Related Rates: Conical Pile of Sand | Vampire_thirst Q&A

[MarkFL]

Here is the question:

Another related rates problem?


Sand falls from a conveyor belt at a rate of 30 m^3/min onto the top of a conical pile. The height of the pile is always the same as the base diameter. Answer the following.

a.) How fast is the height changing when the pile is 8 m high?
Answer = m/min.

b.) How fast is the radius changing when the pile is 8 m high?
Answer = m/min.

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

 

[MarkFL]

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:

$$\frac{dr}{dt}=\frac{1}{2\pi \left(4\text{ m} \right)^2}\left(30\frac{\text{m}^3}{\text{min}} \right)=\frac{15}{16\pi}\frac{\text{m}}{\text{min}}\approx0.298415518297304\frac{\text{m}}{\text{min}}$$