Question
A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is $10 \ cm.$

Answer

Since, $V = \frac{4}{3}\pi {x^3}$
$\therefore \frac{{dV}}{{dx}} = \frac{d}{{dt}}\left( {\frac{4}{3}\pi {r^3}} \right)$
$= \frac{4}{3}\pi .3{r^2} = 4\pi {r^2}$
At $x = 10 \ cm$
$\Rightarrow \frac{{dV}}{{dx}} = 4\pi {\left( {10} \right)^2} = 400\pi $
Therefore, the volume is increasing at the rate of $400\pi \ cm^3/\sec.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free