# Checking to make sure I understand the English in this math problem.

Let $g$ be a function given by $g\left(t\right) = 100 + 20 \sin \left(\frac{\pi t}{2}\right) + 10 \cos \left(\frac{\pi t}{6}\right)$ For $0\leq t \leq 8$, is $g$ decreasing most rapidly when $t =$.

I am confused about the decreasing most rapidly, does it mean

A. when the 1st derivative is equal to zero,

B. When the 2nd derivative is equal to zero,

I understand that I need to take the derivative. Unsure upon which one. I know that the first derivative is velocity and the second is acceleration, if the starting fuction is a position function and differentatie in reguards to time.

My work

$g^{\prime\prime}\left(t\right) = -5\,\sin \left( 1/2\,\pi \,t \right) {\pi }^{2}-{\frac {5}{18}}\,\cos \left( 1/6\,\pi \,t \right) {\pi }^{2}$

$t=0$ at 2.017406169, 4.017973737, 5.964620094

I made a graph of the 2nd derivative (red) and the 3rd derivative (blue)

The change that would be most rapid at the maximum of the graph in blue, which is the 3rd derivative of $g(t)$

-
For the sake of clarity: The first derivative of a position function with respect to time is velocity and the second derivative is acceleration. The original function must be position (of some object) as a function of time and the derivatives must be taken with respect to time for those to be true. –  Todd Wilcox Dec 17 '12 at 1:13
Decreasing most rapidly means that $g'$ is the most negative. You can set $g''=0$ to work this out.