Answer:
[tex]\frac{dV}{dt}[/tex] = 1017.87 in³/min
[tex]\frac{dV}{dt}[/tex] = 17203.35 in³/min
Step-by-step explanation:
given data
radius r of a sphere is increasing at a rate = 3 inches per minute
[tex]\frac{dr}{dt}[/tex] = 3
solution
we know volume of sphere is V = [tex]\frac{4}{3} \pi r^3[/tex]
so [tex]\frac{dV}{dt} = \frac{4}{3} \pi r^2 \frac{dr}{dt}[/tex]
and when r = 9
so rate of change of the volume will be
rate of change of the volume [tex]\frac{dV}{dt} = \frac{4}{3} \pi (9)^2 (3)[/tex]
[tex]\frac{dV}{dt}[/tex] = 1017.87 in³/min
and
when r = 37 inches
so rate of change of the volume will be
rate of change of the volume [tex]\frac{dV}{dt} = \frac{4}{3} \pi (37)^2 (3)[/tex]
[tex]\frac{dV}{dt}[/tex] = 17203.35 in³/min
