Question

The radius of a right circular cylinder increases at a constant rate. Its altitude is a linear function of the radius and increases three times as fast as radius$.$When the radius is $1cm$ the altitude is $6cm.$ When the radius is $6cm,$ then volume is increasing at the rate of $1c{m}^3/sec.$ When the radius is $36cm,$ the volume is increasing at a rate of $nc{m}^3/sec.$ The value of $n$ is equal to$:$

• A. $12$
• B. $22$
• C. $30$
• D. $33$

Answer 1

The correct option is D

$33$

It is given that radius increases at constant rate$.$

So $\frac{dr}{dt}=k$

The altitude is a linear function of the radius and increases three times as fast as radius$.$

$\frac{dh}{dt}=3\frac{dr}{dt}$

integrate both sides

$h=3r+c$

Given at $r=1,h=6$

$6=3+c$

$c=3$

So $h=3r+3$

Volume of right circular cylinder $=\pi r^{2}h$

$=\pi r^2(3r+3)$

$=3\pi (r^3+r^2)$

$\frac{dV}{dt}=3\mathrm{\pi }\left(3{\mathrm{r}}^2\frac{\mathrm{dr}}{\mathrm{dt}}+2\mathrm{r}\frac{\mathrm{dr}}{\mathrm{dt}}\right)$$.......\left(1\right)$

At $r=6,\frac{dV}{dt}=1$

$1=3\mathrm{\pi }\left(3\times 6^2\frac{\mathrm{dr}}{\mathrm{dt}}+2\times 6\frac{\mathrm{dr}}{\mathrm{dt}}\right)$

$1=3\mathrm{\pi }\left(108\frac{\mathrm{dr}}{\mathrm{dt}}+12\frac{\mathrm{dr}}{\mathrm{dt}}\right)$

$1=360\mathrm{\pi }\frac{\mathrm{dr}}{\mathrm{dt}}$

$\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{360\mathrm{\pi }}$

At $r=36,$

$\frac{dV}{dt}=3\pi (3\times {36}^2\frac{dr}{dt}+2\times 36\frac{dr}{dt})$

$=3\pi \times 3960\times \frac{1}{360\mathrm{\pi }}$

$=33$

So, the value of $n$is $33$

Hence, option $\left(D\right)$ is correct.