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ANSWER:
The rate of rising of water when depth is half of a foot is [tex]\frac{\sqrt{3}}{4}[/tex] 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 [tex]\frac{\mathrm{h}^{2}}{\sqrt{3}}[/tex]
Volume = (cross sectional area) [tex]\times[/tex] (height of trough)
[tex]V=(12) \frac{\mathrm{h}^{2}}{\sqrt{3}}[/tex]
Now, take the derivative of both sides with respect to time using the chain rule,
[tex]\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}[/tex]
Now, substitute [tex]\frac{d V}{d t}=3[/tex] (water pumping rate), h = 0.5(depth of water)
[tex]\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}[/tex]
Hence the rate of rising of water when depth is half of a foot is [tex]\frac{\sqrt{3}}{4}[/tex] feet per minute
To answer this question it is necessary to find the equation of the volume (V) of the trough as a function of the height "h"
The solution is:
dh/dt = 1.30 f/min
The bisectrix of a vertex of an equilateral triangle of side L, divides the opposite side into 2 sides of equal size, forming two interior isosceles triangles.
In one of such triangles, we find: By Pithagoras´theorem
h = √ ( L)² - ( L/2)² ⇒ h = √(3/4)×L²
h = (√3)/2 ×L or L = 2×h/√3
Now the area of one of these triangles is:
A(it) =(1/2)× b×h = (1/2) × L/2 × h
as L/2 = h/√3
A(it) = (1/4×√3)×h²
as the whole area is 2×A(it)
then the area of the equilateral triangle as a function of h is:
A(h) = (1/2×√3)×h² and the volume of the trough is
V(h) = (6/√3)×h²
Now tacking derivatives with respect to time on both sides of the equation we get:
dV/dt = (12/√3)×2×h×dh/dt
We are looking for dh/dt = ?
3 = (24/√3)×(1/2)× dh/dt
3 = (12/√3)× dh/dt
dh/dt = 1.30 f/min
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