Planar motion example: acceleration vector | Advanced derivatives | AP Calculus BC | Khan Academy

A particle moves in the XY plane so that at any time ( T ) is greater than or equal to zero, its position vector is given. They provide us the X component and the Y component of our position vectors, and they're both functions of time. What is the particle's acceleration vector at time ( T = 3 )?

All right, so our position, let's denote that it's a vector-valued function. It's going to be a function of time; it is a vector. They already told us that the X component of our position is ( -3T^3 + 4T^2 ) and the Y component is ( T^3 + 2 ). So you give me any time greater than or equal to zero, I put it in here, and I can give you the corresponding X and Y components.

This is one form of notation for a vector. Another way of writing this, you might be familiar with engineering notation, it might be written like:

[
\mathbf{R}(T) = -3T^3 \mathbf{i} + 4T^2 \mathbf{j}
]

or sometimes people write this as unit vector notation:

[
-3T^3 \mathbf{u_x} + 4T^2 \mathbf{u_y}
]

This is just denoting the same thing. This is the X component; this is the Y component. This is a component in the horizontal direction; this is a component in the vertical direction, or the Y component.

Now, the key realization is if you have the position vector, well, the velocity vector is just going to be the derivative of that. So, ( \mathbf{V}(T) ) is just going to be equal to ( \mathbf{R}'(T) ), which is going to be equal to... well, you just have to take the corresponding derivatives of each of the components.

So let's do that. If we want to take the derivative of the X component here with respect to time, we're just going to use the power rule a bunch. So it's ( 3 \times -3 ), so it's ( -9T^2 ) and then plus ( 2 \times 4 = 8 ), so plus ( 8T ).

Then, over here for the Y component, the derivative of ( T^3 ) with respect to ( T ) is ( 3T^2 ), and the derivative of 2 is just zero. So actually, I have space to write that: ( 3T^2 ).

All right, and if we want to find the acceleration function, or the vector-valued function that gives us acceleration as a function of time, well, that's just going to be the derivative of the velocity function with respect to time.

So, this is going to be equal to... let me give myself some space. The X component, well, I just take the derivative of the X component again. Let me find a color I haven't used yet; I'll use this green.

So let's see: ( 2 \times 9 = 18T ) raised to the 1st power plus 8. The derivative of ( 8T ) is just 8 if we're taking the derivative with respect to ( T ). And then here in the orange, the derivative of ( 3T^2 ) using the power rule here over and over again gives us ( 2 \times 3 = 6T ).

So, we've just been able to find the acceleration function by taking the derivative of this position vector-valued function twice. Now, I just have to evaluate it at ( T = 3 ).

So, our acceleration at ( T = 3 ) is equal to: in green, it's going to be ( -9 \times 3^2 + 8 ), and then we're going to have ( 6 \times 3 ).

So what does this simplify to? Well, this is going to be equal to... let's see: ( -9 \times 3^2 = -81 ) and ( -81 + 8 = -73 ). Then for the Y component, we have ( 6 \times 3 = 18 ).

Did I do that arithmetic right? So this is ( -81 + 8 ), which would be ( -73 ), and ( 18 ) stays the same.

Yep, there you have it: the acceleration vector at ( T = 3 ) is:

[
(-73, 18)
]

That is its acceleration. That is its acceleration vector at ( T = 3 ).