Q.The volume of a cube is increasing at the rate of $8 \text{ cm}^3/\text{s}$. How fast is the surface area increasing when the length of an edge is $12 \text{ cm}$?

Concept: Related Rates — we connect the rate of change of volume to the rate of change of surface area through the edge length.

Let the edge be $x$ cm. Volume $V = x^3$, surface area $S = 6x^2$.

Given $\frac{dV}{dt} = 8$ cm³/s. Differentiate $V$ with respect to $t$:

Now differentiate $S$:

This is a classic related rates problem. The key idea: when one quantity changes with time, other quantities linked to it also change. Here, volume and surface area both depend on the edge length $s$, and we know how fast volume is increasing. We want how fast surface area is increasing at a specific moment.

The chain rule is our tool. If two quantities $A$ and $B$ are both functions of $s$, and $s$ itself changes with time, then $\frac{dA}{dt} = \frac{dA}{ds} \cdot \frac{ds}{dt}$ and similarly for $B$. So we can connect $\frac{dV}{dt}$ to $\frac{dS}{dt}$ through $\frac{ds}{dt}$.

Let’s work through it step by step.

1. Write the formulas.

   Let the edge length be $s$ cm.

   Volume: $V = s^3$

   Surface area: $S = 6s^2$ (a cube has 6 faces, each of area $s^2$).

2. What we know.

   $\frac{dV}{dt} = 8 \text{ cm}^3/\text{s}$ (given, constant rate).

   We want $\frac{dS}{dt}$ when $s = 12$ cm.

3. Differentiate volume with respect to time.

   Using the chain rule:

So $8 = 3s^2 \cdot \frac{ds}{dt}$.

4. Solve for $\frac{ds}{dt}$ at the given edge length. When $s = 12$:

So the edge is growing slowly — about 0.0185 cm each second.

5. Differentiate surface area with respect to time.

6. Plug in $s = 12$ and $\frac{ds}{dt} = \frac{1}{54}$.

Method: Related Rates Through an Intermediate Variable

Use this pattern whenever the GIVEN rate and the WANTED rate are both expressed in terms of a common third variable (here, the edge length) rather than directly in terms of each other.

Steps

Step 1: Express every relevant quantity in terms of the common variable

Let the edge length be $x$. Write down both formulas that involve $x$: the one connected to the given rate (volume $V=x^3$) and the one connected to the wanted rate (surface area $S=6x^2$).

Step 2: Differentiate the KNOWN relation with respect to time to find the intermediate rate

Substitute the given $\frac{dV}{dt}$ and the given value of $x$ to solve for $\frac{dx}{dt}$ — this intermediate rate is not asked for directly, but it's the bridge to the answer.

Step 3: Differentiate the TARGET relation with respect to time

Step 4: Substitute the known $x$ and the $\frac{dx}{dt}$ found in Step 2

Plug both numbers in to get $\frac{dS}{dt}$ in the correct units.

Alternative shortcut: dividing the two differentiated relations directly (eliminating $\frac{dx}{dt}$) gives $\frac{dS/dt}{dV/dt}=\frac{4}{x}$, letting you skip solving for $\frac{dx}{dt}$ explicitly.

Common Mistakes

Mistake 1: Substituting the edge length before differentiating

Plugging $x=12$ into $V=x^3$ to get $V=1728$ and then trying to differentiate that constant gives zero — the formula must be differentiated FIRST (while $x$ is still a variable), and the specific edge length substituted only afterward.

Mistake 2: Using the given rate directly in the surface-area formula

$\frac{dV}{dt}=8$ cm³/s cannot be plugged directly into $\frac{dS}{dt}=12x\frac{dx}{dt}$ — the two formulas share $\frac{dx}{dt}$, not $\frac{dV}{dt}$ itself, so the intermediate step of solving for $\frac{dx}{dt}$ from the volume relation cannot be skipped.

Mistake 3: Mixing up which formula belongs to volume and which to surface area

Confusing $V=x^3$ (so $\frac{dV}{dt}=3x^2\frac{dx}{dt}$) with $S=6x^2$ (so $\frac{dS}{dt}=12x\frac{dx}{dt}$) — writing the wrong power or the wrong constant multiplier changes the answer completely.