Question

gravel is being dumped from a conveyor belt at a rate of 40 cubic feet per minute. it forms a pile in the shape of a right circular cone whose base diameter and height are always equal. (remember, the radius is half the diameter, so ( r = 1/2h ). you can use substitution to simplify.) how fast is the height of the pile increasing, when the pile is 25 feet high? recall that the volume of a right circular cone with height ( h ) and radius of the base ( r ) is given by: ( v=\frac{1}{3}pi r^{2}h ). ( \frac{dh}{dt}=\frac{ft}{min} ) question help: video message instructor submit question jump to answer

Explanation:

Step1: Substitute \(r=\frac{1}{2}h\) into the volume formula

The volume formula of a cone is \(V = \frac{1}{3}\pi r^{2}h\). Substituting \(r=\frac{1}{2}h\) gives \(V=\frac{1}{3}\pi(\frac{1}{2}h)^{2}h=\frac{1}{12}\pi h^{3}\).

Step2: Differentiate \(V\) with respect to \(t\)

Using the chain - rule \(\frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}\). Differentiating \(V = \frac{1}{12}\pi h^{3}\) with respect to \(h\) gives \(\frac{dV}{dh}=\frac{1}{4}\pi h^{2}\). So \(\frac{dV}{dt}=\frac{1}{4}\pi h^{2}\frac{dh}{dt}\).

Step3: Solve for \(\frac{dh}{dt}\)

We know that \(\frac{dV}{dt} = 40\) (the rate of change of volume) and \(h = 25\). Substituting these values into \(\frac{dV}{dt}=\frac{1}{4}\pi h^{2}\frac{dh}{dt}\), we get \(40=\frac{1}{4}\pi(25)^{2}\frac{dh}{dt}\).
Then \(\frac{dh}{dt}=\frac{40\times4}{\pi\times625}=\frac{160}{625\pi}=\frac{32}{125\pi}\approx\frac{32}{125\times3.14}\approx\frac{32}{392.5}\approx0.0815\).

Answer:

\(\frac{32}{125\pi}\approx0.0815\)