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# 2015 AP Calculus BC 2a

## Video transcript

at time T is greater than equal to zero a particle moving along a curve in the XY plane has position X of T and Y of T so it's x 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 0.5 T at T equals 1 the particle is at the point 3 comma 5 is that the point 3 comma 5 alright find the x coordinate of the position of the particle at time T is equal to 2 all right so how do we think about this well you could view the x coordinate at time T equals 2 so let's say so we could say X at time 2 which they don't give it they don't give that to us directly but we could say that's going to be X of X of 1 plus some change in X as we go from T equals 1 to T equals 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 the X component of velocity is a function of T is cosine of T 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 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 1 to T equals 2 remember this is the change in X from from t equals 1 to t is equal to 2 so what we have right over here we can say that x of 2 which is what we're trying to solve is going to be X of 1 and they give that at equals one the particle is at the point 3 comma 5 its x-coordinate is 3 so this this right over here is 3 and then our change in X from T equals 1 to T equals 2 is going to be this integral the integral from T equals 1 to T equals 2 of cosine of T squared DT and just to make sure we understand what's going on here remember how much how much are we moving over a very small DT well you take your velocity in that dimension times DT it'll give you displacement in that dimension and then we sum them all up from T equals 1 to T equals 2 and in this part of the AP test we are allowed to use calculators and so let's use 1 all right so there's my calculator and I can evaluate so let's see I want to evaluate 3 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 which that's what you should assume so cosine let's say 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 cosine of x squared DX but will give the same value comma from 1 until 2 and now let the calculator munch on it a little bit and I get approximately two point five five seven so this is approximately two point five five two point five five seven did I let me make sure that I added the three three plus that indefinite integral from one to two two point five five seven and I just rounded that so there you go
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