2015 AP Calculus BC 2a | AP Calculus BC solved exams | AP Calculus BC | Khan Academy

At time ( T ) is greater than or equal to zero, a particle moving along a curve in the XY plane has position ( X(T) ) and ( Y(T) ). So, its x-coordinate is given by the parametric function ( X(T) ) and y-coordinate by the parametric function ( Y(T) ).

With the velocity vector ( V(T) ) equal to, and the x-component of the velocity vector is ( \cos(T^2) ), and the y-component of the velocity vector is ( e^{0.5T} ). At ( T=1 ), the particle is at the point ( (3, 5) ).

All right, find the x-coordinate of the position of the particle at time ( T=2 ). All right, so how do we think about this? Well, you could view the x-coordinate at time ( T=2 ). So, let's say, we could say ( X(2) ), which they don't give to us directly. But we could say that's going to be ( X(1) ) plus some change in x as we go from ( T=1 ) to ( T=2 ).

But what is this going to be? Well, we know what the velocity is, and so the velocity, especially the x-component, we can really focus on the x-component for this first part because we only want to know the x-coordinate of the position of the particle. Well, we know we're going—we know the x-component of velocity is a function of ( T ): ( \cos(T^2) ).

If you take your velocity in a certain dimension and then multiply it times 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=1 ) to ( T=2 ).

Remember this is the change in x from ( T=1 ) to ( T=2 ). So what we have right over here, we can say that ( X(2) ), which is what we're trying to solve, is going to be ( X(1) ). They give that at time ( T=1 ), the particle is at the point ( (3, 5) ). Its x-coordinate is three, so this right over here is three.

Then, our change in x from ( T=1 ) to ( T=2 ) is going to be this integral: the integral from ( T=1 ) to ( T=2 ) of ( \cos(T^2) dT ).

Just to make sure we understand what's going on here, remember how much we are moving over a very small ( dT ). Well, you take your velocity in that dimension times ( dT ), it'll give you your displacement in that dimension, and then we sum them all up from ( T=1 ) to ( T=2 ).

In this part of the AP test, we are 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 want to 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, which that's what you should assume—so ( \cos(T^2) ).

Now, I'll use ( x ) as my variable of integration, so I'll say ( \cos(x) ) of ( x^2 ), and my variable of integration is ( x ). I'm really integrating ( \cos(x^2) , dx ) but it'll give the same value from 1 until 2.

Now, I let the calculator munch on it a little bit, and I get approximately 2.557. So this is approximately 2.55. Did I—let me make sure that I added the three? Yeah, three plus that definite integral from ( 1 ) to ( 2 ) is 2.55, and I just rounded that. So there you go.