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## Solving motion problems using parametric and vector-valued functions

Current time:0:00Total duration:4:35

# Planar motion example: acceleration vector

AP Calc: FUN‑8 (EU), FUN‑8.B (LO), FUN‑8.B.1 (EK)

## Video transcript

- [Voiceover] A particle
moves in the xy plane so that at any time t is greater
than or equal to zero its position vector is,
and they give 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 equals three? Alright, so or position, let's denote that it's a vector valued function, it's gonna be a function of time. It is a vector. And they already told
us that the x component of our position is negative
three t to the third power plus four t-squared, and the y component is t to the third power plus two. And 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. And 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, or sometimes people write this as unit vector notation. Negative three to the
third plus four t-squared times the unit vector in
the horizontal direction, plus t to the third plus
two, times the unit vector in the vertical direction. This is just denoting the same thing. This is the x component,
this is the y component. This is the component in
the horizontal direction, this is the component in
the vertical direction, or the y component. Now the key realization is if you have the position vector, well
the velocity vector's just going to be the derivative of that. So V of t is just going to
be equal to r prime of 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. So if we wanna take the derivative of the x component here
with respect to time, we're just gonna use
the power rule a bunch. So it's three times negative three, so it's negative nine t-squared and then plus two times four is eight, so plus eight t to the first. So plus eight t. And then over here for the y component. So the derivative of t to the third with respect to t is three t-squared. Three t-squared, and then the derivative of two is just a zero, so I actually have space to write that
three t-squared even bigger. Three t-squared, alright. And if we wanna 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, this is going to be equal to, let me give myself some space. And so the x component, well I just take the derivative of the x component again, and let me find a color
I haven't used yet, I'll use this green, so let's see. Two times negative nine, negative 18 times t to the first power plus eight, derivative of eight t is just eight if we're taking the derivative with respect to t. And then here in the orange, derivative of three t-squared, so it's
two power rule here, over and over again. Two times three is six t to
the first power, just six t. So this is, we've just been able to by taking the derivative of this position vector
valued function twice, I'm able to find the
acceleration function. And now I just have to
evaluate it at t equals three. So our acceleration at t is equal to three is equal to, so in green it's going to be negative eighteen times three plus eight, comma, and then we're gonna have six times three, six times three. And so what does this simplify to? Well this is going to be equal to, let's see, negative 18 times three is negative 54, negative 54 plus eight is negative 46. Negative 46, and then six
times three is eighteen. Did I do that arithmetic right? So this is negative 54 plus eight. So negative 54 plus four
would be negative 50, plus another four would be negative 46. Yep, there you have it. Negative 46, comma eighteen. That is its acceleration vector at t is equal to three.