The question is as follows:
Jamie is riding a Ferris wheel that takes fifteen seconds for each complete revolution. The diameter of the wheel is $10$ meters and its center is $6$ meters above the ground. When is Jamie rising most rapidly? At what rate?
I know from the given the information that it takes fifteen seconds for each complete revolution that the ferris wheel is moving $24$ degrees/second. A parametric description of the movement of the ferris wheel is $(x, y) = (5\cos(24t), 5\sin(24t) + 6)$.
I am thinking that graphing an equation may be useful, but I don't know what equation to graph and find the maximum rate of increase. (I am not completely sure of the graphing method either.) Any help will be greatly appreciated.
Let $\vec r = <5\cos(\omega t),5\sin(\omega t)+6>$ be the vector that define the position of the wheel where the angular frequency $\omega = \dfrac{2\pi}{15}$.
The velocity is then:
$$\vec v = \dfrac{d\vec r}{dt} = <-5\omega\cdot \sin(\omega t), 5\omega\cdot \cos(\omega t)>$$
Jamie is rising only when the wheel is going upwards or in other words, the velocity vector is pointing straight up. For that, the $x$ component of velocity vector must be 0.
Let $-5\omega\sin(\theta) = 0$ where $0 \leq \theta \leq \pi$. $$-5\omega\sin(\theta) = 0$$ $$\theta = \arcsin(0) = 0, \pi$$ Hence $$\dfrac{2\pi}{15}t = \pi$$ $$t = 7.5s$$
Notice that the magnitude of the velocity vector is constant: $$\sqrt{(-5\omega\sin(\omega t))^2+(5\omega\cos(\omega t))^2} = 5\omega$$
So he would be rising with a rate of $\frac{2\pi}{3}$ms^-1.