Posted by **W** on Tuesday, February 7, 2012 at 11:50pm.

A block of ice is exposed to heat in such a way that the block maintains a similar shape as it metls. The block of ice is initially 2' wide, 2 'high, and 3' long. If the change in the width of the ice is -1/3 ft/hr, find: the amount of time it will tak efor the block of the ice to completely melt.

I found that the rate of change in the volume of the ice block is -3/2 ft^3/hr. Do you just divide the volume of the cube (12 ft^3) by the rate of change of the volume (-3/2 ft^3/hr)?

- Calculus -
**Anonymous**, Wednesday, February 8, 2012 at 11:04am
v = w*w*3w/2 = 3/2 w^3

dv/dt = 9/2 w^2 dw/dt

= 9/2 * -1/3 = -3/2

your dv/dt is correct.

as for your question, consider the units:

time = ft^3 / ft^3/hr = 12/(3/2) = 8 hr

- Calculus -
**W**, Thursday, February 9, 2012 at 10:08pm
Thank you! That was helpful :)

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