Respuesta :
Given:
The rate of change of volume of the cone V = 26 cubic feet per minute.
We need to find the rate of change of height at height = 50 feet.
The height = twice the radius.
[tex]h=2r[/tex]Divide both sides by 2, we get
[tex]\frac{h}{2}=\frac{2r}{2}[/tex][tex]r=\frac{h}{2}[/tex]Consider the volume of the cone.
[tex]V=\frac{1}{3}h\pi r^2[/tex][tex]\text{Substitute }r=\frac{h}{2}\text{ in the formula.}[/tex][tex]V=\frac{1}{3}h\pi(\frac{h}{2}_{})^2[/tex][tex]V=\frac{1}{3}h\pi\frac{h^2}{4}[/tex][tex]V=\frac{\pi}{12}h^3[/tex]Differentiate with respect to t.
[tex]\frac{dV}{dt}=\frac{\pi}{12}\times3h^2\times\frac{dh}{dt}[/tex][tex]\text{Substitute }\frac{dV}{dt}=26\text{ and }h=50\text{ in the equation.}[/tex][tex]26=\frac{\pi}{12}\times3(50)^2\times\frac{dh}{dt}[/tex][tex]26=1962.5\times\frac{dh}{dt}[/tex]Dividing both sides by 1962.5, we get
[tex]\frac{26}{1962.5}=\frac{dh}{dt}[/tex][tex]\frac{dh}{dt}=0.0132\text{ f}eet\text{ per minute}[/tex]The height of the pile is increasing by 0.01 feet per minute at the instant when the pile is 50 feet high.
