Question

How fast is the volume changing with respect to time when the radius is changing with respect to time when the radius is changing at a rate of dr/dt=1.5 feet per second and r=2 feet?

Answer 1

`18.84955592153876\ \text{cubic ft. per sec}`

Explanation:

The volume of cone at instant of time `t` is `V=\frac{\pi}{3}r^3`

Now, differentiating volume V w.r.t. time `t`, the rate of change of volume `(\frac{dV}{dt})` of cone is
`\frac{dV}{dt}=\frac{d}{dt}(\frac{\pi}{3}r^3)`
`=\frac{\pi}{3}(3r^2)\frac{dr}{dt}`
`=\pi r^2\frac{dr}{dt}`
`=\pi (2)^2(1.5)\quad (\text{setting} \ r=2, \frac{dr}{dt}=1.5 )`
`=18.84955592153876\ \text{cubic ft. per sec`