Problem 65 - Calculus

Question

A spherical balloon is growing with radius \(r=3 t+1\) in centimeters, for time \(t\) in seconds. Find the volume of the balloon at 3 seconds.

Step-by-Step Solution

Answer

At 3 seconds, the balloon's volume is approximately \(4188.79\) cm³.

Step 1: Formula for Volume of a Sphere

The formula to find the volume of a sphere is given by \(V = \frac{4}{3} \pi r^3\), where \(r\) is the radius of the sphere. We will use this formula to find the volume of the balloon once we have the radius at 3 seconds.

Step 2: Calculate Radius at 3 Seconds

The radius \(r\) of the balloon is given by the equation \(r = 3t + 1\). Plug in \(t = 3\) seconds to calculate the radius: \(r = 3(3) + 1 = 9 + 1 = 10\) cm.

Step 3: Substitute Radius into Volume Formula

Now substitute \(r = 10\) cm into the volume formula: \[V = \frac{4}{3} \pi (10)^3\].

Step 4: Calculate Volume of the Balloon

Calculate \((10)^3 = 1000\) and substitute back into the formula: \[V = \frac{4}{3} \pi (1000) = \frac{4000}{3} \pi\].

Step 5: Simplify the Volume Expression

The volume of the balloon at 3 seconds can be expressed as \(V = \frac{4000}{3} \pi\) cubic centimeters. We can calculate this value using the approximation \(\pi \approx 3.1416\).

Key Concepts

Calculus Problem

When dealing with calculus problems, we often encounter scenarios where variables change over time or due to certain stimuli. For example, in this exercise, the change in the balloon's radius over time describes a real-world calculus application. Calculus helps us understand how these changes occur and allows us to calculate certain values at specific times.
The given problem involves a dynamic situation where the radius of the balloon changes as time progresses. To solve such problems:

• Identify the expressions or formulas related to the problem—like the volume of a sphere in this case.
• Determine the function describing the change, such as the growth of the balloon's radius over time.
• Apply these functions at specified points to find out the required values, as was done to calculate the balloon's volume at a specific time.

By approaching calculus problems systematically, we can effectively solve them using derivatives and integrals as needed to understand the rates and overall changes.

Rate of Change

The concept of rate of change is crucial in understanding how one quantity varies relative to another. Here, the rate of change is represented by the expression for the radius of the balloon, given as function of time, \(r = 3t + 1\).
In simpler terms, this equation tells us:

• The balloon's radius increases by 3 cm for each second that passes.
• At time zero, the balloon already has a radius of 1 cm.

Understanding the rate of change allows us to predict how the balloon's size evolves over time. It also becomes a foundational element when solving problems involving differentiation, where we calculate the instantaneous rates of change to understand how variables behave at specific points in time. In this exercise, knowing the rate of change allows us to effectively compute the radius at any given time and further derive other measurements like the volume.

Spherical Geometry

Spherical geometry is the study of figures on the surface of a sphere, which extends to understanding volumes and surface areas. For spheres, the volume formula \(V = \frac{4}{3} \pi r^3\) is crucial when considering changes in dimensions. This is because:

• It relates directly to the radius of the sphere.
• \(\pi\) is a constant representing the ratio of the circumference of any circle to its diameter.
• The exponent 3 on \(r\) indicates that volume is related to the cube of the radius, affecting how rapidly the volume changes with the radius.

By understanding these relationships, you can approach problems involving spheres more confidently. For instance, in this exercise, as the balloon expands, its volume dramatically increases due to the cubic relationship with its radius. Knowing how to apply the spherical volume formula is essential in determining the correct measurements at any given point as geometrical dimensions change.