Question

The radius of a soap bubble is increasing at the rate of 0.2 cm/sec. At what rate the surface area of a bubble increasing when the radius is 7 cm and at what rate the volume of a bubble increasing when the radius is 5 cm ?

Answer 1

Bubble is spherical.

$\left(I\right)\Rightarrow$ surface area $=4\pi r^2=5$

Rate of change is given by :

$\frac{ds}{dt}=4\pi \left(2r\right)\frac{dr}{dt}$

at $r=7cm$

$\frac{dr}{dt}=0.2cm/se$

$\frac{ds}{dt}=8\times \frac{22}{7}\times 7\times 0.2$

$=8\times 22\times \frac{2}{10}=\frac{352}{10}$

$=35.2c{m}^2/sec.$

$\left(II\right)Volume=v=\frac{4}{3}\pi r^3$

Rate of change is :

$\frac{du}{dt}=\frac{4}{3}\pi \left(3r^2\right)\frac{dr}{dt}$

$=4\pi r^2\left(\frac{dr}{dt}\right)$

at $r=5cm,\frac{dr}{dt}=0.2cm/sec$

$\Rightarrow \frac{dr}{dt}=4\times \frac{22}{7}\times 25\times 0.2$

$=62.857c{m}^3/sec.$