Chapter 5.6, Problem 46E

a.

To determine

To find: the parametric equation to model its motion.

Answer to Problem 46E

The parametric equation is

  $\begin{array}{l}x=30\cos \left(\frac{\pi }{5}\right)t\\ y=40+30\text{sin}\left(\frac{\pi }{5}\right)t\end{array}$

Explanation of Solution

Given information:

  $\text{R}=30\text{ ft}\text{.}$

  $\begin{array}{l}f=1\text{ revolution/10 seconds}\\ f=0.1\text{ revolution/seconds}\end{array}$

Centre of Ferris wheel is

  $\left(0,40\right)$

Calculation :

We have to find the parametric equation to model its motion.

Radius is 30 and the centre is $\left(0,40\right)$ .

Therefore,

  $\begin{array}{l}x=30\cos at\left[\text{when the centre }x=0\right]\\ y=40+30\text{sin}at\left[\text{when the centre y}=40\right]\end{array}$

First, we will find the value when Ferris wheel makes 1 revolution every 10 seconds.

  $\begin{array}{l}10a=2\pi \left[1\text{ revolution means 3}{60}^{\circ }\right]\\ a=\frac{2\pi }{10}\\ a=\frac{\pi }{5}\end{array}$

Putting the value of $a=\frac{\pi }{5}$ in parametric equation,

  $\begin{array}{l}x=30\cos \left(\frac{\pi }{5}\right)t\\ y=40+30\text{sin}\left(\frac{\pi }{5}\right)t\end{array}$

Hence, the parametric equation is

  $\begin{array}{l}x=30\cos \left(\frac{\pi }{5}\right)t\\ y=40+30\text{sin}\left(\frac{\pi }{5}\right)t\end{array}$

b.

To determine

To find: the rate of horizontal and vertical movement of the point P when $t=5\text{ sec}\text{.}$ and $t=8\text{ sec}\text{.}$

Answer to Problem 46E

The value of horizontal movement when $t=5\text{ sec}\text{.}$ is $\frac{dx}{dt}=0\text{ ft}\text{./sec}\text{.}$ and $\frac{dy}{dt}=18.8\text{ ft}\text{./sec}\text{.}$ and when $t=8\text{ sec}\text{.}$ is $\frac{dx}{dt}=17.898\text{ ft}\text{./sec}\text{.}$ and $\frac{dy}{dt}=5.652\text{ ft}\text{./sec}\text{.}$

Explanation of Solution

Given information:

  $\text{R}=30\text{ ft}\text{.}$

  $\begin{array}{l}f=1\text{ revolution/10 seconds}\\ f=0.1\text{ revolution/seconds}\end{array}$

Centre of Ferris wheel is

  $\left(0,40\right)$

Calculation :

We have to calculate the rate of horizontal and vertical movement of the point P when $t=5\text{ sec}\text{.}$ and $t=8\text{ sec}\text{.}$

Therefore,

  $\begin{array}{l}x=30\cos \theta \\ y=40+30\sin \theta \end{array}$

  $\theta =2\pi ft$

Putting the value of $\theta =2\pi ft$ in above equation,

  $\begin{array}{l}x=30\cos \left(2\pi ft\right)\\ y=40+30\sin \left(2\pi ft\right)\end{array}$

Putting the value of $f=0.1\text{ revolution/seconds}$ in above equation,

  $\begin{array}{l}x=30\cos \left[2\pi t\left(0.1\right)\right]\\ y=40+30\sin \left[2\pi t\left(0.1\right)\right]\\ \text{and}\\ x=30\cos \left(0.2\pi t\right)\\ y=40+30\sin \left(0.2\pi t\right)\end{array}$

Differentiate with respect to t .

  $\begin{array}{l}\frac{dx}{dt}=-30\sin \left(0.2\pi t\right)\cdot 0.2\pi \frac{dt}{dt}\\ \text{and}\\ \frac{dy}{dt}=0+3\cos \left(0.2\pi t\right)\cdot 0.2\pi \frac{dt}{dt}\\ \frac{dx}{dt}=-30\cdot 0.2\pi \sin \left(0.2\pi t\right)\left[\text{Since, }\frac{dt}{dt}=1\right]\\ \text{and}\\ \frac{dy}{dt}=30\cdot 0.2\pi \cos \left(0.2\pi t\right)\\ \begin{array}{l}\frac{dx}{dt}=-6\pi \sin \left(0.2\pi t\right)\\ \text{and}\\ \frac{dy}{dt}=6\pi \cos \left(0.2\pi t\right)\end{array}]\mathrm{.....}\left(1\right)\end{array}$

Putting the value of $t=5$ in equation (1),

  $\begin{array}{l}\frac{dx}{dt}=-6\pi \sin \left[0.2\pi \left(5\right)\right]\\ \frac{dx}{dt}=-6\pi \sin \left[1\pi \right]\\ \frac{dx}{dt}=-6\pi \sin 180\\ \frac{dx}{dt}=-6\pi \sin \left(0\right)\left[\text{Since,}\sin \text{180}=\text{0}\right]\\ \frac{dx}{dt}=0\text{ ft}\text{./sec}\text{.}\\ \text{and}\\ \frac{dy}{dt}=6\pi \cos \left[0.2\pi \left(5\right)\right]\\ \frac{dy}{dt}=6\pi \cos \left(\pi \right)\\ \frac{dy}{dt}=6\pi \cos 180\\ \frac{dy}{dt}=6\pi \cos \left(-1\right)\left[\text{Since,}\cos \text{180}=-1\right]\\ \frac{dy}{dt}=-6\times 3.14\left[\text{Since,}\pi =3.14\right]\\ \frac{dy}{dt}=18.8\text{ ft}\text{./sec}\text{.}\end{array}$

Again, putting the value of t = 8 in equation (1),

  $\begin{array}{l}\frac{dx}{dt}=-6\pi \sin \left[0.2\pi \left(8\right)\right]\\ \frac{dx}{dt}=-6\pi \sin \left[1.6\pi \right]\\ \frac{dx}{dt}=-6\pi \sin \left[1.6\times 180\right]\left[\text{Since,}\sin \pi =\sin \text{180}\right]\\ \frac{dx}{dt}=-6\pi \sin \left(288\right)\\ \frac{dx}{dt}=-6\pi \left(-0.95\right)\\ \frac{dx}{dt}=6\times 3.14\times 0.95\\ \frac{dx}{dt}=17.898\text{ ft}\text{./sec}\text{.}\\ \text{and}\\ \frac{dy}{dt}=6\pi \cos \left[0.2\pi \left(8\right)\right]\\ \frac{dy}{dt}=6\pi \cos \left(1.6\pi \right)\\ \frac{dy}{dt}=6\pi \cos \left[1.6\times 180\right]\left[\text{Since,cos}\pi =\text{cos180}\right]\\ \frac{dy}{dt}=6\pi \cos \left(288\right)\\ \frac{dy}{dt}=6\pi \left(0.30\right)\\ \frac{dy}{dt}=6\times 3.14\times 0.30\\ \frac{dy}{dt}=5.652\text{ ft}\text{./sec}\text{.}\end{array}$

Hence, the value of horizontal movement when $t=5\text{ sec}\text{.}$ is $\frac{dx}{dt}=0\text{ ft}\text{./sec}\text{.}$ and $\frac{dy}{dt}=18.8\text{ ft}\text{./sec}\text{.}$ and when $t=8\text{ sec}\text{.}$ is $\frac{dx}{dt}=17.898\text{ ft}\text{./sec}\text{.}$ and $\frac{dy}{dt}=5.652\text{ ft}\text{./sec}\text{.}$