Chapter 1.4, Problem 74E

a.

To determine

Estimates the force $F$ (in tons) of water against the face of a dam.

Answer to Problem 74E

  $\begin{array}{cccccc}y& 5& 10& 20& 30& 40\\ F\left(y\right)& 26454.21& 149653.19& 846558.24& 2332844.47& 4788868.99\end{array}$

Explanation of Solution

Given information:

The function $F\left(y\right)=149.76\sqrt{10}{y}^{5/2}$ estimates the force $F$ (in tons) of water against the face of a dam, where $y$ is the depth of the water (in feet).

Complete the table. What can you conclude from the table?

  $\begin{array}{cccccc}y& 5& 10& 20& 30& 40\\ F\left(y\right)& & & & & \end{array}$

Calculation:

The function $F\left(y\right)=149.76\sqrt{10}{y}^{5/2}$ estimates the force $F$ (in tons) of water against the face of a dam, where $y$ is the depth of the water (in feet).

To calculate the force $F$ put $y=5,10,20,30,40$ in $F\left(y\right)=149.76\sqrt{10}{y}^{5/2}$ .

  $\begin{array}{l}F\left(5\right)=149.76\sqrt{10}{\left(5\right)}^{5/2}\\ =149.76\times 3.16\times 55.9\\ =26454.21\end{array}$

  $\begin{array}{l}F\left(10\right)=149.76\sqrt{10}{\left(10\right)}^{5/2}\\ =149.76\times 3.16\times 316.23\\ =149653.19\end{array}$

  $\begin{array}{l}F\left(20\right)=149.76\sqrt{10}{\left(20\right)}^{5/2}\\ =149.76\times 3.16\times 1788.85\\ =846558.24\end{array}$

  $\begin{array}{l}F\left(30\right)=149.76\sqrt{10}{\left(30\right)}^{5/2}\\ =149.76\times 3.16\times 4929.5\\ =2332844.47\end{array}$

  $\begin{array}{l}F\left(40\right)=149.76\sqrt{10}{\left(40\right)}^{5/2}\\ =149.76\times 3.16\times 10119.29\\ =4788868.99\end{array}$

The table can be complete as.

  $\begin{array}{cccccc}y& 5& 10& 20& 30& 40\\ F\left(y\right)& 26454.21& 149653.19& 846558.24& 2332844.47& 4788868.99\end{array}$

b.

To determine

Estimates the force $F$ (in tons) of water against the face of a dam.

Answer to Problem 74E

  $\boxed{y=20\&22}$

Explanation of Solution

Given information:

The function $F\left(y\right)=149.76\sqrt{10}{y}^{5/2}$ estimates the force $F$ (in tons) of water against the face of a dam, where $y$ is the depth of the water (in feet).

  $\begin{array}{cccccc}y& 5& 10& 20& 30& 40\\ F\left(y\right)& 26454.21& 149653.19& 846558.24& 2332844.47& 4788868.99\end{array}$

Use the table to approximate the depth at which the force against the dam is $1,000,000$ tons.

Calculation:

The force value $1,000,000$ is between $846558.24\&2332844.47$ lies for value of $y=20\&30$

Hence, the force value $1,000,000$ is nearer to $846558.24$ while $y=20\&22$ .

c.

To determine

Estimates the force $F$ (in tons) of water against the face of a dam.

Answer to Problem 74E

  $\boxed{y=21.37ft}$

Explanation of Solution

Given information:

The function $F\left(y\right)=149.76\sqrt{10}{y}^{5/2}$ estimates the force $F$ (in tons) of water against the face of a dam, where $y$ is the depth of the water (in feet).

Find the depth at which the force against the dam is $1,000,000$ tons algebraically.

Calculation:

To find depth $y$ put force $F=1,000,000$ in function $F\left(y\right)=149.76\sqrt{10}{y}^{5/2}$ .

  $\begin{array}{l}1,000,000=149.76\sqrt{10}{\left(y\right)}^{5/2}\\ \frac{1,000,000}{149.76\sqrt{10}}={\left(y\right)}^{5/2}\\ y={\left(2111.5636\right)}^{2/5}\\ y=21.37ft\end{array}$

Hence the depth at which the force against the dam is $1,000,000$ tons is

  $\boxed{y=21.37ft}$