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Find the average rate of change of the volume of the balloon in Example 2 as the radius increases from (a) 2 to 5 inches (b) 4 to 8 inches

Short Answer

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Question: Calculate the average rates of change in the volume of a balloon for the following changes in radius: (a) from 2 inches to 5 inches, and (b) from 4 inches to 8 inches. Answer: (a) The average rate of change of the volume of the balloon when the radius increases from 2 inches to 5 inches is \(\frac{468\pi}{9}\). (b) The average rate of change when the radius increases from 4 inches to 8 inches is \(\frac{448\pi}{3}\).

Step by step solution

01

1. Formula for volume of a balloon

The volume formula for a sphere (which is the shape of the balloon in this case) is given as: V = \frac{4}{3}\pi r^3 where V is the volume and r is the radius of the balloon.
02

2. Calculate the volume for each radius

(a) For radius increasing from 2 inches to 5 inches: At r = 2 inches, calculate the volume (V1): V1 = \frac{4}{3}\pi(2)^3 = \frac{32\pi}{3} At r = 5 inches, calculate the volume (V2): V2 = \frac{4}{3}\pi(5)^3 = \frac{500\pi}{3} (b) For radius increasing from 4 inches to 8 inches: At r = 4 inches, calculate the volume (V3): V3 = \frac{4}{3}\pi(4)^3 = \frac{256\pi}{3} At r = 8 inches, calculate the volume (V4): V4 = \frac{4}{3}\pi(8)^3 = \frac{2048\pi}{3}
03

3. Calculate the average rate of change

To calculate the average rate of change of the volume, we will subtract the initial volume from the final volume and then divide by the change in radius. (a) For radius increasing from 2 inches to 5 inches: Average Rate of Change = \frac{V2 - V1}{5 - 2} = \frac{\frac{500\pi}{3} - \frac{32\pi}{3}}{3} = \frac{468\pi}{9} (b) For radius increasing from 4 inches to 8 inches: Average Rate of Change = \frac{V4 - V3}{8 - 4} = \frac{\frac{2048\pi}{3} - \frac{256\pi}{3}}{4} = \frac{448\pi}{3} Therefore, the average rates of change of the volume of the balloon are: (a) \frac{468\pi}{9} when the radius increases from 2 to 5 inches (b) \frac{448\pi}{3} when the radius increases from 4 to 8 inches.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume of a Sphere
The volume of a sphere is a fundamental concept in geometry, crucial for understanding various real-life applications such as calculating the amount of air in a balloon or the capacity of a spherical container. The formula to determine the volume of a sphere is expressed as:
\[ V = \frac{4}{3}\pi r^3 \]
In this expression, \( V \) represents the volume and \( r \) is the radius of the sphere. The cubed radius \( r^3 \) indicates that the volume of a sphere scales rapidly with an increase in its radius—not linearly but cubically, which means that small changes in the radius can result in significant changes in the volume.
Precalculus
Precalculus serves as a bridge between algebra and calculus, providing students with a deeper understanding of functions, graphs, and limits. In the context of understanding the volume of a sphere, precalculus helps students grasp how the volume changes with the radius by studying functions like \( r^3 \) and learning about rates of change, which are foundational concepts for calculus.
It's essential for students to have a solid understanding of precalculus to handle problems involving rates of change and to prepare for the more advanced concepts they'll encounter in calculus.
Rate of Change Calculation
The rate of change tells us how one quantity changes in relation to another. In precalculus, understanding the average rate of change is pivotal to analyzing real-world problems. To find the average rate of change for the volume of a sphere as the radius changes, you subtract the initial volume from the final volume and divide by the change in the radius:
\[ \text{Average Rate of Change} = \frac{V_{\text{final}} - V_{\text{initial}}}{r_{\text{final}} - r_{\text{initial}}} \]
This formula gives us a 'per unit' change, enabling us to understand the relationship between two variables in a meaningful way. Applying this to volume changes of a sphere, we see how sensitively the volume responds to changes in radius.
Sphere Radius
The radius of a sphere is the distance from the center of the sphere to any point on its surface. It is a critical measurement that not only defines the size of the sphere but also influences calculations related to the sphere, such as surface area and volume. In the discussed exercise, students must consider the change in the radius to determine how the volume of a sphere changes. Since the volume depends on the cubic power of the radius, even small increases in the radius result in large increases in volume, which is seen in the dramatic rate of change for the volume as the balloon’s radius grows.

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Most popular questions from this chapter

Use a calculator and the Horizontal Line Test to determine whether or not the function \(f\) is one-to-one. $$f(x)=.1 x^{3}-.1 x^{2}-.005 x+1$$

Use the Round-Trip Theorem on page 223 to show that \(g\) is the inverse of \(f\) $$f(x)=x+1, \quad g(x)=x-1$$

Verify that \((f \circ g)(x)=x\) and \((g \circ f)(x)=x\) for every \(x\) $$f(x)=9 x+8, \quad g(x)=\frac{x-8}{9}$$

Example \(11(b)\) showed how we create a table of values for a function when you get to choose all the values of the inputs. The technique presented does not work for Casio calculators. This exercise is designed for users of Casio calculators. \(\cdot \quad\) Enter an equation such as \(y=x^{3}-2 x+3\) in the equation memory. This can be done by selecting TABLE in the MAIN menu. \- Return to the MAIN menu and select LIST. Enter the numbers at which you want to evaluate the function as List 1 \(\cdot \quad\) Return to the MAIN menu and select TABLE. Then press SET-UP [that is, 2nd MENU] and select LIST as the Variable; on the LIST menu, choose List 1. Press EXIT and then press TABL to produce the table. \- Use the up/down arrow key to scroll through the table. If you change an entry in the X column, the corresponding \(y_{1}\) value will automatically change. (a) Use this technique to duplicate the table in Example \(11(\mathrm{b})\) (b) Change the number -11 to \(10,\) and confirm that you've obtained \(10^{3}-2(10)+3\)

Give an example of two different functions \(f\) and \(g\) that have all of the following properties: $$ f(-1)=1=g(-1) \quad \text { and } \quad f(0)=0=g(0) $$ and \(\quad f(1)=1=g(1)\)

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