Q.A balloon, which always remains spherical on inflation, is being inflated by pumping in $900 \text{ cubic centimetres}$ of gas per second. Find the rate at which the radius of the balloon increases when the radius is $15 \text{ cm}$.

Concept: Related Rates — we connect the rate of change of volume $\frac{dV}{dt}$ to the rate of change of radius $\frac{dr}{dt}$ using the formula for the volume of a sphere.

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

Differentiate both sides with respect to time $t$:

Step 2: We are given $\frac{dV}{dt} = 900 \text{ cm}^3/\text{s}$ and $r = 15 \text{ cm}$. Substitute:

$900 = 4\pi (15)^2 \frac{dr}{dt}$ …

The key idea here is that the balloon’s volume and radius are linked by a fixed geometric relationship — the volume of a sphere. As gas is pumped in, the volume changes at a known rate, and we want to know how fast the radius changes at a particular instant. This is a related rates problem: we connect the rates of change of two quantities through their relationship.

Why this works:

If two quantities are related by an equation, then their rates of change (derivatives with respect to time) are also related. Differentiate the equation implicitly with respect to time, then plug in the known values to solve for the unknown rate.

1. Write the relationship between volume and radius. For a sphere,

This holds at every instant during inflation.

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

This formula directly connects the rate of change of volume to the rate of change of radius.

3. Identify what we know and what we need.

   • $\frac{dV}{dt} = 900$ cm³/s (given).
   • We want $\frac{dr}{dt}$ when $r = 15$ cm.

4. Substitute the known values and solve for $\frac{dr}{dt}$.

   $900 = 4\pi (15)^2 \cdot \frac{dr}{dt}$ …

Method: Related Rates via the Sphere Volume Formula

This method applies to any problem where a sphere-shaped object (balloon, bubble, raindrop) is inflating or shrinking, you are given the rate at which one of volume/radius changes, and asked for the rate of the other at a specific instant.

Steps

Step 1: Identify the two quantities that vary with time

Here it is the volume $V$ and the radius $r$ of the sphere, both functions of time $t$. Write down what is given (a rate, in units per second) and what is asked (the other rate, at a stated instant).

Step 2: Write the fixed geometric relation connecting them

For any sphere,

This equation holds at every instant, no matter how $V$ and $r$ are changing.

Step 3: Differentiate both sides with respect to time $t$

Since both $V$ and $r$ are functions of $t$, differentiate using the chain rule — the power rule on $r^3$ brings down a factor of $\frac{dr}{dt}$:

This single equation is the bridge between the two rates for a sphere of ANY radius — never substitute numbers before this step. …

Common Mistakes

Mistake 1: Substituting $r=15$ into the volume formula before differentiating

Why it's wrong: if you plug $r=15$ into $V=\frac{4}{3}\pi r^3$ first, $r$ becomes a fixed number, and its derivative $\frac{dr}{dt}$ silently disappears from the equation — you lose the very unknown you're solving for. Correct approach: differentiate $V=\frac{4}{3}\pi r^3$ with respect to $t$ FIRST, obtaining the general relation $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$, and only then substitute the specific values of $r$ and $\frac{dV}{dt}$.

Mistake 2: Misapplying the power rule on $r^3$ …