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

`(dV)/(dt)=0.5 cm^3/min`

Explanation:

We are given that

`(dS)/(dt)=0.1 cm62/min`

Diameter=d=10 cm

Radius,r=
`(d)/(2)=(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

`(dS)/(dt)=8\pi r(dr)/(dt)`

Substitute the values

`0.1=8\pi (10)(dr)/(dt)`

`(dr)/(dt)=(0.1)/(80\pi)` cm/min

Volume of sphere,V=
`(4)/(3)\pi r^3`

`(dV)/(dt)=4\pi r^2(dr)/(dt)=4\pi(10)^2* (0.1)/(80\pi)=0.5 cm^3/min`

Hence,
`(dV)/(dt)=0.5 cm^3/min`