I have little time more before I have to go for work. I just want to practice on this. I'm groping right now, rusty.
To find: rate of decrease of diameter
So dD/dt, where D means diameter
Sphere:
V = (4/3)(pi)(r^3)
in terms of D,
V = (4/3)(pi)[(D/2)^3]
V = (1/6)(pi)(D^3)
Differentiate both sides with respect to time t,
dV/dt = (pi/6)[3 D^2 dD/dt]
dV/dt = (pi/2)(D^2) dD/dt ----------**
So when D = 10cm, and since dV/dt = -1 cu.cm,
-1 = (pi/2)(10^2) dD/dt
-1 = 50pi dD/dt
dD/dt = -1/(50pi)
Therefore, when the diameter is 10cm, the diameter is decreasing at the rate of 1/(50pi) cm/sec. ---------------answer.
No time to review. If there are mistakes, I'll rectify after work---about 11 hrs from now.