a spherical snowball is melting symmetrically at the rate of 4pi cubic cm per hr. how fast is the diameter changing when it is 20 cm?

V = (4/3)πr³ 

dV/dt = 4πr².dr/dt

so, when the diameter is 20 cm we have:

-4π = 4πr².dr/dt

Hence, dr/dt = -1/r²

i.e. dr/dt = -1/10² => -1/100

Now, d = 2r

so, d(d)/dt = 2.dr/dt

=> d(d)/dt = -1/50 cm/hour

or, -0.2 mm/hour

:)>

r: radius  and  d: diammeter
Volume, V = (4/3)πr³   ⇒   V = (4/3)π(d/2)³   ⇒   V = (1/6)πd³

dV/dt = d[(1/6)πd³]/dt
dV/dt = (1/2)πd²(dd/dt)

Now, dV/dt = -4π cm³/hr  and  d = 20 cm:
-4π =(1/2)π(20)²(dd/dt)
dd/dt = -0.02 cm/hr

The diameter DECREASES at a rate of 0.02 cm/hr = 0.2 mm/hr

V = (4/3)πr^3 = (π/6)D^3
dV/dt = dV/dD * dD/dt
-4π = (π/2)D^2* dD/dt, (where V is in cubic cm)
dD/dt = -8/D^2, and when D = 20
dD/dt = -8/400 = -1/50 cm/hour
dD/dt = - 0.2 mm/hour