Chapter 4.5, Problem 85E

a.

To determine

Find the period of the model.

Answer to Problem 85E

  $\boxed{20\sec onds}$

Explanation of Solution

Given information:

The height h (in feet) above ground of a seat on a Ferris wheel at time t (in seconds) is modelled by

  $\text{h(t) = 53 + 50 sin}\left(\frac{\pi }{10}t-\frac{\pi }{2}\right).$

Find the period of the model. What does the period tell you about the ride?

Calculation:

The height $h$ of a seat from the ground on the Ferris wheel is given by the function:

  $\text{h(t) = 53 + 50 sin}\left(\frac{\pi }{10}t-\frac{\pi }{2}\right).$

It is known that for a function of the form $y=d+a\sin \left(bx-c\right)$ , the period and amplitude is given by:

Amplitude= $|a|$

Period= $\frac{2\pi }{b}$

The shift is $d$ units upwards for $d>0$ and $d$ units downward for $d<0$

The period of $h\left(t\right)=53+50\sin \left(\frac{\pi }{10}t-\frac{\pi }{2}\right)$ is,

Period $=\frac{2\pi }{b}$

  $=\frac{2\pi }{\left(\frac{\pi }{10}\right)}$

  $=20$

Hence, The period is found to be $\boxed{20}$ seconds. It means a Ferris wheel complete one ride in $20$ seconds.

b.

To determine

Find the amplitude of the model

Answer to Problem 85E

  $\boxed{50feet}$

Explanation of Solution

Given information:

The height h (in feet) above ground of a seat on a Ferris wheel at time t (in seconds) is modelled by

  $\text{h(t) = 53 + 50 sin}\left(\frac{\pi }{10}t-\frac{\pi }{2}\right).$

Find the amplitude of the model. What does the amplitude tell you about the ride?

Calculation:

The amplitude of $h\left(t\right)=53+50\sin \left(\frac{\pi }{10}t-\frac{\pi }{2}\right)$ is,

Amplitude= $|50|$

  $=50$

Hence, the amplitude is found to be $\boxed{50}$ feet above the ground. It means a Ferries wheel has a ride of height $50feet$ .

c.

To determine

Use a graphing utility to graph one cycle of the model.

Answer to Problem 85E

  

Explanation of Solution

Given information:

The height h (in feet) above ground of a seat on a Ferris wheel at time t (in seconds) is modelled by

  $\text{h(t) = 53 + 50 sin}\left(\frac{\pi }{10}t-\frac{\pi }{2}\right).$

Use a graphing utility to graph one cycle of the model.

Calculation:

To graph this function in $TI-83$ , proceed as below:

Press the button $\boxed{Y=}$ and type $Y=53+50\sin \left(\frac{\pi }{10}\times -\frac{\pi }{2}\right)$

  

Press the button $\boxed{WINDOW}$ and adjust the window as shown :

  

Press the button $\boxed{TRACE}$ or $\boxed{GRAPH}$ to obtain the graph

  

Hence, the result is shown in utility graph.