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

Answer:

$\dfrac{dA}{dt}=25\pi$

Explanation:

given,

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

volume at certain time = $\dfrac{32\pi}{3}$

calculate the rate of change of surface area = ?

volume of sphere

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

$\dfrac{32\pi}{3} = \dfrac{4}{3}\pi r^3$

    r³ = 8

    r = 2 m

surface are of the sphere

 A = 4 π r²

$\dfrac{dA}{dr}=\dfrac{d}{dr}(4\pi r^2)$

$\dfrac{dA}{dr}=8\pi r$

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

$\dfrac{dA}{dr}=4\pi r^2$

now,

$\dfrac{dA}{dt}=\dfrac{dA}{dr}\dfrac{dr}{dV}\dfrac{dV}{dt}$

$\dfrac{dA}{dt}=8\pi r \dfrac{1}{4\pi r^2}\dfrac{dV}{dt}$

$\dfrac{dA}{dt}=\dfrac{2}{r}\dfrac{dV}{dt}$

$\dfrac{dA}{dt}=\dfrac{2}{2}\times (25\pi)$

$\dfrac{dA}{dt}=25\pi$