Question

The volume of a sphere is increasing at a rate of 25\pi25π25, pi cubic meters per hour. At a certain instant, the volume is \dfrac{32\pi}{3} 3 32π ​ start fraction, 32, pi, divided by, 3, end fraction cubic meters.

Answer 1

`(dA)/(dt)=25\pi`

Step-by-step explanation:

given,

rate of increase of volume of sphere = 25π m³/h

volume at certain time =
`(32\pi)/(3)`

calculate the rate of change of surface area = ?

volume of sphere

`V = (4)/(3)\pi r^3`

`(32\pi)/(3) = (4)/(3)\pi r^3`

r³ = 8

r = 2 m

surface are of the sphere

A = 4 π r²

`(dA)/(dr)=(d)/(dr)(4\pi r^2)`

`(dA)/(dr)=8\pi r`

`(dV)/(dr)=(d)/(dr)((4)/(3)\pi r^3)`

`(dA)/(dr)=4\pi r^2`

now,

`(dA)/(dt)=(dA)/(dr)(dr)/(dV)(dV)/(dt)`

`(dA)/(dt)=8\pi r (1)/(4\pi r^2)(dV)/(dt)`

`(dA)/(dt)=(2)/(r)(dV)/(dt)`

`(dA)/(dt)=(2)/(2)* (25\pi)`

`(dA)/(dt)=25\pi`