Notyo's question at Yahoo Answers regarding related rates

[MarkFL]

Here is the question:

Gravel is being dumped from a conveyor belt at a rate of 30 ft3/min, and its coarseness is such that it...?

Gravel is being dumped from a conveyor belt at a rate of 30 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 12 ft high? (Round your answer to two decimal places.)


___3.9, 12

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Notyo,

The statement:

"Gravel is being dumped from a conveyor belt at a rate of 30 ft3/min"

tells us regarding the rate of change of the volume of the pile of gravel:

\(\displaystyle \frac{dV}{dt}=30\frac{\text{ft}^3}{\text{min}}\)

That is, the volume of the pile is increasing at a rate of 30 cubic feet per minute.

The statement:

"it forms a pile in the shape of a cone whose base diameter and height are always equal"

tells us:

\(\displaystyle V=\frac{1}{3}\pi \left(\frac{h}{2} \right)^2(h)=\frac{1}{12}\pi h^3\)

This comes from the formula for the volume of a cone, where the base radius is equal to half the height.

Now, if we implicitly differentiate this equation with respect to time $t$, we obtain:

\(\displaystyle \frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}\)

Since we are asked to find how fast the height of the pile is increasing, we want to solve for \(\displaystyle \frac{dh}{dt}\):

\(\displaystyle \frac{dh}{dt}=\frac{4}{\pi h^2}\frac{dV}{dt}\)

Now, using the given data:

\(\displaystyle \frac{dV}{dt}=30\frac{\text{ft}^3}{\text{min}},\,h=12\text{ ft}\)

we find:

\(\displaystyle \frac{dh}{dt}=\frac{4}{\pi \left(12\text{ ft} \right)^2}\left(30\frac{\text{ft}^3}{\text{min}} \right)=\frac{5}{6\pi}\frac{\text{ft}}{\text{min}}\approx0.265258238486492\frac{\text{ft}}{\text{min}}\)