Problem 180 - A Complete Resource Book in Mathematics for JEE Main

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 theradius of the balloon decreases 49 minutes after the leakage began is (A) \(\frac{9}{7}\) (B) \(\frac{7}{9}\) (C) \(\frac{2}{9}\) (D) \(\frac{9}{2}\)

Step-by-Step Solution

Answer

The rate is \(-\frac{2}{9}\) meters per minute, corresponding to option (C).

Step 1: Understand the Volume Formula

The volume of a sphere is given by the formula \[ V = \frac{4}{3} \pi r^3 \]where \( V \) is the volume and \( r \) is the radius of the sphere.

Step 2: Differentiate with Respect to Time

We need the rate at which the radius changes, so we differentiate the volume formula with respect to \( t \). This gives:\[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \]Where \( \frac{dr}{dt} \) is the rate we are solving for.

Step 3: Determine Volume After 49 Minutes

Initially, the volume is \( 4500 \pi \) cubic meters. After 49 minutes, with leakage at \( 72 \pi \) cubic meters per minute, the volume is:\[ V = 4500\pi - 72\pi \times 49 = 990\pi \]

Step 4: Solve for Radius at Volume \(990\pi\)

Using the volume formula \( V = \frac{4}{3} \pi r^3 = 990 \pi \), solving for \( r \) gives:\[ r^3 = \frac{3 \times 990}{4} = 742.5 \]\[ r = (742.5)^{1/3} \approx 9 \text{ meters} \]

Step 5: Solve for \( \frac{dr}{dt} \)

Substitute \( r \approx 9 \) meters and \( \frac{dV}{dt} = -72 \pi \) (for decreasing volume) into the differentiated formula:\[ -72\pi = 4\pi \times 9^2 \times \frac{dr}{dt} \]\[ -72\pi = 324\pi \times \frac{dr}{dt} \]Divide both sides by \(324\pi\):\[ \frac{dr}{dt} = -\frac{72}{324} = -\frac{2}{9} \text{ meters per minute} \]

Step 6: Conclusion

After substituting all known values and solving, the rate at which the radius of the balloon decreases 49 minutes after the leak began is \(-\frac{2}{9}\) meters per minute. Therefore, the correct answer is option (C).

Key Concepts

Rate of Change

In calculus, the concept of rate of change is vital for understanding how one quantity changes in relation to another. For this exercise, we looked at the rate at which the sphere's volume—and subsequently its radius—changes over time.

• Rate of change indicates how fast or slow a quantity, like radius, changes over a specific time period.
• In our scenario, the leak causes the volume to decrease at a constant rate of \(72\pi\) cubic meters per minute. This is a negative rate because the volume is diminishing.

Knowing the rate of change enables us to predict how long it will take for the balloon to deflate or how the sphere's physical dimensions evolve. It's a practical real-world application of calculus that gives insights into dynamic systems.

Differentiation

Differentiation is a fundamental concept in calculus, dealing with finding the derivative of functions. Here, we utilized differentiation to determine how the changing volume of the sphere impacts its radius.

• Differentiate with respect to time when monitoring changes over time. This provides us with the instantaneous rate of change of one variable compared to time.
• In the exercise, the volume formula for the sphere was \( V = \frac{4}{3} \pi r^3 \). To find out how the radius changes, we used the derivative \( \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \).

Differentiation simplifies the process of understanding how changes in one quantity can impact another. It is key to solving various problems in physics, engineering, economics, and more.

Sphere Volume

The volume of a sphere is a measure of how much space is contained within the sphere. It's calculated using the formula \( V = \frac{4}{3} \pi r^3 \).

• This formula helps us understand the initial size of the balloon when it was filled with helium gas. With a given volume, you can figure out the corresponding radius.
• During the leakage, we used this formula to see how the volume decreased over time and how it affected the size of the balloon. By substituting the current volume into the equation, we computed the new radius.

Understanding the volume of a sphere and how it's affected by changes in radius is essential in fields like geometry and physics, where the shapes of objects often need precise calculation.