Question

The radius of a right circular cone is increasing at a rate of 2 inches per second and its height is decreasing at a rate of 2 inches per second. At what rate is the volume of the cone changing when the radius is 40 inches and the height is 20 inches? NOTE: The volume of a cone with base radius r and height h is given by V=13πr2h.

Answer 1

Answer:

Rate of change of volume when the radius is 40 inches and the height is 20 inches = 0 inch³/s

Step-by-step explanation:

We have

         $\texttt{Volume = }\frac{1}{3}\pi r^2h$

Taking derivative with respect to time,

         $V=\frac{1}{3}\pi r^2h\\\\\frac{dV}{dt}=\frac{1}{3}\pi \frac{d}{dt}\left (r^2h \right )\\\\\frac{dV}{dt}=\frac{1}{3}\pi \left ( 2r\frac{dr}{dt}\times h+r^2\frac{dh}{dt}\right)$

Given that

          r = 40 inches

          h = 20 inches

         $\frac{dr}{dt}=2inch/s\\\\\frac{dh}{dt}=-2inch/s$

Substituting

        $\frac{dV}{dt}=\frac{1}{3}\pi \left ( 2r\frac{dr}{dt}\times h+r^2\frac{dh}{dt}\right)\\\\\frac{dV}{dt}=\frac{1}{3}\pi \left ( 2\times 40\times 2\times 20+40^2\times (-2)\right)\\\\\frac{dV}{dt}=\frac{1}{3}\pi \left (3200-3200\right)=0inch^3/s$

Rate of change of volume when the radius is 40 inches and the height is 20 inches = 0 inch³/s