A cubical block of ice with edges 20cm long starts to melt . The edges decrease at 2.5cm/hr. What is the rate of change of volume of the block two hours later?

Question

Calculus 12 word problem im stuck on

Answer 1

Hello Jessie,

This is a problem involving implicit differentiation. First we need a formula to work from. We have a cubical block and are concerned about volume. The volume of a cube is

V = x^3 where x is the length of each side.

Next, we take the derivative with respect to time t

dV/dt = 3x^2 dx/dt

Last, we are told dx/dt = -2.5 cm/hr. (Note that the derivative is negative because the edges decrease, not increase.) After two hours, the edges have decreased from 20 cm long to 20 - 2.5(2) = 15 cm long. So x = 15.

Just plug in these values for your solution

dV/dt = 3(15)^2 (-2.5) = -1,687.5

So, the ice's volume is decreasing at 1,687.5 cubic cm/hr

Answer 2

Let x = length of a side of the ice block at time t and let V = volume at time t.

Given: dx/dt = -2.5 cm/hr. Find dV/dt when t = 2 hours

V = x³

dV/dt = 3x²(dx/dt) When t = 2, x = 20 - 2(2.5) = 15 cm

So, dV/dt = 3(15cm)²(-2.5 cm/hr) = -1687.5 cm³/hr

When t = 2 hr, the volume of the ice block is decreasing at the rate of 1687.5 cm³/hr.