Hope this helps! :)
SA = 4πr^2
d(SA)/dt = 8πr dr/dt
subbing in our values
d(SA)/dt = 8π(3)(-2) = -48π
The radius of a spherical snowball is decreasing at a rate of 2 inches per hour. How fast is the snowball melting in terms of its volume when its diameter is 8 inches?
2 answers
It asked for the rate of change in terms of its VOLUME.
V = (4/3)π r^3
dV/dt = 4π r^2 dr/dt
= 4π (16)(-2)
= -128π cubic inches
V = (4/3)π r^3
dV/dt = 4π r^2 dr/dt
= 4π (16)(-2)
= -128π cubic inches
1 answer