Answer:
[tex]\dfrac{dA}{dt}=25\pi[/tex]
Explanation:
given,
rate of increase of volume of sphere = 25π m³/h
volume at certain time = [tex]\dfrac{32\pi}{3}[/tex]
calculate the rate of change of surface area = ?
volume of sphere
[tex]V = \dfrac{4}{3}\pi r^3[/tex]
[tex]\dfrac{32\pi}{3} = \dfrac{4}{3}\pi r^3[/tex]
r³ = 8
r = 2 m
surface are of the sphere
A = 4 π r²
[tex]\dfrac{dA}{dr}=\dfrac{d}{dr}(4\pi r^2)[/tex]
[tex]\dfrac{dA}{dr}=8\pi r[/tex]
[tex]\dfrac{dV}{dr}=\dfrac{d}{dr}(\dfrac{4}{3}\pi r^3)[/tex]
[tex]\dfrac{dA}{dr}=4\pi r^2[/tex]
now,
[tex]\dfrac{dA}{dt}=\dfrac{dA}{dr}\dfrac{dr}{dV}\dfrac{dV}{dt}[/tex]
[tex]\dfrac{dA}{dt}=8\pi r \dfrac{1}{4\pi r^2}\dfrac{dV}{dt}[/tex]
[tex]\dfrac{dA}{dt}=\dfrac{2}{r}\dfrac{dV}{dt}[/tex]
[tex]\dfrac{dA}{dt}=\dfrac{2}{2}\times (25\pi)[/tex]
[tex]\dfrac{dA}{dt}=25\pi[/tex]
