A perfect cube shaped ice cube melts so that the length of its sides are decreasing at a rate of 2 mm/sec. Assume that the block retains its cube shape as it melts. At what rate is the volume of the ice cube changing when the sides are 2 mm each?

Question

Cameron Alexander · Accepted Answer

The volume is decreasing at a rate of #24 "mm"^3"/s"#.

Let the length of the cube be denoted as #l#. The volume of the cube, #V#, is given by #V = l^3# Since the sides are decreasing at a rate of #2"mm/s"#, we write #frac{"d"l}{"d"t} = -2"mm/s"#, where #t# represents time. The negative sign is there as #l# is decreasing with time. To find the rate at which the volume change, #frac{"d"V}{"d"t}#, we can use the chain rule #frac{"d"V}{"d"t} = frac{"d"V}{"d"l} frac{"d"l}{"d"t}# From simple differentiation, we know that #frac{"d"V}{"d"l} = frac{"d"}{"d"l}(l^3) = 3l^2# Therefore, #frac{"d"V}{"d"t} = 3l^2 frac{"d"l}{"d"t}# #= 3 xx (2"mm")^2 xx (-2"mm/s")# #= -24 "mm"^3"/s"# The volume is changing at a rate of #-24 "mm"^3"/s"#.

Christopher Allers · Answer

undefined The rate at which the volume of the ice cube is changing when the sides are 2 mm each is $ -24 \, \text{mm}^3/\text{sec} $.