Answer:
[tex]\dfrac{dV}{dt} =424.12\ cm^3/min[/tex]
Explanation:
given,
radius increasing rate = 0.6 cm/minute
the volume changing when the radius = 7.5 cm
volume of the sphere= [tex]\dfrac{4}{3}\pi r^3[/tex]
[tex]\dfrac{dr}{dt} = 0.6 cm/min[/tex]
[tex]V =\dfrac{4}{3}\pi r^3[/tex]
[tex]\dfrac{dV}{dt} =\dfrac{4}{3}\pi (3r^2)\dfrac{dr}{dt}[/tex]
[tex]\dfrac{dV}{dt} =4\pi r^2\dfrac{dr}{dt}[/tex]
[tex]\dfrac{dV}{dt} =4\pi 7.5^2\times 0.6[/tex]
[tex]\dfrac{dV}{dt} =424.12\ cm^3/min[/tex]
