A = 4 pi r^2
r = D/2
so
A = 4 pi D^2/4 = pi D^2
dA/dD = 2 pi D
dA/dt = 2 pi dD/dt
1 = 2 pi dD/dt
dD/dt = 1/(2pi)
A spherical snowball melts in such a way that its surface area decreases at a rate of 1 cm2/min. Find the rate at which its diameter is decreasing when the diameter is 10 cm
1 answer