cube side length increasing at 1/3 inch per second volume rate

Expert intro: This is a standard related rates problem. The key is to write volume in terms of side length, differentiate with respect to time, and then substitute the given values.

Detailed walkthrough

Let the side length of the cube be $s(t)$ and the volume be $V(t)$.

We know

$V=s^3.$

Given:

$\frac{ds}{dt}=\frac13 \text{ inch/sec}, \qquad s=2 \text{ inches}.$

Step 1: Differentiate the volume formula

$\frac{dV}{dt}=3s^2\frac{ds}{dt}.$

Step 2: Substitute the given values

$\frac{dV}{dt}=3(2^2)\left(\frac13\right).$

$\frac{dV}{dt}=3\cdot 4\cdot \frac13=4.$

Answer

The volume is increasing at

$\boxed{4\text{ cubic inches per second}}.$

💡 Pitfall guide

A frequent error is to differentiate $V=s^3$ and then forget to multiply by $\frac{ds}{dt}$. In related rates, every variable depending on time needs chain rule.

🔄 Real-world variant

If the side length were changing at a different rate, say $\frac{ds}{dt}=k$, then the volume rate at $s=2$ would be

$\frac{dV}{dt}=3(2^2)k=12k.$

So the setup stays the same; only the rate changes.

🔍 Related terms

related rates, chain rule, volume of a cube