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### Course: AP®︎/College Calculus BC > Unit 11

Lesson 3: AP Calculus BC 2011- 2011 Calculus BC free response #1a
- 2011 Calculus BC free response #1 (b & c)
- 2011 Calculus BC free response #1d
- 2011 Calculus BC free response #3a
- 2011 Calculus BC free response #3 (b & c)
- 2011 Calculus BC free response #6a
- 2011 Calculus BC free response #6b
- 2011 Calculus BC free response #6c
- 2011 Calculus BC free response #6d

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# 2011 Calculus BC free response #1a

Velocity and acceleration vectors for particle. Created by Sal Khan.

## Want to join the conversation?

- In this problem you evaluate "cos 9" with the calculator. For the AP test, are all trig values assumed to be in radians or degrees, or was it specified somewhere not shown in problem? Obviously cosine of 9 degrees is very different than cosine of 9 radians.(11 votes)
- When you're simply given a number like this 99.9% of the time you're working with radians, especially if they want you to differentiate or integrate.

If they write something like theta = x degrees or there's a degrees sign you'll know you're working with degrees.(3 votes)

- are polar coordinates, parametric equations, and conics on the Calculus BC AP exam?(4 votes)
- Yes, on the BC exam, no for the AB. Usually only a small portion though.(2 votes)

- What if I showed this as my work, instead of Sal's engineering-style vectors:

Speed = sqrt( x' ^2 + y' ^2)

Would that be okay?(4 votes)- Yes. If you directly applied the formula Speed = sqrt( x' ^2 + y' ^2), you would get the same result Sal derived in the video.(1 vote)

- The answer to the acceleration vector problem would actually be <4, -5.467>, because it's asking for a vector, not a definite solution (like the speed or distance formula). That was simply found by taking the 2nd Derivative of x and y: <x''(3), y''(3)>(2 votes)
- <4, -5.467> and 4 i + -5.467 j are 2 equally valid ways of expressing an vector because they clearly denote the x and y component of the vector.(4 votes)

- Is it not acceptable to give the hypotenuse of the triangle demarcated by the x and y vectors, and then use inverse tangent to find the angle? That's how I was always taught to leave vectors in physics.(1 vote)
- @4:26, and other places in the video, Sal refers to the unit vectors as "i" and "j". Shouldn't he be saying "i hat" and "j hat"?(1 vote)
- Sure, he should do that to insure clarity. Note that he did state that he that he was using engineering notation. @1:00he even calls out the î as the "i unit vector" and continues when writing ĵ , the "j unit vector."(1 vote)

- i dont understand anything on "cos" i need to learn the basics(0 votes)
- Good idea. Have a look at some of the "basic trigonometry" videos on this site: https://www.khanacademy.org/math/trigonometry/basic-trigonometry(2 votes)

## Video transcript

Problem number 1: at time t, a particle moving in the xy-plane is at position (x(t), y(t)), where x(t) and y(t) are not explicitly given. For t is greater than or equal to 0, the derivative of x with respect to t is 4t+1 the derivative of y with respect to t is sin(t^2) and at time t=0, x(0)=0 and y(0)=-4 Fair enough, let's do part a. Find the speed of the particle at time t=3, and find the acceleration vector of the particle at time t is equal to 3 as well. So, the speed is really just the magnitude of the velocity vector. So what's the velocity vector? Our velocity vector as a function of time is going to be equal to the derivative of our x position as a function of time, or we could say the velocity in the x-direction times the i unit vector, plus the velocity in the y-direction times the j unit vector. I just wrote this in engineering notation, obviously there are many ways that you could specify a vector. And, in this case, x'(t), they've already given it to us: x'(t) is 4t+1 they just wrote it as dx/dt. So this over here is 4t+1 times the i unit vector, and then this here, this y'(t), they gave it to us the derivative of y with respect to t is sin(t^2). So, plus sin(t^2) times the j unit vector. That is our velocity vector. So what is our velocity- or this is our velocity vector as a function of time, So what is our velocity at time t=3? Well, we just have to substitute 3 for t, so 4(3)+1 is 13i. Plus 3 squared- 3 squared is 9 -sine of 9. Plus sin9j. Our speed, what they want us to figure out is going to be equal to- or maybe I should write it as our speed at time 3 -our speed is going to be equal to the magnitude of our velocity vector at time 3, which is equal to the magnitude of 13i+(sin9)j. And then this we literally, this is, we use essentially the Pythagorean Theorem to find the magnitude. This is going to be equal to the square root of 169 (that's 13 squared) plus (sin9)^2. So let me get my TI-85 out. This part of the exam we are allowed to use calculators. So we get the square root of 169+(sin9)^2. Alright, that's 13 squared plus sin9 squared. And, we get- we get 13.00- I'll just round it to 7, just to show that we didn't get just straight up 13, that the sin9 squared did do something. So 13.007. So this is approximately equal to 13.007. So we've answered the first part. The speed of the particle at time t=3 is 13.007. Then they want us to find the acceleration vector of the particle at time t=3. So the acceleration vector as a function of time is just going to be the second derivative of x with respect to t, or you'd say the acceleration in the x-direction as a function as a function of time times i, plus the acceleration in the y-direction. So it's the second derivative of y with respect to t in the j-direction. And this, of course, is just going to be equal to the derivative of x of- the derivative of x- the second derivative of x(t) is just the derivative of the first derivative. So the first derivative is 4t+1, you take the derivative of that, you just get 4. So, we get 4i plus... and then the derivative- the second derivative, or the second derivative of y with respect to t is just the derivative of the first derivative. The derivative- the first derivative is sin(t^2), so derivative of that is going to be 2t*cos(t^2), and of course we have our j. We have our j right over there and I just did this from the chain rule. The derivative of t^2 with respect to t is 2t, derivative of sin(t^2) with respect to t^2 is cos(t^2). And so that is the acceleration vector as a function of t but they care at time t is equal to 3. So, 4 still stays 4, then you have 2(3)+cos(9). So let me write this, this is going to be our acceleration vector our acceleration vector at time 3 is going to be 4i plus... and let's just put, if this is 2 times 3, which is 6... (6cos9)j. Let's get the calculator out so we actually get a value there since they allow us to use or almost want us to use a calculator here. So, 6cos9 is negative 5.47, or 5.467 if we want. So, plus negative- oh, what was I doing, this is j -so plus -5.467j. So that is our acceleration vector. And once again, I wrote it in engineering notation (which I'm sure the advanced placement exam people wouldn't mind) but you could also write it in kind-of the paramiterized vector notation or your ordered-pair notation as a vector you could say that the acceleration vector at time 3 you could write it this way too: <4, -5.467> These are really specifying the same thing. This is the x-component of the vector, this is the y-component of the vector. This is the x-component of the vector, this is the y-component of the vector.