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 $1\text{cm},$ the altitude is $6\text{cm}.$ When the radius is $6\text{cm},$ the volume is increasing at the rate of $1\text{cu. cm/sec}.$ When the radius is $36\text{cm},$ then the rate of increase of the volume $\left(\text{in cu. cm/sec}\right)$ is

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

Answer 1

The correct option is D $33$

Let the radius, altitude, volume be $r,h,V$ respectively.
$\frac{dr}{dt}=c$ (constant) and $h=ar+b$
Also, $\frac{dh}{dt}=3\frac{dr}{dt}$
$\therefore a\frac{dr}{dt}=3\frac{dr}{dt}\Rightarrow a=3$
Hence, $h=3r+b$

When $r=1;h=6$
$\Rightarrow 6=3+b\Rightarrow b=3$
$\therefore h=3(r+1)$

$V=\pi r^{2}h=3\pi r^2(r+1)=3\pi (r^3+r^2)$
$\frac{dV}{dt}=3\pi (3r^2+2r)\frac{dr}{dt}$
When $r=6;\frac{dV}{dt}=1$ cu. cm/sec
$\therefore 1=3\pi (108+12)\frac{dr}{dt}$
$\Rightarrow 360\pi \frac{dr}{dt}=1$
$\Rightarrow \frac{dr}{dt}=\frac{1}{360\pi }$

Again when $r=36,$
$\frac{dV}{dt}=3\pi ({(3\times 36)}^2+2\times 36)\frac{dr}{dt}$
$=3\pi \times 36\left(110\right)\times \frac{1}{360\pi }$
$=33$ cu. cm/sec