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## Calculus, all content (2017 edition)

### Unit 8: Lesson 2

AP Calculus BC questions- 2015 AP Calculus BC 2a
- 2015 AP Calculus BC 2b
- 2015 AP Calculus BC 2c
- 2015 AP Calculus BC 2d
- 2015 AP Calculus BC 5a
- 2015 AP Calculus BC 5b
- 2015 AP Calculus BC 5c
- 2015 AP Calculus BC 5d
- 2015 AP Calculus BC 6a
- 2015 AP Calculus BC 6b
- 2015 AP Calculus BC 6c
- AP Calculus BC exams: 2008 1 a
- AP Calculus BC exams: 2008 1 b&c
- AP Calculus BC exams: 2008 1 c&d
- AP Calculus BC exams: 2008 1 d
- Calculus BC 2008 2 a
- Calculus BC 2008 2 b &c
- Calculus BC 2008 2d
- 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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# 2015 AP Calculus BC 2a

x coordinate of particle at a certain time.

## Video transcript

- [Voiceover] "At time t is
greater than or equal to zero, "a particle moving along
a curve in the xy-plane "has position x of t and y of t." So its coordinate is given
by the parametric function x of t, and y-coordinate by the
parametric function, y of t. "With velocity vector v of t is equal to," and the x component of the velocity vector is cosine of t squared, and the y component of the velocity vector is e to the 0.5t. "At t equals one, "the particle is at
the point 3, comma, 5." Is at the point 3, comma, five. All right. "Find the x-coordinate of
the position of the particle "at time t is equal to two." All right, so how do we think about this? Well, you could view the
x-coordinate at time t equals two. So let's say, so we
could say x at time two, which they don't give that to us directly, but we could say that's
going to be x of one plus some change in x as we go from t equals
one to t equals two. But what is this going to be? Well, we know what the velocity is. And so the velocity,
especially the x component, and we can really focus on the x component for this first part
'cause we only wanna know the x-coordinate of the
position of the particle. Well, we know the x component of velocity as a function of t is cosine of t squared. And if you take your velocity
in a certain dimension and then multiply it times
a very small change in time, in a very small change in time, dt, this would give you your
very small change in x. If you multiply velocity
times change in time, it'll give you a displacement. But what we can do is we can
sum up all of the changes in time from t equals one to t equals two. Remember, this is the change in x from t equals one to t is equal to two. So what we have right over here, we can say that x of two, which is what we're trying to solve, is going to be x of one, and they give that at time equals one, the particle is at the
point three, comma, five. Its x-coordinate is three. So this right over here is three. And then our change in x from t equals one to t equals two is going
to be this integral. The integral from t
equals one to t equal two of cosine of t squared, dt. And just to make sure we
understand what's going on here, remember, how much are we
moving over a very small dt? Well, you take your
velocity in that dimension times dt, it'll give you a
displacement in that dimension, and then we sum 'em all
up from t equals one to t equals two. And in this part of the AP test, we're allowed to use calculators and so let's use one. All right. So there's my calculator. And I can evaluate. So let's see, I wanna evaluate three plus the definite integral. I click on Math, and then I can scroll down to function integral, right there. The definite integral of, and I make sure I'm in radian mode, that's what you should assume, unless they tell you otherwise. Cosine of t squared, now
I'll use x as my variable of integration, so I'll say cosine of x, cosine of x squared. And then my variable of integration is x, so I'm really integrating
of x squared, dx, but it'll give the same value, comma, from one until two, and now let the calculator
munch on it a little bit, and I get approximately 2.557. So this is approximately 2.55, 2.557. Let me make sure that I added the three. Yeah, three plus that definite integral from one to two. 2.557. And I just rounded that. So there you go.