gravel is being dumped from a conveyor belt at a rate of 20 cubic feet per minute. it forms a pile in the…

gravel is being dumped from a conveyor belt at a rate of 20 cubic feet per minute. it forms a pile in the shape of a right circular cone whose base diameter and height are always equal. how fast is the height of the pile increasing when the pile is 11 feet high? recall that the volume of a right circular cone with height h and radius of the base r is given by $v = \frac{1}{3}pi r^{2}h$

gravel is being dumped from a conveyor belt at a rate of 20 cubic feet per minute. it forms a pile in the shape of a right circular cone whose base diameter and height are always equal. how fast is the height of the pile increasing when the pile is 11 feet high? recall that the volume of a right circular cone with height h and radius of the base r is given by $v = \frac{1}{3}pi r^{2}h$

Answer

Explanation:

Step1: Express radius in terms of height

Since the base diameter (d) and height (h) are equal ((d = h)), and (d = 2r), then (r=\frac{h}{2}).

Step2: Substitute (r) into the volume formula

Substitute (r = \frac{h}{2}) into (V=\frac{1}{3}\pi r^{2}h). [ \begin{align*} V&=\frac{1}{3}\pi(\frac{h}{2})^{2}h\ &=\frac{1}{3}\pi\frac{h^{2}}{4}h\ &=\frac{\pi}{12}h^{3} \end{align*} ]

Step3: Differentiate (V) with respect to time (t)

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

Step4: Solve for (\frac{dh}{dt})

We know that (\frac{dV}{dt} = 20) (cubic feet per minute). When (h = 11), substitute into (\frac{dV}{dt}=\frac{\pi}{4}h^{2}\cdot\frac{dh}{dt}): [ \begin{align*} 20&=\frac{\pi}{4}(11)^{2}\cdot\frac{dh}{dt}\ \frac{dh}{dt}&=\frac{80}{\pi\times121}\ &=\frac{80}{121\pi} \end{align*} ]

Answer:

(\frac{80}{121\pi})