Calculus - rates of change

Question

A cube of ice is melting at a uniform rate. The initial volume of the cube is 20 cm^3, and the volume after five minutes is 15 cm^3. Find the rate at which the edge of the cube is decreasing after two minutes.

Answer: 0.04853 cm/min

Answer 1

V = a³

 

dV/dt = 3a² da/dt

 

da/dt = (dV/dt) / (3a²)

 

dV/dt = constant = (15 - 20)/5 cm³/min = -1 cm³/min

 

Uniform rate of change in V means that

 

V(t) = V₀ + (dV/dt)t = 20 - t,

 

with t in minutes, V in cm³.

 

V(2) = 20 - 2 = 18 cm³

 

a(2) = [V(2)]^1/3 = 18^1/3 ≅ 2.621 cm

 

At t = 2 min,

 

da/dt = (dV/dt) / [3a(2)]² = -1/[3(2.621)²] ≅ - 0.0485 cm/min