Answer:
Rate of increase in height =[tex]\frac{dh}{dt}=0.3156ft/min[/tex]
Step-by-step explanation:
we know that volume of a cone is given by
[tex]V=\frac{1}{12}\pi d^{2}h[/tex]
It is Given that diameter equals height thus we have
[tex]V=\frac{1}{12}\pi h^{2}h\\\\V=\frac{1}{12}\pi h^{3}[/tex]
Differentiating both sides with respect to time we get
[tex]\frac{dV}{dt}=\frac{1}{12}\pi \frac{dh^{3}}{dt}\\\\\frac{dV}{dt}=\frac{1}{12}\pi(3h^{2}\frac{dh}{dt})[/tex]
Applying values and solving for [tex]\frac{dh}{dt}[/tex] we get
[tex]\frac{dh}{dt}=\frac{12\frac{dV}{dt}}{3\pi h^{2}}\\\\\frac{dh}{dt}=0.3156ft/min[/tex]
