Chapter 4.6, Problem 46E

(a)

To determine

To find: Write the parametric equation to model the motion of Ferris wheel at $\left(0, 40\right)$

Answer to Problem 46E

The parametric equation to model the motion of Ferris wheel is $x=30\cos \left(\pi t\right),y=40+30\sin \left(\pi t\right)$

Explanation of Solution

Given information:

Given that a Ferris wheel with radius 30 ft. makes one revolution every 10 seconds.

The center of the wheel is at $\left(0,40\right)$ and a point $p(30,40)$ is on wheel at $t=0$ .

Calculation:

The general form of parametric equation is:

  $x=r\cos \theta ,y=r\sin \theta$

The parametric equation of the circle of radius 30will be given as:

  $x=30\cos \theta ,y=30\sin \theta$

The wheel makes a revolution in 2 seconds.

period=2 seconds.

It is known that $n=\frac{2\pi }{Period}$ .

  $\begin{array}{l}n=\frac{2\pi }{2}\\ n=\pi \end{array}$

  $\theta =nt$

Now the parametric equation of the wheel when there is a point on the Ferris wheel is located at $\left(30, 40\right)$ will be:

  $x=30\cos \left(\pi t\right),y=40+30\sin \left(\pi t\right)$

This is the required parametric equation of motion.

(b)

To determine

To find: Find the rate of horizontal movement, and rate of vertical movement,

Answer to Problem 46E

The rate of horizontal movement and rate of vertical movement, at $t=5$ are $\frac{dx}{dt}=0,\frac{dy}{dt}=-30\pi$ respectively.

The rate of horizontal movement and rate of vertical movement, at $t=8$ are $\frac{dx}{dt}=0,\frac{dy}{dt}=30\pi$ respectively.

Explanation of Solution

Given information:

Given that a Ferris wheel with radius 30 ft. makes one revolution every 10 seconds.

The center of the wheel is at $\left(0,40\right)$ and a point $p(30,40)$ is on wheel at $t=0$ .

Calculation:

The parametric equation of the circle of radius 30will be given as: $x=30\cos \left(\pi t\right),y=40+30\sin \left(\pi t\right)$

In order to find the rate of horizontal movement, and rate of vertical movement, of the point $p(30,40)$ .Differentiate the equations with respect to $t$ .

Differentiate $x=30\cos (\pi t)$ with respect to $t$ :

  $\frac{dx}{dt}=-30\sin (\pi t)\cdot \pi \mathrm{....}(1)$

Differentiate $y=40+30\sin (\pi t)$ with respect to $t$ :

  $\frac{dy}{dt}=30\cos (\pi t)\cdot \pi \mathrm{....}(2)$

Now substitute $t=5$ in equation (1) gives:

  $\begin{array}{l}\frac{dx}{dt}=-30\sin (\pi \cdot 5)\cdot \pi \\ \frac{dx}{dt}=0\sin (n\pi )=0\end{array}$

Now substitute $t=5$ in equation (2) gives:

  $\begin{array}{l}\frac{dy}{dt}=30\cos \left(5\pi \right)\cdot \pi \\ \frac{dy}{dt}=-30\pi \cos \left(5\pi \right)=-1\end{array}$

Now substitute $t=8$ in equation (1) gives:

  $\begin{array}{l}\frac{dx}{dt}=-30\sin (\pi \cdot 8)\cdot \pi \\ \frac{dx}{dt}=0\sin (n\pi )=0\end{array}$

Now substitute $t=8$ in equation (2) gives:

  $\begin{array}{l}\frac{dy}{dt}=30\cos \left(8\pi \right)\cdot \pi \\ \frac{dy}{dt}=30\pi \cos \left(8\pi \right)=1\end{array}$

The rate of horizontal movement and rate of vertical movement, at $t=5$ are $\frac{dx}{dt}=0,\frac{dy}{dt}=-30\pi$ respectively.

The rate of horizontal movement and rate of vertical movement, at $t=8$ are $\frac{dx}{dt}=0,\frac{dy}{dt}=30\pi$ respectively.