Q.A balloon, which always remains spherical, has a variable diameter $\frac{3}{2} (2x+1)$. Find the rate of change of its volume with respect to $x$.

Concept: Related Rates — we differentiate the volume formula with respect to $x$, using the given diameter to express the radius.

Step 1: Radius $r$ is half the diameter:

Step 2: Volume of a sphere:

Step 3: Simplify and differentiate with respect to $x$: …

This is a classic related rates problem, but with a twist: instead of time, the independent variable is $x$. The core idea is simple — when one quantity (here, the diameter) changes with $x$, anything that depends on it (like volume) also changes. The chain rule is our bridge.

Let’s walk through it.

1. Write what’s given. The diameter $D$ is:

Since the balloon is spherical, the radius $r$ is half the diameter:

2. Recall the volume of a sphere.

This is the fundamental relation. Our goal is $\frac{dV}{dx}$, not $\frac{dV}{dr}$ — so we need to connect $V$ to $x$ through $r$.

3. Differentiate using the chain rule. We have $V$ as a function of $r$, and $r$ as a function of $x$. So:

First, $\frac{dV}{dr}$:

Next, $\frac{dr}{dx}$:

4. Multiply them.

But $r$ is still in terms of $x$ — we must substitute back:

So: …

Method: Two-Link Chain Rule (Volume Depends on Radius, Radius Depends on $x$)

This method applies whenever the quantity you must differentiate (like volume) is not given directly as a function of the target variable (like $x$), but only indirectly, through an intermediate variable (like radius or diameter).

Steps

Step 1: Express the intermediate variable in terms of the target variable

If the diameter $D$ is given as a function of $x$, first get the radius: $r = \frac{D}{2}$, still written in terms of $x$.

Step 2: Write the volume formula in terms of the intermediate variable

Step 3: Apply the chain rule through the intermediate variable

Since $V$ depends on $r$, and $r$ depends on $x$, the chain rule gives

Compute $\frac{dV}{dr}$ and $\frac{dr}{dx}$ separately, then multiply.

Step 4: Substitute $r$ (still in terms of $x$) back into the result

The product from Step 3 will contain $r^2$ — replace this with the expression for $r$ from Step 1, so the final derivative is entirely in terms of $x$, matching what the question asked for. …

Common Mistakes

Mistake 1: Using the diameter formula directly as the radius

Why it's wrong: the problem gives the DIAMETER as $\frac{3}{2}(2x+1)$, not the radius — using this expression directly as $r$ in the volume formula doubles the radius used and gives an answer $8\times$ too large after cubing. Correct approach: always halve the given diameter first, $r=\frac{1}{2}\cdot\frac{3}{2}(2x+1)=\frac{3}{4}(2x+1)$, before substituting into $V=\frac{4}{3}\pi r^3$.

Mistake 2: Leaving the final derivative expressed in terms of $r$ instead of $x$

Why it's wrong: after computing $\frac{dV}{dr}\cdot\frac{dr}{dx}$, the result still contains $r^2$ — reporting the answer with $r$ still present doesn't answer "rate of change with respect to $x$", since $x$ was the variable asked for. Correct approach: substitute $r=\frac{3}{4}(2x+1)$ back in so the final expression is purely a function of $x$. …