Question

the edges of a cube increase at a rate of 4 cm/s. how fast is the volume changing when the length of each edge is 20 cm? write an equation relating the volume of a cube, v, and an edge of the cube, a. differentiate both sides of the equation with respect to t. the rate of change of the volume is (simplify your answer.)

Explanation:

Step1: Write volume formula

The volume formula of a cube is $V = a^{3}$.

Step2: Differentiate with respect to time

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

Step3: Substitute given values

We know that $\frac{da}{dt}=4$ cm/s and $a = 20$ cm. Substitute these values into the derivative formula: $\frac{dV}{dt}=3\times(20)^{2}\times4$.

Step4: Calculate the result

$\frac{dV}{dt}=3\times400\times4=4800$ $cm^{3}/s$.

Answer:

The equation relating volume $V$ and edge $a$ is $V = a^{3}$.
Differentiating with respect to $t$ gives $\frac{dV}{dt}=3a^{2}\frac{da}{dt}$.
The rate of change of the volume is $4800$ $cm^{3}/s$.