Question

The radius of a cylinder is increasing at a rate of $3$ $\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$ and its altitude is decreasing at a rate of $4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$.

The rate of change of volume when radius is $4m$ and altitude is $6m$ is

• A. $80\mathrm{\pi }\raisebox{1ex}{${\mathrm{m}}^3$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.$
• B. $144\mathrm{\pi }\raisebox{1ex}{${\mathrm{m}}^3$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.$
• C. $80\raisebox{1ex}{${\mathrm{m}}^3$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.$
• D. $60\raisebox{1ex}{${\mathrm{m}}^3$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.$

Answer 1

The correct option is A

$80\mathrm{\pi }\raisebox{1ex}{${\mathrm{m}}^3$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.$

Explanation for the correct option:

Given that,

• The radius of a cylinder is increasing at a rate of $3$ $\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$ $\Rightarrow \frac{dr}{dt}=$ $3$ $\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$
• Altitude is decreasing at a rate of $4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$$\Rightarrow \frac{dh}{dt}=-$$4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$
• Radius is $r=4m$ and altitude is $h=6m$

We know that Volume of cylinder is given by $V={\mathrm{\pi r}}^2\mathrm{h}$

Differentiating both sides with respect to $t$,

$\begin{array}{crcll}\Rightarrow & \frac{dV}{dt}& =& \frac{d}{dt}\left({\mathrm{\pi r}}^2\mathrm{h}\right)& \\ & & =& \mathrm{\pi }\left[\mathrm{h}\left(\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{r}}^2\right)+{\mathrm{r}}^2\frac{\mathrm{dh}}{\mathrm{dt}}\right]& \left[\because \frac{d}{dx}\left(U·V\right)=V\frac{dU}{dx}+U\frac{dV}{dx}\right]\\ \Rightarrow & \frac{dV}{dt}& =& \mathrm{\pi }\left[2\mathrm{rh}\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)+{\mathrm{r}}^2\frac{\mathrm{dh}}{\mathrm{dt}}\right]& \left[\because \frac{d}{dt}{x}^n=n{x}^{n-1}\frac{dx}{dt}\right]\\ \Rightarrow & {\left[\frac{dV}{dt}\right]}_{r=4,h=6}& =& \mathrm{\pi }\left[2\left(4\right)\left(6\right)\left(3\right)+{\left(4\right)}^2\left(-4\right)\right]& \left[\because \frac{dr}{dt}=3,\frac{dh}{dt}=-4\right]\\ & & =& \mathrm{\pi }\left[144-64\right]& \\ \Rightarrow & {\left[\frac{dV}{dt}\right]}_{r=4,h=6}& =& 80\mathrm{\pi }\raisebox{1ex}{${\mathrm{m}}^3$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.& \end{array}$

Hence, the correct option is option $\left(A\right)$ i.e. $80\mathrm{\pi }\raisebox{1ex}{${\mathrm{m}}^3$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.$