A spherical balloon is filled with 4500π cubic meters of helium gas. If a leak in the balloon causes

Q: Wait, why is the rate of change of volume negative? Shouldn't it be positive?

That is a fantastic question! In calculus, the sign of the derivative tells us the direction of change. Since the gas is leaking out, the volume of the balloon is decreasing over time. Therefore, the rate of change of volume must be negative. If you use a positive value, you are calculating the rate at which the balloon is inflating, not leaking!

Q: Can I just use the initial radius to find the rate of change?

Not quite! The balloon is shrinking, so the radius is constantly changing. The radius at the start ($t=0$) is different from the radius at the moment of interest ($t=49$). You must first calculate the volume at $t=49$ and then find the corresponding radius at that specific instant before you can find the rate of change.

Q: Why do we use the Chain Rule here? Can't we just solve for $r$ and plug it in?

You are absolutely right that you need to find $r$ at that instant, but the Chain Rule is essential because we are dealing with rates of change. We have the rate of change of volume, but we need the rate of change of the radius. The Chain Rule is the mathematical mechanism that connects these two different rates through the geometry of the sphere.

Question

A spherical balloon is filled with $4500\pi$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $72\pi$ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases $49$ minutes after the leakage began is

Options:

(A) 9/7

(B) 7/9

(C) 2/9

(D) 9/2

Sigma Percentile · Accepted Answer

### The Dance of the Leaking Sphere: A Calculus Journey

Imagine you are standing in a field, holding a massive, perfectly spherical balloon filled with helium. It is a beautiful, sunny day, but there is a problem: your balloon has a tiny, persistent leak.

As the gas escapes, the balloon slowly shrinks. This is not just a simple geometry problem; it is a story of change, a classic 'related rates' problem that sits at the very heart of JEE Advanced physics and mathematics.

Let us break this down, not as a dry calculation, but as a dynamic process.

### The Geometric Anchor

First, we must ground ourselves in the geometry of the situation. We know the balloon is a sphere, and we know its initial volume is $V_0 = 4500\pi 	ext{ m}^3$.

The fundamental relationship between the volume $V$ and the radius $r$ of any sphere is given by the elegant formula:

$$V = \frac{4}{3}\pi r^3$$

This formula is our anchor. It tells us that the volume is not just a number; it is a cubic function of the radius. If the radius changes, the volume changes in a very specific, non-linear way.

### The Reality of the Leak

Now, let us introduce the element of time. The gas is escaping at a constant rate of $72\pi 	ext{ m}^3/	ext{min}$.

In the language of calculus, this is the derivative of volume with respect to time, $\frac{dV}{dt}$. Because the gas is leaving the balloon, the volume is decreasing. Therefore, we must define our rate as:

$$\frac{dV}{dt} = -72\pi 	ext{ m}^3/	ext{min}$$

Never forget that negative sign! It is the physical manifestation of the gas escaping. If you miss it, your final answer will be positive, implying the balloon is growing, which would be physically impossible in this scenario.

### The Snapshot at $t=49$

The problem asks us to analyze the balloon exactly $49$ minutes after the leak began. We need to know the state of the balloon at this specific moment.

First, let us calculate the total volume lost:

$$	ext{Total volume leaked} = 49 	imes 72\pi = 3528\pi 	ext{ m}^3$$

Subtracting this from our initial volume, we find the remaining volume $V_{49}$:

$$V_{49} = 4500\pi - 3528\pi = 972\pi 	ext{ m}^3$$

Now, we need the radius at this exact moment. We return to our geometric anchor:

$$972\pi = \frac{4}{3}\pi r^3$$

The $\pi$ terms cancel out beautifully. Multiplying by $3$ and dividing by $4$, we get $r^3 = 729$. Taking the cube root, we find that at $t=49$, the radius is exactly $r = 9 	ext{ m}$.

### The Calculus of Change

Now, we arrive at the core of the problem. We need to find the rate at which the radius is decreasing, which is $\frac{dr}{dt}$.

We have the volume formula $V = \frac{4}{3}\pi r^3$. To relate the rate of change of volume to the rate of change of radius, we differentiate both sides with respect to time $t$.

Using the Chain Rule, we get:

$$\frac{dV}{dt} = \frac{4}{3}\pi \cdot (3r^2) \cdot \frac{dr}{dt}$$

Simplifying this, the $3$s cancel out, leaving us with:

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

This equation is powerful. It tells us that the rate of change of volume depends on the current surface area ($4\pi r^2$) and the rate at which the radius is changing.

### The Final Resolution

We have all the pieces of the puzzle. We know $\frac{dV}{dt} = -72\pi$ and we know that at our specific moment, $r = 9$.

Let us plug these values into our derivative equation:

$$-72\pi = 4\pi (9)^2 \frac{dr}{dt}$$

The $\pi$ cancels out again. Calculating $9^2$, we get $81$. So:

$$-72 = 4 \cdot 81 \cdot \frac{dr}{dt}$$

$$-72 = 324 \frac{dr}{dt}$$

Finally, we isolate $\frac{dr}{dt}$:

$$\frac{dr}{dt} = -\frac{72}{324}$$

Both numbers are divisible by $36$. Since $72 = 2 	imes 36$ and $324 = 9 	imes 36$, the fraction simplifies to:

$$\frac{dr}{dt} = -\frac{2}{9} 	ext{ m/min}$$

The negative sign confirms our intuition: the radius is decreasing. Thus, the rate at which the radius decreases is **$\frac{2}{9} 	ext{ m/min}$**.

You have just navigated a complex related rates problem by connecting geometry, algebra, and calculus. Keep this mindset—always visualize the physical reality behind the equations—and you will master any problem the JEE throws at you.

Select Answer:

Visualized Solution

Initial Setup: $V_{0} = 4500 π$

Rate of Leakage: $\frac{d V}{d t}$

Volume at $t = 49 min$

Radius at $t = 49 min$

Differentiating Volume: $\frac{d V}{d t}$

Substituting Known Values

Isolating $\frac{d r}{d t}$

Final Rate of Decrease

The Sigma Insight: Tangents, Normals and Rate Measure

The Dance of the Leaking Sphere

The Geometric Anchor

The Reality of the Leak

The Snapshot at $t = 49$

The Calculus of Change

The Final Resolution

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