Question

Suppose a spherical balloon is being inflated, and the surface area of the balloon is increasing at a rate of 0.1 cm2/min when the diameter is 10 cm. At what rate is the volume of the balloon increasing when the diameter is 10 cm?

Answer 1

Answer:

$\frac{dV}{dt}=0.5 cm^3/min$

Step-by-step explanation:

We are given that

$\frac{dS}{dt}=0.1 cm62/min$

Diameter=d=10 cm

Radius,r=$\frac{d}{2}=\frac{10}{2}=5 cm$

We have to find the volume of the balloon increasing when the diameter is 10 cm.

Area of balloon=$S=4\pi r^2$

Differentiate w.r.t t

$\frac{dS}{dt}=8\pi r\frac{dr}{dt}$

Substitute the values

$0.1=8\pi (10)\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{0.1}{80\pi}$ cm/min

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

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=4\pi(10)^2\times \frac{0.1}{80\pi}=0.5 cm^3/min$

Hence, $\frac{dV}{dt}=0.5 cm^3/min$