Chapter 1.4, Problem 70E

i.

To determine

To calculate: The function $F\left(y\right)=149.76\sqrt{10}{y}^{\frac{5}{2}}$ estimate 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?

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

Answer to Problem 70E

  $\begin{array}{cccccc}y& 5& 10& 20& 30& 40\\ F\left(y\right)& 26473.93& 149759.15& 847165.66& 2334514.04& 4792292.65\end{array}$

From table it can be conclude that as depth increases, the force also increases

Explanation of Solution

Given information: The given function is $F\left(y\right)=149.76\sqrt{10}{y}^{\frac{5}{2}}$

At $y=5,F\left(y\right)=149.76\sqrt{10}\left(5^{\frac{5}{2}}\right)=473.58\left(5^{\frac{5}{2}}\right)=26473.93$

At $y=10,F\left(y\right)=149.76\sqrt{10}\left({10}^{\frac{5}{2}}\right)=473.58\left({10}^{\frac{5}{2}}\right)=149759.15$

At $y=20,F\left(y\right)=149.76\sqrt{10}\left({20}^{\frac{5}{2}}\right)=473.58\left({20}^{\frac{5}{2}}\right)=847165.66$

At $y=30,F\left(y\right)=149.76\sqrt{10}\left({30}^{\frac{5}{2}}\right)=473.58\left({30}^{\frac{5}{2}}\right)=2334514.04$

At $y=40,F\left(y\right)=149.76\sqrt{10}\left({40}^{\frac{5}{2}}\right)=473.58\left({40}^{\frac{5}{2}}\right)=4792292.65$

Conclusion:

  $\begin{array}{cccccc}y& 5& 10& 20& 30& 40\\ F\left(y\right)& 26473.93& 149759.15& 847165.66& 2334514.04& 4792292.65\end{array}$

From the table it can be conclude that as depth increases, the force also increases

ii.

To determine

To calculate: The function $F\left(y\right)=149.76\sqrt{10}{y}^{\frac{5}{2}}$ estimate the force F (in tons) of water against the face of a dam, where y is the depth of the water (in feet). Use the table to approximate the depth at which the force against the dam is 1,000,000 tons.

Answer to Problem 70E

Approximately 21 feet

Explanation of Solution

Given information:

Using above table it can be seen that at $847165.66$ tons, the height is 20 feet and it is increasing rapidly. So using this pattern at 1,000,000 tons the approximate height will be 21 feet.

iii.

To determine

To calculate: The function $F\left(y\right)=149.76\sqrt{10}{y}^{\frac{5}{2}}$ estimate 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.

Answer to Problem 70E

The depth is 21.37 feet

Explanation of Solution

Given information: The given function is $F\left(y\right)=149.76\sqrt{10}{y}^{\frac{5}{2}}$ here $F\left(y\right)=1000000$

Calculation: $F\left(y\right)=149.76\sqrt{10}{y}^{\frac{5}{2}}$

  $\begin{array}{l}1000000=149.76\sqrt{10}{y}^{\frac{5}{2}}\\ \frac{1000000}{149.76\sqrt{10}}=y^{\frac{5}{2}}\\ {\left(\frac{1000000}{149.76\sqrt{10}}\right)}^{\frac{2}{5}}=y^{\frac{5}{2}\times \frac{2}{5}}\\ y=21.3718\end{array}$

Conclusion: Depth is 21.37 feet