Question

The radius r of a cone is increasing at a rate of 4in/min. The height h of the cone is always 4 times the radius. Find the rate of change of the volume V when the radius is 7 inches. Hint: V = 1/3 pi *r^2*h

Answer 1

$\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}\qquad h=4r\implies V=\cfrac{\pi r^2 \cdot 4r}{3}\implies V=\cfrac{4\pi r^3}{3}\\\\ -----------------------------\\\\ \cfrac{dv}{dt}=\cfrac{4\pi }{3}\cdot 3r^2\cdot \cfrac{dr}{dt}\implies \cfrac{dv}{dt}=4\pi r^2\cdot \cfrac{dr}{dt}\qquad \begin{cases} \frac{dr}{dt}=4\\\\ r=7 \end{cases} \\\\\\ \cfrac{dv}{dt}=4\pi (7)^2\cdot 4$