A hypothetical cube shrinks at a rate of 8 m³/min. At what rate are the sides of the cube changing when the sides are 3 m each?

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Answer 1 · 2016-10-12T13:33:06

When the sides are $3$ $m$ long, they are decreasing at a rate of $8/27$ $m$ / $min$ .

Identify the Variables
The units $m^3$ / $min$ are units for volume over time. We are also asked about the sides of the cube. The variables are:

$V$ = the volume of the cube

$x$ = the length of a side of the cube

$t$ = time in minutes

Identify the Rates of Change

The volume of the cube is decreasing at 8 $m^3$ / $min$ , so

$(dV)/dt = -8$ $m^3$ / $min$ ,.

We are asked to find the rate at which the sides are changing, so we want to

find $dx/dt$ when $x = 3$ $m$

Find an Equation Relating the Variables

The volume of a cube is given by the equation

$V = x^3$

Differentiate To find the equation relating the variables and their rates of change.

$(dV)/dt = 3x^2 dx/dt$

Plug in what you know and solve for what you're looking for.

$-8 =3 (3^2) dx/dt$

$27 dx/dt = -8$

$dx/dt = -8/27$

Answer the question

When the sides are $3$ $m$ long, they are decreasing at a rate of $8/27$ $m$ / $min$ .

If you prefer to use units all the way through:

$-8 m^3/min=3 (3m)^2 dx/dt$

$27 m^2 dx/dt = -8 m^3/min$

$dx/dt = -8/27 m/min$