How Fast is the Height of a Gravel Cone Increasing?

[karush]

Gravel is being dumped from a conveyor belt at a rate of
$30\displaystyle\frac{ ft}{min}$
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
$10 ft$ high?

$$V=\frac{1}{3}\pi r^2h$$

 

[MarkFL]

Gravel is being dumped from a conveyor belt at a rate of \(\displaystyle 30\,\frac{\text{ft}^3}{\text{min}}\)

This tells us, if we let all measures of length be in feet, and all measures of time be in minutes, that we may write:

\(\displaystyle \d{V}{t}=30\)

and its coarseness is such that it forms a pile in the shape of a cone:

This tells is we may write:

\(\displaystyle V=\frac{\pi}{3}r^2h\)

whose base diameter and height are always equal:

This tells is we may write:

\(\displaystyle 2r=h\implies r=\frac{h}{2}\)

How fast is the height of the pile increasing when the pile is 10 ft high?

This tells us we want to find \(\displaystyle \d{h}{t}\) when \(h=10\).

I would begin by expressing the volume \(V\) of the cone as a function of \(h\):

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

Now, implicitly differentiate w.r.t \(t\):

\(\displaystyle \d{V}{t}=\frac{\pi}{4}h^2\d{h}{t}\)

Can you finish?

 

[karush]

$\displaystyle \d{V}{t}=\frac{\pi}{4}h^2\d{h}{t} $
since
$\d{V}{t}=30$and $h=10$
then$\displaystyle 30=\frac{\pi}{4}10^2\d{h}{t} $
so far