Problem 50 - A Complete Resource Book in Mathematics for JEE Main 2017

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 40 minutes after the leakage began is (A) \(9 / 7\) (B) \(7 / 9\) (C) \(2 / 9\) (D) \(9 / 2\)

Step-by-Step Solution

Answer

The radius decreases at \(\frac{2}{9}\) meters per minute.

Step 1: Determine the Initial Volume

The initial volume of the spherical balloon is given as \(4500\pi\) cubic meters. This volume will be used to determine the initial radius of the balloon using the formula for the volume of a sphere.

Step 2: Apply Volume Formula to Find Initial Radius

The formula for the volume of a sphere is \(V = \frac{4}{3}\pi r^3\). We set \(V = 4500\pi\) to find the initial radius:\[4500\pi = \frac{4}{3}\pi r^3\]Canceling \(\pi\) from both sides, we simplify to:\[4500 = \frac{4}{3} r^3\]Multiply both sides by \(\frac{3}{4}\) to get:\[r^3 = 3375\]Taking the cube root gives:\[r = 15 \text{ meters}\]

Step 3: Calculate Remaining Volume After 40 Minutes

Since helium is escaping at a rate of \(72\pi\) cubic meters per minute, in 40 minutes it will escape:\[40 \times 72\pi = 2880\pi \text{ cubic meters}\]The volume after 40 minutes is:\[4500\pi - 2880\pi = 1620\pi \text{ cubic meters}\]

Step 4: Find the Radius at t = 40 Minutes

Use the reduced volume \(V = 1620\pi\) and the sphere volume formula to find the radius at 40 minutes:\[1620\pi = \frac{4}{3}\pi r^3\]Cancel \(\pi\) and solve for \(r^3\):\[1620 = \frac{4}{3} r^3\]Multiply by \(\frac{3}{4}\) to get:\[r^3 = 1215\]Cube root gives:\[r = 11 \text{ meters}\]

Step 5: Relate the Change of Volume to Radius Using Chain Rule

Differentiate the volume formula \(V = \frac{4}{3}\pi r^3\) with respect to time:\[\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\]

Step 6: Solve the Differential Equation for \(\frac{dr}{dt}\)

Given \(\frac{dV}{dt} = -72\pi\) and \(r = 11\) meters at 40 minutes, substitute into the differential equation:\[-72\pi = 4\pi (11)^2 \frac{dr}{dt}\]Simplify and solve for \(\frac{dr}{dt}\):\[-72 = 4 \times 121 \frac{dr}{dt}\]\[-72 = 484 \frac{dr}{dt}\]\[\frac{dr}{dt} = \frac{-72}{484} = -\frac{18}{121} = -\frac{2}{9} \text{ meters per minute}\]

Step 7: Conclusion: Determine the Correct Answer Choice

The rate at which the radius decreases is \(-\frac{2}{9}\), corresponding to answer choice (C) with the rate \(\frac{2}{9}\) when considering the magnitude.

Key Concepts

Differentiation in Mathematics

In mathematics, differentiation is a fundamental concept used to determine how a quantity changes with respect to another. In simple terms, differentiation helps you find the rate at which something is changing.

For example, if you want to understand how fast a balloon is shrinking as air leaks out, differentiation comes into play. When applying differentiation to problems involving real-world applications, such as physics or engineering tasks, it's often essential to find the derivative of a function.

**Key Points about Differentiation:**

• Differentiation involves taking the derivative of a function.
• It helps calculate the rate of change of one variable with respect to another.
• In our balloon example, we used differentiation to relate changes in volume to changes in radius.
• The chain rule is crucial when variables are dependent on each other, as it allows us to differentiate composite functions.

Spherical Volume Formula

The volume of a sphere can be calculated using the spherical volume formula: \( V = \frac{4}{3} \pi r^3 \).

This formula offers a way to understand how the volume of a spherical object, such as a balloon, changes based on its radius. When you know the volume of the sphere, you can also work backward to find the radius, which is crucial in solving related rates problems.

**Applying the Spherical Volume Formula:**

• To find the volume, you need the radius of the sphere.
• To find the radius when the volume is given, rearrange the formula: \( r = \sqrt[3]{\frac{3V}{4\pi}} \).
• This formula aids in simplifying how changes in volume correlate directly to changes in the radius.
• In our example, knowing the initial volume allowed us to compute the initial radius before the helium began to escape.

Related Rates Problems

Related rates problems involve finding a rate at which one quantity changes concerning another, commonly using calculus. These problems often occur in scenarios where multiple variables are interdependent, and a change in one affects the others.

Take the spherical balloon problem: as the volume decreases due to a leak, the radius decreases. To find the rate at which the radius changes, we rely on related rates methods to relate volume reduction and radius decrease.

**Steps to Solve Related Rates Problems:**

• Identify the known rates of change and the rate you need to find.
• Express all related variables using appropriate formulas, like the spherical volume formula.
• Differentiate the equations with respect to time using the chain rule, if necessary.
• Substitute known values and solve for the required rate of change.

In the exercise, the distinct steps, including determining the rate of helium escape and applying differentiation, allowed for the calculation of the decrease in the balloon’s radius over time.