Mathematics, 13.10.2019 05:01 maddieb1011

Awater 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?

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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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Answer from: taylorb9893

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)

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