the edges of a cube increase at a rate of 4 cm/s. how fast is the volume changing when the length of each…

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.)

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.)

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 $4800$ $cm^{3}/s$.