Question

The radius of a circular soap bubble is increasing at the rate of 0.2 cm . Find the rate of increase of its volume when the radius is 4cm.

Answer 1

Answer:

The rate of change of volume increasing is $12.8\pi$ cubic cm / min

Step-by-step explanation:

Let r and V be the radius and volume of the spherical soap bubble at time t.

Given:

$\frac{dr}{dt}=0.2\:cm/min$

Volume of the spherical bubble

$V=\frac{4}{3}*\pi*r^3$

Differentiate with respect to t

$\frac{dV}{dt}=\frac{4}{3}*\pi*3r^2*\frac{dr}{dt}$

$\frac{dV}{dt}=\frac{4}{3}*\pi*3(4)^2*\frac{dr}{dt}$

$\frac{dV}{dt}=4*\pi*(4)^2*(0.2)$

$\frac{dV}{dt}=4*\pi*16*(0.2)$

$\frac{dV}{dt}=64\pi*(0.2)$

$\implies\:\frac{dV}{dt}=12.8\pi\:cm^3/min$

Answer 2

The rate of increase of its volume when the radius is 4 cm is $12.8\pi\ cm^3/sec$

Explanation:

Let r be the radius , t be time and V be the volume of the  circular soap bubble.

Given : Rate of increasing the radius of a circular soap bubble: $\dfrac{dr}{dt}= 0.2 cm/ sec$

Volume of sphere = $V=\dfrac{4}{3}\pi r^3$

Differentiate both sides with respect to x , we get

$\dfrac{dV}{dt}=\dfrac{4}{3}\pi(3r^2)\dfrac{dr}{dt}$

$\dfrac{dV}{dt}=4\pi(r^2)\dfrac{dr}{dt}$

Put value of $\dfrac{dr}{dt}= 0.2 cm/ sec$ and r= 4 , we get

$\dfrac{dV}{dt}=4\pi(4)^2(0.2)=12.8\pi$

Hence, the rate of increase of its volume when the radius is 4 cm is $12.8\pi\ cm^3/sec$

# Learn more :

The radius of an air bubble is increasing at the rate of 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

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