Problem 180 - 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 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, which is option (C).

Step 1: Define the Volume Formula and Derive its Differential

The volume of a sphere is given by the formula \( V = \frac{4}{3} \pi r^3 \). To find how the radius changes over time as the volume changes, we'll differentiate the volume formula with respect to time \( t \). This is done using implicit differentiation: \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \).

Step 2: Determine the Rate of Change of Volume

We're given that the volume of the gas is decreasing at the rate of \( 72\pi \) cubic meters per minute. Thus, \( \frac{dV}{dt} = -72\pi \) since the volume is decreasing.

Step 3: Calculate the Volume of the Balloon after 49 Minutes

After 49 minutes, the volume of the balloon is reduced from \( 4500\pi \) by \( 49 \times 72\pi \) cubic meters. So, the volume is \( 4500\pi - 3528\pi = 972\pi \) cubic meters.

Step 4: Solve for the Radius 49 Minutes After Leakage Began

Using the volume of the balloon \( V = 972\pi \), substitute back into the volume formula: \[ \frac{4}{3} \pi r^3 = 972\pi \] Cancel \( \pi \) from both sides and solve for \( r^3 \): \[ \frac{4}{3} r^3 = 972 \] \[ r^3 = \frac{3 \times 972}{4} = 729 \] Thus, \( r = \sqrt[3]{729} = 9 \) meters.

Step 5: Solve for the Rate of Change of the Radius

We know from the differential equation \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \) that: \[ -72\pi = 4\pi \times 9^2 \times \frac{dr}{dt} \] First, simplify: \[ -72 = 4 \times 81 \times \frac{dr}{dt} \] \[ -72 = 324 \times \frac{dr}{dt} \] Solve for \( \frac{dr}{dt} \): \[ \frac{dr}{dt} = \frac{-72}{324} = -\frac{2}{9} \text{ meters per minute} \].

Step 6: Conclusion: Identify the Correct Option

The rate of decrease of the balloon's radius 49 minutes after the leakage began is \(-\frac{2}{9}\) meters per minute, which matches option (C).

Key Concepts

Volume of a Sphere

Understanding the volume of a sphere is crucial in many areas of mathematics and real-world applications. A sphere is a perfectly round three-dimensional shape, and its volume is represented by the formula: \[ V = \frac{4}{3} \pi r^3 \] Where \( V \) is the volume and \( r \) is the radius. The formula comes from the integration of the surface area of a sphere. What's interesting is how the volume changes considerably with just a small change in the radius due to the cubic factor \( r^3 \). In the context of the given problem, the volume formula is used to determine how full the balloon is at any given time. It's critical to understand that any change in volume affects the whole sphere, not just the outer edge. By calculating how much volume is left after the gas has been leaking, we have a more tangible grasp of how large or small the sphere actually is at various times.

Implicit Differentiation

When dealing with variables that are interdependent in equations, implicit differentiation becomes very useful. Unlike regular differentiation where equations have one dependent variable, implicit differentiation handles situations where variables get intertwined, like in the volume formula for a sphere. When the volume \( V \) of a sphere changes, it affects its radius \( r \). By taking the derivative of both sides of the volume formula with respect to time \( t \), we implicitly find how \( V \) and \( r \) are connected as they change together: \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \] This derivative represents a chain of effects: a change in volume leads to a change in radius. Understanding this concept allows us to find relationships between different rates in complex shapes like spheres, without first solving for one in terms of the others.

Rate of Change

In differential calculus, "rate of change" refers to how a quantity increases or decreases over a specific period. For the spherical balloon problem, this concept connects how fast volume decreases to the changing size of the radius. The problem states the volume decreases at \( 72 \pi \) cubic meters per minute. Since the volume is getting smaller, \( \frac{dV}{dt} = -72 \pi \). The negative sign illustrates that the quantity is decreasing. We use this rate to find out how fast the radius is decreasing by substituting into the implicit differentiation formula: \[ -72\pi = 4\pi \times 9^2 \times \frac{dr}{dt} \] Solving for \( \frac{dr}{dt} \) involves basic algebra, revealing this to be \( -\frac{2}{9} \text{ meters per minute} \). Thus, understanding rate changes gives insight into how quickly or slowly elements of a system like a balloon change over time. This is particularly relevant in physics and engineering, where dynamic systems are often analyzed.