Question

the volume of a cube decreases at a rate of 0.2 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: Recall volume formula for cube

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

Step2: Differentiate with respect to time $t$

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

Step3: Substitute given values

We know that $\frac{dV}{dt}=- 0.2$ (negative because the volume is decreasing) and $s = 11$. Substitute these values into the equation $\frac{dV}{dt}=3s^{2}\frac{ds}{dt}$:
$-0.2=3\times(11)^{2}\times\frac{ds}{dt}$.

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

First, calculate $3\times(11)^{2}=3\times121 = 363$. Then, $\frac{ds}{dt}=\frac{-0.2}{363}\approx - 0.00055$ ft/min.

Answer:

$\frac{ds}{dt}\approx - 0.00055$ ft/min