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?

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Answer 1 · 2016-08-20T10:11:27

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"#.