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 trough at a rate of 10 cubic feet per second. How fast is the water level rising when the depth of the water is 1/2 foot? ( Hint: First, what is the height h of an equilateral triangle of side length s? Next, what is the area of an equilateral triangle in terms of the side length s? Then write the area in terms of h. The volume of the water in the trough at time t is the product of the cross-sectional area with water and the length of the trough. ) a) What is the height h of an equilateral triangle of side length s?

Answer 1

Answer:

The water level rises at 11.76 $ft^{2} /s$

$h=\sqrt{3/2} s$ let be h: height and s: side of an equilateral triangle

Explanation:

The picture shows a diagram of the situation, first we have to determine the height of the trough as follows:

With the Pytagorean Theorem we can find out that:

$h^{2} =s^{2} -s^{2} /4 =\sqrt{3} /2 s$

Then, the area of an equilateral triangle, as any triangle, is a half of its base times its height:

$A=\frac{1}{2} hs=\frac{1}{2} \frac{\sqrt{3} }{2} s^{2} =\frac{\sqrt{3} }{4} s^{2}$

Replacing values, we have:

$A=1.73ft^{2}$

That is the total area of the trough, but the problem specifies that it has been filled until 0.5 ft. Therefore, we have to find the cross section area of the water flow by substracting the total area minus the unfilled area of the trough:

$h_{u}: height of the unfilled triangle\\s_{u} : side of the unfilled triangle\\A_{u}: area of the unfilled triangle\\h_{u}=\sqrt{3} -0.5 =1.23\\s_{u}= \frac{2}{\sqrt{3} } 1.23=1.42\\A_{u}=\frac{\sqrt{3} }{4} 1.42^{2} =0.87ft^{2} A_{u}$

Then, the cross section area of water flow is

$A_{s} =1.73ft^{2} -0.87ft^{2} =0.85ft^{2}$

Finally, to determine the speed of water flow at this point we solve for v, the  flow formula:

$Q: water flow\\v: speed\\Q=vA\\v=Q/A =\frac{10ft^{3}/s }{0.85ft^{2} } =11.76 \frac{ft^{2}}{s}$