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

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.

[tex](\frac{s}{2}) ^{2} +h^{2}=s^{2}[/tex]

Clearing the expression, we obtain: [tex]h= \frac{\sqrt{3}}{2}s[/tex]

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

[tex]V=A*h[/tex]

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

[tex]A=\frac{base*height}{2}[/tex]

Base=s=[tex]\frac{2h}{\sqrt{3}}[/tex]

Therefore, [tex]A=\frac{\frac{2h}{\sqrt{3}} *h}{2} =\frac{h^{2} }{\sqrt{3}}[/tex]

[tex]V=7 ft*\frac{h^{2}}{\sqrt{3}}[/tex]

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

[tex]\frac{dV}{dt} =2*7*\frac{h}{\sqrt{3}} (\frac{dh}{dt} )[/tex]

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

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

Step-by-step explanation:

jtellezd jtellezd · Answer 2 · 2021-10-10T23:50:42+02:00

To answer this question we get the volume of the trough as a function of "h" and after that, with the help of derivatives we solve the problem

The solution is:

dh/dt = 6.92 f³/sec

The volume of the trough is:

V = Area of equilateral triangle × height of the triangle

To calculate the area of the lateral triangle we need to find "h" as a function of the side s

h divided the equilateral triangle in 2 isosceles triangles of the same area

h = √ s² - (s/2)² ⇒ h = √ (4×s² - s²)/4

h = (1/2) ×√ 3×s² ⇒ h = (√3/2)×s or s = (2/√3)×h

Then the area of each isosceles triangle is:

A(it) = [(√3/2)×s] × s/4

A(it) = √3/8 * s²

And the area of the equilateral triangle is

A(et) = 2×√3/8 * s² ⇒ A(et) = (√3/4)× s²

Then the volume of the trough as a function of h is

V(et) = (√3/4)×[(2/√3)×h]²

V(h) = (√3/4)× ( 4/3)×h²

V(h) = (1/√3 )× h²

Now getting derivatives with respect to time on both sides of the equation:

dV/ dt = (1/√3 )× 2 × h ×dh/dt

In that equation we know

dV/dt = 4 f³/sec h = 0.5 = 1/2 then by substitution

4 f³/sec = ( 1 /√3 ) ×2 × (1/2) × dh/dt

dh/dt = 4×√3

dh/dt = 6.92 f³/sec

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