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 8 ft? write an equation relating the volume of a cube, v, and an edge of the cube, s.

Explanation:

Step1: Recall volume formula

The volume formula of a cube is $V = s^{3}$, where $V$ is the volume and $s$ is the side - length.

Step2: Differentiate with respect to time

Differentiate both sides of $V = s^{3}$ with respect to time $t$ using the chain - rule. $\frac{dV}{dt}=3s^{2}\frac{ds}{dt}$.

Step3: Substitute given values

We know that $\frac{dV}{dt}=- 0.4$ (negative because the volume is decreasing) and $s = 8$. Substitute these values into the equation $\frac{dV}{dt}=3s^{2}\frac{ds}{dt}$:
$-0.4=3\times8^{2}\times\frac{ds}{dt}$.
$-0.4 = 3\times64\times\frac{ds}{dt}$.
$-0.4=192\times\frac{ds}{dt}$.

Step4: Solve for $\frac{ds}{dt}$

$\frac{ds}{dt}=\frac{-0.4}{192}=-\frac{1}{480}\text{ ft/min}\approx - 0.002083\text{ ft/min}$.

Answer:

$-\frac{1}{480}\text{ ft/min}$