Expert: Abe Mantell - 7/3/2011

Question
A cubical box with sides of length 15cm is covered with ice of uniform thickness. The ice is melting at the rate of 80 cm/hr. How fast is the thickness changing when the ice is 1 cm thick?

Answer
Hello Dana,

Let x=the thickness of the ice formation.
Then, the volume of ice is the volume of a cube
with sides 15+2x minus the volume of a cube with
sides 15...V=(15+2x)^3-15^3, which reduces to
V=1350x+180x^2+8x^3, now differentiate with respect
to t (time):
dV/dt=1350(dx/dt)+360x(dx/dt)+24x^2(dx/dt)
Solving for dx/dt, gives:
dV/dt=[1350+360x+24x^2](dx/dt), so
dx/dt=(dV/dt)/[1350+360x+24x^2]

Now substitute dV/dt=-80 cm^3/hr (units sould be cubic cm per hr)
and 1 cm for x...which gives:
dx/dt=-80/[1350+360+24]
= -80/1734 or approx. -0.046136 cm/hr (the negative indicates the
thickness is decreasing)

OK?

Abe