2015 AP Calculus BC 2a

- [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.