Answer:
[tex]\mathbf{\dfrac{dV}{dt} = \pi \bigg ( \dfrac{{t}+2}{4\sqrt{t}}\bigg)}[/tex]
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:
[tex]V = \pi r^2 h[/tex]
[tex]V = \pi(t+6) \bigg ( \dfrac{\sqrt{t}}{6}\bigg)[/tex]
[tex]V = \pi \bigg( \dfrac{t^{3/2}}{6}+ \sqrt{t} \bigg)[/tex]
[tex]\dfrac{dV}{dt} = \pi \bigg ( \dfrac{3 \sqrt{t}}{12}+ \dfrac{1}{2\sqrt{t}} \bigg)[/tex]
[tex]\dfrac{dV}{dt} = \pi \bigg ( \dfrac{3t}{12 \sqrt{t}}+ \dfrac{6}{12\sqrt{t}} \bigg)[/tex]
[tex]\dfrac{dV}{dt} = \pi \bigg ( \dfrac{3 {t}+6}{12\sqrt{t}}\bigg)[/tex]
[tex]\mathbf{\dfrac{dV}{dt} = \pi \bigg ( \dfrac{{t}+2}{4\sqrt{t}}\bigg)}[/tex]
