Question

A water trough is 7 feet long, and its cross section is an equilateral triangle with sides 4 feet long. Water is pumped into the trough at a rate of 4 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?
b) The water level is rising at a rate of

Answer 1

Answer:

a. An equilateral triangle main characteristic is that all the sides lenght are the same (s). To find the height (h), we could divide the triangle (as seen in the picture) and apply Pytagorean theorem.

$(\frac{s}{2}) ^{2} +h^{2}=s^{2}$

Clearing the expression, we obtain: $h= \frac{\sqrt{3}}{2}s$

b. Knowing the rate at which the volume is changing 4 $ft^{3}$, we can find the relation between the change in the volume and the height.

$V=A*h$

As we want to express the volume in terms of the height, we have to find the area in terms of height

$A=\frac{base*height}{2}$

Base=s=$\frac{2h}{\sqrt{3}}$

Therefore, $A=\frac{\frac{2h}{\sqrt{3}} *h}{2} =\frac{h^{2} }{\sqrt{3}}$

$V=7 ft*\frac{h^{2}}{\sqrt{3}}$

Therefore the change in the volume with the height, will be the derivate of this expression

$\frac{dV}{dt} =2*7*\frac{h}{\sqrt{3}} (\frac{dh}{dt} )$

Knowing dV/dt=4 cubic feet per second, and h=1/2 foot, we can know dh/dt

$\frac{dh}{dt}=\frac{\frac{dV}{dt}*\sqrt{3}}{7*2*h} =\frac{4*\sqrt{3} }{14*\frac{1}{2} }=0.98 ft/sec$

Step-by-step explanation: