Question

The radius of a right circular cylinder increases at the rate of 0.1 cm per minute, and the height decreases at the rate of 0.2 cm per minute. Find the rate of change of Volume of the cylinder in $c{{m}^{3}}$ per minute, when the radius is 2cm and the height is 3cm.

Answer 1

Hint: Now we know that the volume of cylinder is given by $V=\pi {{r}^{2}}h$ differentiating the function with respect to time and then using the formula $\left( uv \right)'=u'v+v'u$ we will get an differential equation. Now we know the radius of a right circular cylinder increases at the rate of 0.1 cm per minute, and the height decreases at the rate of 0.2 cm per minute. Hence substituting this we will again get an equation in r and h. Now we want to find the change in volume when radius is 2cm and the height is 3cm. hence we substitute the values to find the change in volume.

Complete step by step answer:
Now we know that if h is the height of cylinder and r is the radius of the cylinder then the volume of cylinder is given by $V=\pi {{r}^{2}}h$ .
Now let us differentiate the whole equation with respect to time.
Hence we have $\dfrac{dV}{dt}=\dfrac{d\left( \pi {{r}^{2}}h \right)}{dt}$
Now we know that $\pi $ is a constant value and hence we can take it out of the differentiation. Hence we get,
$\dfrac{dV}{dt}=\pi \dfrac{d\left( {{r}^{2}}h \right)}{dt}$
Now we know that differentiation of u.v is given by $u'v+v'u$ . Hence using this we get,
$\dfrac{dV}{dt}=\pi \left( \dfrac{hd\left( {{r}^{2}} \right)}{dt}+{{r}^{2}}\dfrac{d\left( h \right)}{dt} \right)$
Now again we know differentiation of $f\left( g\left( x \right) \right)$ is $f'\left( g\left( x \right) \right).g'\left( x \right)$ .
 Hence we have $\dfrac{d\left( {{r}^{2}} \right)}{dt}=2r\dfrac{dr}{dt}$ now substituting this in the above equation we get,
$\dfrac{dV}{dt}=\pi \left( 2hr\dfrac{dr}{dt}+{{r}^{2}}\dfrac{dh}{dt} \right)$
Now we are given that the radius of a right circular cylinder increases at the rate of 0.1 cm per minute, and the height decreases at the rate of 0.2 cm per minute.
Hence we have $\dfrac{dh}{dt}=-0.2$ and $\dfrac{dr}{dt}=0.1$ . Now substituting this in the above equation we get,
$\dfrac{dV}{dt}=\pi \left( 2hr\left( 0.1 \right)+{{r}^{2}}\left( -0.2 \right) \right)$
Now we want to calculate the change in volume when r = 2 and h = 3. Hence substituting the values in the above equation we get,
$\begin{align}
  & \dfrac{dV}{dt}=\pi \left( 2\left( 2 \right)\left( 3 \right)\left( 0.1 \right)-\left( {{2}^{2}} \right)\left( 0.2 \right) \right) \\
 & \Rightarrow \dfrac{dV}{dt}=\pi \left( 1.2-0.8 \right) \\
 & \Rightarrow \dfrac{dV}{dt}=0.4\pi \\
\end{align}$

Hence the change in volume is $0.4\pi c{{m}^{3}}$ per minute.

Note: Now note that since we have the height is decreasing, we have used negative signs while substituting the values. Also note that since we are differentiating with respect to t ${{r}^{2}}$ becomes an implicit function as r is itself a function of time t.