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?

Question

William Abdalla · Accepted Answer

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#

Ezekiel Alexander · Answer

undefined To find the rate at which the sides of the cube are changing when each side is 3 meters long, we can use the formula for the volume of a cube, $V = s^3$, where $V$ is the volume and $s$ is the length of a side.

Differentiating both sides with respect to time $t$, we get:

$\frac{{dV}}{{dt}} = 3s^2 \frac{{ds}}{{dt}}$

Given that $ \frac{{dV}}{{dt}} = -8 \, m^3/min$ and $s = 3 \, m$, we can solve for $\frac{{ds}}{{dt}}$:

$-8 = 3(3^2) \frac{{ds}}{{dt}}$

$-8 = 27 \frac{{ds}}{{dt}}$

$\frac{{ds}}{{dt}} = \frac{{-8}}{{27}} \, m/min$

So, when the sides of the cube are each 3 meters long, they are decreasing at a rate of $\frac{{-8}}{{27}}$ meters per minute.