Question

the volume of a cube decreases at a rate of 0.4 ft³/min. what is the rate of change of the side length when the side lengths are 11 ft? write an equation relating the volume of a cube, v, and an edge of the cube, s.

Explanation:

Step1: Write volume - side length equation

The volume $V$ of a cube with side - length $s$ is given by the formula $V = s^{3}$.

Step2: Differentiate with respect to time $t$

Using the chain - rule, $\frac{dV}{dt}=3s^{2}\frac{ds}{dt}$.

Step3: Solve for $\frac{ds}{dt}$

We can rewrite the equation as $\frac{ds}{dt}=\frac{1}{3s^{2}}\frac{dV}{dt}$.

Step4: Substitute given values

We know that $\frac{dV}{dt}=- 0.4$ (negative because the volume is decreasing) and $s = 11$.
Substituting these values into the equation for $\frac{ds}{dt}$, we get $\frac{ds}{dt}=\frac{-0.4}{3\times(11)^{2}}=\frac{-0.4}{3\times121}=\frac{-0.4}{363}\approx - 0.0011$ ft/min.

Answer:

The equation relating the volume $V$ and the edge $s$ of the cube is $V = s^{3}$, and the rate of change of the side - length when $s = 11$ ft is $\frac{ds}{dt}\approx - 0.0011$ ft/min.