Question

A water trough is 12 feet long and its cross section is an equilateral triangle with sides 2 feet long. Water is pumped into the triangle at a rate of 3 cubic feet per minute. How fast is the water level rising when the depth is half of a foot?

Answer 1

ANSWER:

The rate of rising of water when depth is half of a foot is  $\frac{\sqrt{3}}{4}$ feet per minute

SOLUTION:

Given, a water trough is 12 feet long and its cross section is an equilateral triangle with sides 2 feet long.

Water is pumped into the triangle at a rate of 3 cubic feet per minute.

We need to find how fast is the water level rising when the depth is half of a foot.  

The volume of water in the trough is equal to the cross-sectional area submerged times the length of the trough. The cross-sectional area is an equilateral triangle.  

If you are given the height of an equilateral triangle as “h”, then area of triangle is  $\frac{\mathrm{h}^{2}}{\sqrt{3}}$

Volume = (cross sectional area) $\times$ (height of trough)

$V=(12) \frac{\mathrm{h}^{2}}{\sqrt{3}}$

Now, take the derivative of both sides with respect to time using the chain rule,

$\begin{array}{l}{\frac{d V}{d t}=12 \times \frac{d}{d t}\left(\frac{\mathrm{h}^{2}}{\sqrt{3}}\right)} \\\\ {\frac{d V}{d t}=\frac{12}{\sqrt{3}} \times \frac{d\left(\mathrm{h}^{2}\right)}{d t}} \\\\ {\frac{d V}{d t}=\frac{12}{\sqrt{3}} \times 2 \mathrm{h} \times \frac{d h}{d t}}\end{array}$

Now, substitute $\frac{d V}{d t}=3$ (water pumping rate), h = 0.5(depth of water)

$\begin{array}{l}{3=\frac{12}{\sqrt{3}} \times 2(0.5) \times \frac{d h}{d t}} \\\\ {3=\frac{12}{\sqrt{3}} \times 1 \times \frac{d h}{d t}} \\\\ {\frac{d h}{d t}=3 \times \frac{\sqrt{3}}{12}} \\\\ {\frac{d h}{d t}=\frac{\sqrt{3}}{4}}\end{array}$

Hence the rate of rising of water when depth is half of a foot is  $\frac{\sqrt{3}}{4}$ feet per minute