# Ap Calculus BC 2002 Form B Question 1?

## Apr 28, 2018

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 $t = \frac{\pi}{2}$ you would have $x \left(t\right) = \sin \left(\frac{\pi}{2}\right) = 1$ and $y \left(t\right) = 2 \left(\frac{\pi}{2}\right) = \pi$

Thus you would plot the point $\left(1 , \pi\right)$. 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 $y \left(t\right) = 2 t$ 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 $\left\{y | y \in \mathbb{R}\right\}$. 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 $x ' \left(t\right) = 3 \cos \left(3 t\right)$. This will have critical points when $0 = 3 \cos \left(3 t\right)$, and the smallest value that satisfies is when $3 t = \frac{\pi}{2}$, or $t = \frac{\pi}{6}$. We note that at $t = 1$, the derivative is positive, and that at $t = \frac{\pi}{3}$ it's negative, thus $t = \frac{\pi}{6}$ is indeed a maximum.

Recall that speed is different than velocity; speed is given by $s = \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

Thus

$s = \sqrt{{\left(3 \cos \left(3 t\right)\right)}^{2} + {2}^{2}} \mathrm{dt}$

This is at $t = \frac{\pi}{6}$. You don't even need a calculator to see that

$s = \sqrt{4} = 2$

d) Recall that distance travelled is given by $d = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

Thus

$d = {\int}_{- \pi}^{\pi} \sqrt{{\left(3 \cos \left(3 t\right)\right)}^{2} + 4} \mathrm{dt}$

Use a calculator to see that

$d \approx 17.973$

And we can also verify that $5 \pi \approx 15.708$

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

Hopefully this helps!