Question

the edges of a cube increase at a rate of 5 cm/s. how fast is the volume changing when the length of each edge is 40 cm? write an equation relating the volume of a cube, v, and an edge of the cube, a.

Explanation:

Step1: Write volume - edge equation

The volume of a cube is given by the formula $V = a^{3}$, where $V$ is the volume and $a$ is the length of an edge.

Step2: Differentiate with respect to time

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

Step3: Substitute given values

We know that $\frac{da}{dt}=5$ cm/s and $a = 40$ cm. Substitute these values into the equation $\frac{dV}{dt}=3a^{2}\frac{da}{dt}$.
$\frac{dV}{dt}=3\times(40)^{2}\times5$.
First, calculate $(40)^{2}=1600$. Then, $3\times1600\times5 = 24000$ $cm^{3}/s$.

Answer:

The equation relating the volume of a cube $V$ and an edge of the cube $a$ is $V=a^{3}$, and the rate of change of the volume when $a = 40$ cm is $24000$ $cm^{3}/s$.