Q.The volume of a cube is increasing at a rate of $9$ cubic centimetres per second. How fast is the surface area increasing when the length of an edge is $10$ centimetres?

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

Given $\frac{dV}{dt} = 9$ cm³/s. Differentiate $V$ with respect to time:

Now differentiate $S$:

This is a classic related rates problem. The key idea: when one quantity (volume) changes at a known rate, and another quantity (surface area) depends on the same variable (edge length), we can connect their rates using the chain rule. We don't need the edge length's rate directly — we find it as a stepping stone.

Let the edge length be $s$ cm, volume $V$ cm³, and surface area $A$ cm². All are functions of time $t$ (seconds).

1. Write the formulas.

   Volume of a cube: $V = s^3$

   Surface area of a cube (six faces): $A = 6s^2$

2. Differentiate both with respect to time $t$.

   Using the chain rule:

Notice that $\frac{ds}{dt}$ appears in both — that's our bridge.

3. Use the given rate to find $\frac{ds}{dt}$. We know $\frac{dV}{dt} = 9$ cm³/s. At the moment of interest, $s = 10$ cm.

4. Now find $\frac{dA}{dt}$ at $s=10$. Substitute $s=10$ and $\frac{ds}{dt}=0.03$ into the surface area rate equation:

So the surface area is increasing at $3.6$ cm²/s.

Method: Chaining Two Related Rates Through a Common Variable

This method solves related-rates problems where the rate you're given and the rate you want both depend on a third, unmentioned variable — here, the cube's edge length — so you first solve for that variable's rate, then use it as a stepping stone.

Steps

Step 1: Introduce the linking variable

Let the edge length be $s$ (a function of time), and write both quantities of interest in terms of it: $V=s^3$ (volume) and $S=6s^2$ (surface area). Neither formula directly relates $V$ and $S$ to each other — they're both functions of the same underlying $s$.

Step 2: Differentiate both formulas with respect to time

Notice $\frac{ds}{dt}$ appears in both — this is the bridge between the given rate and the wanted rate.

Step 3: Use the given rate to solve for the linking rate

Substitute the known $\frac{dV}{dt}$ and the given instantaneous value of $s$ into the first equation, and solve for $\frac{ds}{dt}$.

Step 4: Substitute the linking rate into the second equation

Plug the same instantaneous $s$ and the just-found $\frac{ds}{dt}$ into $\frac{dS}{dt}=12s\frac{ds}{dt}$ to get the answer.

Step 5: State the answer with correct units and sign

Check whether the surface area is increasing or decreasing (sign of $\frac{dS}{dt}$) and attach the correct area-per-time unit.

Whenever two quantities don't have a direct formula linking them but both depend on a shared third variable, this "solve for the linking rate first, then substitute into the second relation" approach is the standard way through — it generalises beyond cubes to any shape where volume and surface area (or two other quantities) are both functions of one common length.

Common Mistakes

Mistake 1: Assuming $\frac{ds}{dt}$ is a fixed constant, valid at every edge length

After finding $\frac{ds}{dt}=0.03$ cm/s at $s=10$, a student may reuse that same value at a different edge length without recomputing. Why it's wrong: $\frac{ds}{dt}$ depends on $s$ through $\frac{dV}{dt}=3s^2\frac{ds}{dt}$ (since $\frac{dV}{dt}$ is fixed but $s^2$ isn't), so it changes as the cube grows — it is only valid at the specific instant $s=10$. Correct approach: always recompute $\frac{ds}{dt}$ from the given $\frac{dV}{dt}$ at the exact edge length asked about.

Mistake 2: Trying to relate $V$ and $S$ directly without going through $s$

A student might attempt $\frac{dS}{dt} = k\,\frac{dV}{dt}$ for some guessed constant $k$, skipping the edge-length variable altogether. Why it's wrong: $S$ and $V$ are related non-linearly ($S=6V^{2/3}$), so there's no single constant multiplier between their rates — the relationship changes with $s$, and only differentiating each with respect to time through the shared variable $s$ gives a correct, instant-specific answer. Correct approach: always introduce the shape's defining linear dimension as the linking variable rather than guessing a shortcut between the two given quantities.