Chapter 5.4, Problem 56E

Solution

(a)

To show:That the cross-sectional area of the end of the tunnel is $400\sin 2\theta$ .

Given information:

The equation of circle is $x^2+y^2=400$ .

Formula used:

Use the double angle identity of sine.

  $\sin 2u=2\sin u\cos u$

Proof:

From the circle equation,

  $\begin{array}{c}{r}^2=400\\ r=20\end{array}$

In the given figure, use cosine function.

  $\begin{array}{c}\cos \theta =\frac{x}{20}\\ x=20\cos \theta \end{array}$

In the given figure, use sine function.

  $\begin{array}{c}\sin \theta =\frac{y}{20}\\ y=20\sin \theta \end{array}$

The area is as follows.

  $\begin{array}{c}A=wh\\ =\left(2x\right)\left(y\right)\\ =2xy\\ =2\left(20\cos \theta \right)\left(20\sin \theta \right)\end{array}$

Simplify further.

  $\begin{array}{c}A=400\left(2\sin \theta \cos \theta \right)\\ =400\sin 2\theta \end{array}$

Therefore, it is proved that the cross-sectional area of the end of the tunnel is $400\sin 2\theta$ .

b)

To find:The dimensions of the rectangular end of the tunnel.

The dimension of the rectangular end is $w=28.28$ and $y=14.14$ .

Calculation:

The maximum of the area is reached for the angle $\theta$ for which $\sin 2\theta =1$ .

  $\begin{array}{c}\sin 2\theta =1\\ 2\theta =\frac{\pi }{2}\\ \theta =\frac{\pi }{4}\end{array}$

Find the width as follows.

  $\begin{array}{c}2x=2\left(20\cos \theta \right)\\ =40\cos \frac{\pi }{4}\\ =40\cdot \frac{\sqrt{2}}{2}\\ =28.28\end{array}$

Find the height as follows.

  $\begin{array}{c}y=20\sin \frac{\pi }{4}\\ =20\cdot \frac{\sqrt{2}}{2}\\ =14.14\end{array}$

Therefore, the dimension of the rectangular end is $w=28.28$ and $y=14.14$ .