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)

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

Hence the rate of rising of water when depth is half of a foot is $\frac{\sqrt{3}}{4}$ feet per minute

Answer from: Quest

here you hope this .

step-by-step explanation:

hypotenuse = √(15² + 8²)

= √(225 +64)

= √(289) = 17

sin(x) = cos(y) = 8/17

they are identical so long as you are doing the sine of one angle and the cosine of the only other angle that isn't the right angle.

Hurry need your use the image below to answer the following question. find the value of sin x° and

Answer from: Quest

76%

step-by-step explanation:

24% are dogs and the remaining is 76% because 100-24=76

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