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 of the cylinder, V, with respect to time.(Volume(V) of a cylinder with a circular base area (A) and hight (h) is given by V=Ah).

Answer 1

Answer:

The rate change of volume of the cylinder is $\frac{\pi}{4} ( \sqrt t+\frac2{ \sqrt t})$ cubic inch per second.

Step-by-step explanation:

Given that the radius of right circular cylinder is $\sqrt{(t+6)}$  and its height is $\frac16 \sqrt t$ where t is time in second and the dimension are inches.

$\therefore r = \sqrt{(t+6)}$

The base area of the cylinder is A= $\pi r^2$

                                                        $=\pi (\sqrt{t+6})^2$

                                                       $= \pi (t+6)$

$\therefore A= \pi(t+6)$

Differentiating with respect to t

$\frac{dA}{dt}=\pi$

$\therefore h=\frac16\sqrt t$

Differentiating with respect to t

$\frac{dh}{dt}=\frac16 \times \frac12(t)^{\frac12-1}$

$\Rightarrow \frac{dh}{dt}=\frac1{12} (t)^{-\frac12}$

The volume of cylinder is V= Ah

∴V= Ah

Differentiating with respect to t

$\frac{dV}{dt}=A\frac{dh}{dt}+h\frac{dA}{dt}$

    $=\pi (t+6). \frac1{12}t^{-\frac12} +\frac16\sqrt t . \pi$

   $=\pi. \frac1{12}.t^{\frac12}+\pi . 6.\frac1{12} t^{-\frac12} +\pi\frac16 \sqrt t$

   $=\frac{\pi}{4} ( \sqrt t+\frac2{ \sqrt t})$

The rate change of volume of the cylinder is $\frac{\pi}{4} ( \sqrt t+\frac2{ \sqrt t})$ cubic inch per second.