Q.The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.

Concept: Related Rates — we connect the rate of change of volume to the rate of change of surface area using the side length as the intermediate variable.

Let the side of the cube be $x$, volume $V = x^3$, and surface area $S = 6x^2$.

Given $\frac{dV}{dt} = k$ (constant). Differentiate $V$:

Now differentiate $S$:

$\frac{dS}{dt} = 12x \frac{dx}{dt} = 12x \cdot \frac{k}{3x^2} = \frac{4k}{x}.$ …

Why This Works: The Idea of Related Rates

When a quantity changes with time, any other quantity linked to it by geometry also changes. Here, the cube's volume $V$ grows at a fixed rate $\frac{dV}{dt} = k$ (a constant). The surface area $S$ depends on the side length $a$, which itself changes because volume is increasing. The chain rule lets us connect $\frac{dS}{dt}$ to $\frac{dV}{dt}$ through $a$.

The key insight: as the cube gets larger, the same increase in volume produces a smaller increase in surface area per unit time. That's the "inverse" relationship we need to prove.

Step-by-Step Proof

1. Write the geometric formulas.

For a cube of side length $a$:

• Volume: $V = a^3$
• Surface area: $S = 6a^2$

2. Express the given condition.

The volume increases at a constant rate:

3. Differentiate the volume relation with respect to time.

Using the chain rule:

So:

This tells us how fast the side length itself grows — slower for larger cubes.

4. Differentiate the surface area relation.

Similarly:

5. Substitute $\frac{da}{dt}$ from step 3.

Since $k$ is constant, $\frac{dS}{dt}$ is proportional to $\frac{1}{a}$.

6. Interpret the result. …

Method: Related Rates for a Solid — Eliminating the Shared Constant to Prove a Variation Law

Just as a growing 2-D shape has its area and perimeter both governed by one changing length, a growing 3-D solid has its volume and surface area both governed by one changing dimension — the proof technique is the same: express both quantities in terms of that one variable, then eliminate the variable's own (unknown) rate.

Steps

Step 1: Pick the single governing variable.

For a solid like a cube, sphere, or cylinder, identify the one length (side, radius, etc.) that every other measurement — volume, surface area — is built from.

Step 2: Write both quantities of interest as powers of that variable.

For a cube of side $a$: $V=a^3$ and $S=6a^2$. The volume formula is degree $3$ and the surface-area formula is degree $2$ in the same variable — this mismatch in degree is exactly what produces the "varies as" relationship you're asked to prove.

Step 3: Differentiate the volume relation and isolate the governing variable's own rate. …

Common Mistakes

Mistake 1: Trying to relate $\frac{dS}{dt}$ to $\frac{dV}{dt}$ directly, skipping the side length $x$

Why it's wrong: both $V=x^3$ and $S=6x^2$ depend on the common variable $x$, and the chain-rule link between $\frac{dS}{dt}$ and $\frac{dV}{dt}$ only exists through $\frac{dx}{dt}$ — there is no direct formula connecting $V$ and $S$'s rates without passing through $x$. Correct approach: find $\frac{dx}{dt}$ from $\frac{dV}{dt}=k$ first, then substitute it into $\frac{dS}{dt}=12x\frac{dx}{dt}$.

Mistake 2: Treating the side length $x$ as constant while differentiating …