The question is as follows:
Illuminated by the rays of the setting sun, Andy rides alone on a merry-go-round, casting a moving shadow on a wall. The merry-go- round is turning 40 degrees per second. As the top view shows, Andy is 24 feet from its center, and the sun’s rays are perpendicular to the wall. Let N be the point on the wall that is closest to the merry-go- round. What is the speed (feet per second) of Andy’s shadow when it passes N? What is the speed of this shadow when it is 12 feet from N?
I was thinking that the equation that would represent the way the circle is travelling is $(x, y) = (24\cos(40x), \text{ }24\sin(40x))$ assuming that the merry-go-round is moving CCW. The closest point would be when $t = 0$ I suppose, therefore, the coordinates of $A$ would be $(24, 0)$. Is it safe to say that the distance between $A$ (the point on the merry-go-round closest to the wall) and $N$ is $24$ feet? However, what is making me doubt my answer is the time that the merry-go-round is traveling. Any help will be greatly appreciated.
Guide:
$40^\circ$ per seconds, hence $360^\circ$ would take $9$ seconds.
Assuming at time $t=0$, Andy just passes $N$, then we have
$$y =24\sin\left( \frac{t(2\pi)}{9}\right) $$
Hence, given a particular value of $y$, we can compute the corresponding value of $t$.
Also, we are interested in computing $\frac{dy}{dt}$.
Edit:
$$\frac{dy}{dt}=24\cdot \left( \frac{2\pi}{9} \right)\cos\left(\frac{2\pi t}9 \right)$$