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 $-24 "mm"^3"/s"$ .