Chapter 8.4, Problem 24E

To determine

To find the total cost to apply the sealer inside the tunnel.

Answer to Problem 24E

Total cost of apply the sealer inside the tunnel is 38421.6 $

Explanation of Solution

Given information:

The given function is $y=25\cos \left(\frac{\pi x}{50}\right)$

The length of the tunnel is 300 ft. and width is 50 ft.

The cost of sealer is $1.75 per square foot.

Formula used:

If a smooth curve begins at $\left(a,c\right)$ and ends at $\left(b,d\right)$ , $a<b,c<d$ , then the length of the curve

  $\begin{array}{l}L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}\text{ if }y\text{ is smooth function of }x\text{ on }\left[a,b\right]\\ L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}dy}\text{ if }x\text{ is smooth function of }y\text{ on }\left[c,d\right]\end{array}$

Calculation :

From the figure the boundary for x is $-25\le x\le 25$

Evaluate $\frac{dy}{dx}$

  $\frac{dy}{dx}=-25\cdot \frac{\pi }{50}\sin \left(\frac{\pi x}{50}\right)$

Which is continuous on $\left[-25,25\right]$

Calculate the arc length

  $\begin{array}{l}L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}\\ L={\displaystyle \underset{-25}{\overset{25}{\int }}\sqrt{1+{\left(-25\cdot \frac{\pi }{50}\sin \left(\frac{\pi x}{50}\right)\right)}^2}dx}\\ L=73.184\left[\text{use NINT}\right]\end{array}$

Total area to be sealed

  $\begin{array}{l}A=73.184\times 300\\ =21995.2\text{ square foot}\end{array}$

Total cost of sealing the tunnel

  $\begin{array}{l}\text{=21995}\text{.2}\times 1.75\\ =38421.6\text{}\text{ \$}\end{array}$

Therefore,

Total cost of apply the sealer inside the tunnel is 38421.6 $