Q.The radius of an air bubble is increasing at the rate of $\frac{1}{2} \text{ cm/s}$. At what rate is the volume of the bubble increasing when the radius is $1 \text{ cm}$?

Concept: Related Rates — we differentiate the volume formula with respect to time, using the chain rule.

Step 1: Volume of a sphere: $V = \frac{4}{3}\pi r^3$.

Differentiate both sides with respect to $t$:

$\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$. …

This is a classic related rates problem. The key idea: when two quantities (here, radius and volume) are linked by a formula, their rates of change are also linked — through differentiation with respect to time.

You’re told how fast the radius grows. You want how fast the volume grows at a specific moment. The chain rule is your bridge.

1. Write the relationship between volume and radius. For a sphere (an air bubble),

2. Differentiate both sides with respect to time $t$. Since $V$ depends on $r$, and $r$ depends on $t$, use the chain rule:

This is the general rate at which volume changes, for any radius $r$ and any radial speed $\frac{dr}{dt}$.

3. Plug in the given values.

   You are told:

   • $\frac{dr}{dt} = \frac{1}{2}$ cm/s (positive, so radius is increasing).
   • At the instant of interest, $r = 1$ cm.

   So:

   $\frac{dV}{dt} = 4\pi (1)^2 \cdot \frac{1}{2} = 4\pi \cdot \frac{1}{2} = 2\pi.$ …

Method: Related Rates — Given One Rate, Find the Other Directly

This method applies to sphere-growth problems (bubble, balloon, droplet) where the radius's rate of change is given directly, and you must find the resulting volume rate at a stated instant — the same core relation as any sphere related-rates problem, just read in the other direction.

Steps

Step 1: Write the geometric relation between volume and radius

Step 2: Differentiate both sides with respect to time $t$ using the chain rule

This links the two rates for a sphere of any size at any instant — it is not specific to any one radius.

Step 3: Identify which rate is given and which is asked

Here $\frac{dr}{dt}$ (a constant, given speed of growth) and a specific radius $r$ are known; $\frac{dV}{dt}$ at that instant is wanted — this is a direct substitution, with no algebraic rearrangement needed, unlike problems that hand you $\frac{dV}{dt}$ and ask for $\frac{dr}{dt}$.

Step 4: Substitute both the given radius and the given rate into the Step 2 formula …

Common Mistakes

Mistake 1: Forgetting the chain-rule factor of $\frac{dr}{dt}$

Why it's wrong: differentiating $V=\frac{4}{3}\pi r^3$ as though $r$ were the independent variable gives $4\pi r^2$ alone, silently dropping the multiplication by $\frac{dr}{dt}$ that is required because $r$ itself changes with time. Correct approach: always write the full relation $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$ before substituting any numbers.

Mistake 2: Substituting $r=1$ before differentiating …