Question

Ap Calculus BC 2002 Form B Question 1?

Answer 1

a) Although it hasn't happened in recent years, it's always possible that the AP exam will ask you to graph something, perhaps a curve, a set of parametric equations, or even polar curves, so it's good to be ready!

What you have to do here is make a table of values. For instance, when `t = pi/2` you would have `x(t) = sin(pi/2) = 1` and `y(t) = 2(pi/2) = pi`

Thus you would plot the point `(1, pi)`. Repeating this on the interval `-pi ≤ t ≤ pi` (5 is a reasonable number of points in that interval), you should get a graph resembling the following.

Don't forget the direction! Note that the particle is moving up because the y-coordinates are constantly increasing (since `y(t) = 2t` is a uniformly increasing function).

b) The domain will be the domain of the x function, which will be `-1 ≤ x(t) ≤ 1`. The range will be the range of the y function, which will be `{y| y in RR}`. However, we are only considering `-pi ≤ t ≤ pi`, therefore the range is `-2pi ≤ y(t) ≤ 2pi`.

c) We only consider the x equation in this one. The first derivative is given by `x'(t) = 3cos(3t)`. This will have critical points when `0 = 3cos(3t)`, and the smallest value that satisfies is when `3t = pi/2`, or `t= pi/6`. We note that at `t = 1`, the derivative is positive, and that at `t = pi/3` it's negative, thus `t = pi/6` is indeed a maximum.

Recall that speed is different than velocity; speed is given by `s = sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2)dt`

Thus

`s = sqrt((3cos(3t))^2 + 2^2) dt`

This is at `t = pi/6`. You don't even need a calculator to see that

`s = sqrt(4) = 2`

d) Recall that distance travelled is given by `d = int_a^b sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2) dt`

Thus

`d = int_(-pi)^pi sqrt((3cos(3t))^2 +4) dt`

Use a calculator to see that

`d ~~ 17.973`

And we can also verify that `5pi~~15.708`

Therefore, the distance travelled is indeed greater than `5pi`.

Hopefully this helps!