Question

The radius of a right circular cylinder is given by √(t+6) and its height is 1/6√t, where t is time in seconds and the dimensions are in inches. Find the rate of change of the volume with respect to time. and its height is? (in3/s)

Answer 1

`\mathbf{(dV)/(dt) = \pi \bigg ( \frac{{t}+2}{4√(t)}\bigg)}`

Explanation:

Given that:

The radius of a right cylinder is given by √(t+6) and its height is 1/6√t;

The volume of a given circular cylinder is:

`V = \pi r^2 h`

`V = \pi(t+6) \bigg ( (√(t))/(6)\bigg)`

`V = \pi \bigg( (t^(3/2))/(6)+ √(t) \bigg)`

`(dV)/(dt) = \pi \bigg ( (3 √(t))/(12)+ (1)/(2√(t)) \bigg)`

`(dV)/(dt) = \pi \bigg ( (3t)/(12 √(t))+ (6)/(12√(t)) \bigg)`

`(dV)/(dt) = \pi \bigg ( \frac{3 {t}+6}{12√(t)}\bigg)`

`\mathbf{(dV)/(dt) = \pi \bigg ( \frac{{t}+2}{4√(t)}\bigg)}`