# Finding the Point When the Rate is Increasing Rapidly in a Ferris Wheel

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.

• Think about it this way: his speed is constant, but the direction of the velocity vector is always changing. Does it ever point straight up? – amd May 28 '18 at 17:41
• @amd I am still not able to understand the idea that you're trying to express to me. Can you please rephrase that? – geo_freak May 28 '18 at 17:45
• Maximum rise means maximum slope, ideally infinite. Since here, velocity is in the direction of slope, the point with maximum rising rate is the time when the velocity is pointed exactly upwards. – DynamoBlaze May 28 '18 at 19:29

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.