivatives of H with respect to T.

So let's first take the first derivative H prime of T. Well, as you'll see, that's actually quite straightforward. You're just going to take the respective components with respect to T. So the X component with respect to T, if you were to take the derivative, what are you going to get?

Well, we're going to use the power rule right over here: 5 * the negative 1, or time the negative, you're going to get -5 * T^(5 - 1) power, so T^4. The derivative with respect to T of -6, well that's just zero. So that's the rate of change of the X component with respect to T.

Now we go to the Y component, so we're going to do the same thing. The derivative with respect to T is going to be, and once again we just use the power rule. 4 * 4 is 16, T^3. The derivative of 2T is just 2, and then the derivative of a constant, well, that's zero; we've already seen that.

So there you have it. This is the rate of change of the X component with respect to T, this is the rate of change of the Y component with respect to T. One way to do it, and you know a vector can represent many, many, many different things, but the type of a two-dimensional vector like this, you could imagine this being H of T being a position vector in two dimensions.

If you're looking at the rate of change of position with respect to time, well then this would be the velocity vector. If we were to take the derivative of this with respect to time, well, we're going to get the acceleration vector.

So if we say H prime prime of T, what is that going to be equal to? H prime prime of T, well we just apply the power rule again. So 4 * -5 is equal to -20 T^(4 - 1), so T^3. Then we have 3 * 16 is 48 T^2, and then the derivative of 2 is just zero.

So there you have it. For any, if you view T as time, for any time, if you view this one as position, this one as velocity, and this is acceleration, you could, this would now give you the position, velocity, and acceleration. But it's important to realize that these vectors could represent anything of a two-dimensional nature.