Hello vampire_thirst,
The statement:
"Sand falls from a conveyor belt at a rate of 30 m^3/min..."
tells us regarding the rate of change of the volume of the pile of sand:
$$\frac{dV}{dt}=30\frac{\text{m}^3}{\text{min}}$$
That is, the volume of the pile is increasing at a rate of 30 cubic meters per minute.
The statement:
"The height of the pile is always the same as the base diameter"
tells us:
$$h=2r$$
And so, using the formula for the volume of a cone:
$$V=\frac{1}{3}\pi r^2h$$
we find:
(1) The volume as a function of the height $h$:
$$V(h)=\frac{1}{3}\pi \left(\frac{h}{2} \right)^2(h)=\frac{1}{12}\pi h^3$$
(2) The volume as a function of the base radius $r$:
$$V(r)=\frac{1}{3}\pi r^2(2r)=\frac{2}{3}\pi r^3$$
a.) How fast is the height changing when the pile is 8 m high?
If we implicitly differentiate (1) with respect to time $t$, we obtain:
$$\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$$
Since we are asked to find how fast the height of the pile is increasing, we want to solve for $$\frac{dh}{dt}$$:
$$\frac{dh}{dt}=\frac{4}{\pi h^2}\frac{dV}{dt}$$
Now, using the given data:
$$\frac{dV}{dt}=30\frac{\text{m}^3}{\text{min}},\,h=8\text{ m}$$
we find:
$$\frac{dh}{dt}=\frac{4}{\pi \left(8\text{ m} \right)^2}\left(30\frac{\text{m}^3}{\text{min}} \right)=\frac{15}{8\pi}\frac{\text{m}}{\text{min}}\approx0.596831036594608\frac{\text{m}}{\text{min}}$$
b.) How fast is the radius changing when the pile is 8 m high?
If we implicitly differentiate (2) with respect to time $t$, we obtain:
$$\frac{dV}{dt}=2\pi r^2\frac{dr}{dt}$$
Since we are asked to find how fast the base radius of the pile is increasing, we want to solve for $$\frac{dr}{dt}$$:
$$\frac{dr}{dt}=\frac{1}{2\pi r^2}\frac{dV}{dt}$$
Now, using the given data:
$$\frac{dV}{dt}=30\frac{\text{m}^3}{\text{min}},\,r=4\text{ m}$$
we find:
$$\frac{dr}{dt}=\frac{1}{2\pi \left(4\text{ m} \right)^2}\left(30\frac{\text{m}^3}{\text{min}} \right)=\frac{15}{16\pi}\frac{\text{m}}{\text{min}}\approx0.298415518297304\frac{\text{m}}{\text{min}}$$
