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

Answer:

$\mathbf{\dfrac{dV}{dt} = \pi \bigg ( \dfrac{{t}+2}{4\sqrt{t}}\bigg)}$

Step-by-step 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:

$V = \pi r^2 h$

$V = \pi(t+6) \bigg ( \dfrac{\sqrt{t}}{6}\bigg)$

$V = \pi \bigg( \dfrac{t^{3/2}}{6}+ \sqrt{t} \bigg)$

$\dfrac{dV}{dt} = \pi \bigg ( \dfrac{3 \sqrt{t}}{12}+ \dfrac{1}{2\sqrt{t}} \bigg)$

$\dfrac{dV}{dt} = \pi \bigg ( \dfrac{3t}{12 \sqrt{t}}+ \dfrac{6}{12\sqrt{t}} \bigg)$

$\dfrac{dV}{dt} = \pi \bigg ( \dfrac{3 {t}+6}{12\sqrt{t}}\bigg)$

$\mathbf{\dfrac{dV}{dt} = \pi \bigg ( \dfrac{{t}+2}{4\sqrt{t}}\bigg)}$