Problem 14 - Calculus of a Single Variable

Question

A spherical balloon is inflated with gas at the rate of 800 cubic centimeters per minute. How fast is the radius of the balloon increasing at the instant the radius is (a) 30 centimeters (b) 60 centimeters?

Step-by-Step Solution

Answer

(a) The radius is increasing at approximately 0.177 cm/min when the radius is 30 cm. (b) The radius is increasing at approximately 0.044 cm/min when the radius is 60 cm.

Step 1: Identify the Known and Unknown Quantities

The spherical balloon's volume has a formula: \( V = \frac{4}{3} \pi r^3 \), where V is the volume and r is the radius. The problem provides the rate of change of volume (\( \frac{dV}{dt} = 800 \) cubic cm/min), and asks for the rate of change of the radius, \( \frac{dr}{dt} \), when the radius is 30 cm and 60 cm respectively.

Step 2: Differentiate the Volume Equation

Implicitly differentiate the volume equation with respect to \(t\), we get: \( \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} \)

Step 3: Solve for \(\frac{dr}{dt}\)

Rearrange the equation in the previous step and solve for \( \frac{dr}{dt} \), we get: \( \frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^2} \)

Step 4: Solve for Specific Radius Mesurements

(a) For \( r = 30 \) cm, substitute \( r \), \( \frac{dV}{dt} \) into the equation: \( \frac{dr}{dt} = \frac{800}{4\pi (30)^2} \) cm/min. Then calculate. (b) Repeat this process for \( r = 60 \) cm.

Key Concepts

Volume of a Sphere

The concept of volume is crucial when dealing with three-dimensional objects such as spheres. The volume of a sphere determines how much space it occupies. For a sphere, there is a specific formula to calculate the volume: \[ V = \frac{4}{3} \pi r^3 \] where \( V \) is the volume, \( \pi \) is a mathematical constant approximately equal to 3.14159, and \( r \) is the radius of the sphere.

• The formula reveals that the volume is directly related to the cube of the radius.
• This means that even a small change in the radius can lead to a significant change in the volume.

Understanding how to calculate the volume helps us in dealing with real-life problems involving spherical objects.

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the rate at which one quantity changes concerning another. This is especially useful in physics and engineering to study motion and change.

• It helps us determine how a change in one variable, say the radius, affects another variable, like volume.
• The derivative tells us the instantaneous rate of change, which is particularly useful when dealing with problems involving time.

In the case of a spherical balloon, differentiation allows us to understand how changes in the radius and volume relate as the balloon is inflated.

Implicit Differentiation

Implicit differentiation is a technique used when a function is not solved for one variable explicitly. In many real-world applications, relationships between variables are intertwined. This makes implicit differentiation a valuable tool.

• Instead of separating variables, we differentiate both sides of an equation concerning a third variable, often time in related rates problems.
• This is particularly useful in our spherical balloon problem, where we need to find \( \frac{dr}{dt} \) without solving explicitly for \( r \).

By applying implicit differentiation to the volume formula, it enables us to find how the radius changes as the volume changes over time.

Rate of Change

The rate of change is a key concept when dealing with dynamic systems or processes. It's all about understanding how quickly or slowly something happens. In calculus terms, it's about how one quantity responds to changes in another.

• In the balloon problem, the rate of change of volume is given, and we are asked to find the rate of change of the radius.
• Rate of change gives us insight into the speed of expansion or contraction of a system.

For the balloon, knowing \( \frac{dV}{dt} = 800 \) cubic cm/min, we use the concept of rate of change to find \( \frac{dr}{dt} \) at specific radius values, showing how fast the radius increases.