/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 In Example \(2.11\) we solved th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 3

In Example \(2.11\) we solved the equation $$ 62.4 \pi d^{2}\left(2-\frac{d}{3}\right)=436 $$ on the interval from \(d=0\) to \(d=4\). If you graph the function \(62.4 \pi d^{2}(2-d / 3)\) and the constant function 436 on the span from \(d=-2\) to \(d=7\), you will see that this equation has two other solutions. Find these solutions. Is there a physical interpretation of these solutions that makes sense?

Short Answer

Expert verified
The other solutions are approximately \(d = -0.5\) and \(d = 5.5\).

Step by step solution

01

Setting the Problem

We are given the equation \(62.4 \pi d^{2}(2-\frac{d}{3})=436\) and need to find solutions for \(d\) besides the solution on the interval \(d=0\) to \(d=4\). The hint is that there are more solutions when plotting from \(d=-2\) to \(d=7\).
02

Rearranging the Equation

Rearrange the equation to make a function that we can graph: \(62.4 \pi d^{2}(2-\frac{d}{3}) - 436 = 0\). We will denote this function as \(f(d) = 62.4 \pi d^{2}(2-\frac{d}{3}) - 436\).
03

Graph the Functions

Graph the function \(f(d) = 62.4 \pi d^{2}(2-\frac{d}{3}) - 436\) and the line \(y=0\) over the interval \(d=-2\) to \(d=7\) using a graphing calculator or software. Look for points where \(f(d)\) intersects with \(y=0\).
04

Identifying Solutions

From the graph, identify the points where \(f(d)\) intersects the \(x\)-axis, which represent the solutions to the equation. Besides the known solution, you should see two additional intersection points at different values of \(d\).
05

Verifying Solutions

Using rough estimates from the graph, find approximate values of \(d\) for the other solutions. You can use numerical methods or a calculator to get more precise values for these intersection points.
06

Find the Additional Solutions

By examining the graph, we find approximate values for \(d\), which are around \(d = -0.5\) and \(d = 5.5\). Use a numerical method or calculator to refine these solutions if necessary.
07

Interpreting Solutions

Consider whether the additional solutions \(d = -0.5\) and \(d = 5.5\) can have a physical interpretation. Depending on the context of the problem, negative or excessive values of \(d\) may not make physical sense.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Graphing Functions
Graphing functions is a way to visually represent the relationship between variables in an equation. To graph a function like \( f(d) = 62.4 \pi d^{2}(2-\frac{d}{3}) - 436 \), we start by plotting it on a graph where the horizontal axis represents \(d\) and the vertical axis represents \(f(d)\).
  • Identifying Key Features: Look for parts of the graph where it crosses the horizontal axis. These are known as the roots or solutions of the equation.
  • Using Tools: Graphing calculators or software, such as Desmos or GeoGebra, are extremely useful. They help in easily identifying these intersections.
  • Analyzing the Graph: Check for the shape, symmetry, and any other features that might provide insights into the behavior of the function.
When you graph the function \( f(d) \) alongside \( y = 0 \), you're looking for where they meet鈥攖hese meeting points (intersections) indicate the solutions to the equation.
Equation Solving
Equation solving involves finding the values of the variables that satisfy the given equation. In our scenario, we are interested in finding values of \(d\) such that \(f(d) = 0\). The process generally includes:
  • Rearranging the Equation: The first step is often to transform the equation to one side, making it a function \( f(d) \) equal to zero.
  • Graphical Solutions: Using graphical methods, as described, to visually find where the function \( f(d) \) intersects the axis.
  • Verification: Taking rough estimates from the graph, one might need to use a calculator or algebra to verify each potential solution.
Once you've identified these points on the graph, the process isn't over. Verifying them by plugging back into the original equation ensures the solutions are accurate.
Numerical Methods
Numerical methods provide approximations to the roots of equations that might be difficult to solve algebraically. They are especially useful when dealing with complex equations like our function \( f(d) \).Some popular numerical methods include:
  • Bisection Method: It involves dividing the interval repeatedly and selecting subintervals where the function changes sign until the root is found.
  • Newton's Method: This iterative method is useful when you can calculate derivatives. It uses tangents to approximate the root.
  • Secant Method: Similar to Newton's but doesn't require derivatives, instead using successive approximations.
For example, in the context of our exercise, one might use these methods to refine rough graph-based estimates of \(d = -0.5\) and \(d = 5.5\), to ensure they are as accurate as needed. While graphical solutions give a general idea, these iterative numerical approaches provide precise solutions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

City A lies on the north bank of a river that is 1 mile wide. You need to run a phone cable from City A to City B, which lies on the opposite bank 5 miles down the river. You will lay \(L\) miles of the cable along the north shore of the river, and from the end of that stretch of cable you will lay \(W\) miles of cable running under water directly toward City B. (See Figure \(2.111\) on the next page.) You will need the following fact about right triangles. A right triangle has two legs, which meet at the right angle, and the hypotenuse, which is the longest side. An ancient and beautiful formula, the Pythagorean theorem, relates the lengths of the three sides: Length of hypotenuse \(=\sqrt{\text { Length of one leg }}\) a. Find an appropriate right triangle that shows that \(W=\sqrt{1+(5-L)^{2}}\) b. Find a formula for the length of the total phone cable \(P\) from City A to City B as a function of \(L\). c. Make a graph of the total phone cable length \(P\) as a function of \(L\), and explain what the graph is showing. d. What value of \(L\) gives the least length for the total phone cable? Draw a picture showing the least-length total phone cable.

Between the ages of 7 and 11 years, the weight \(w\), in pounds, of a certain girl is given by the formula $$ w=8 t $$ Here \(t\) represents her age in years. a. Use a formula to express the age \(t\) of the girl as a function of her weight \(w\). b. At what age does she attain a weight of 68 pounds? c. The height \(h\), in inches, of this girl during the same period is given by the formula $$ h=1.8 t+40 . $$ i. Use your answer to part b to determine how tall she is when she weighs 68 pounds. ii. Use a formula to express the height \(h\) of the girl as a function of her weight \(w\). iii. Answer the question in part \(i\) again, this time using your answer to part ii. c. The height \(h\), in inches, of this girl during the same period is given by the formula $$ h=1.8 t+40 . $$ i. Use your answer to part b to determine how tall she is when she weighs 68 pounds. ii. Use a formula to express the height \(h\) of the girl as a function of her weight \(w\). iii. Answer the question in part \(i\) again, this time using your answer to part ii.

In this exercise we develop a model for the growth rate \(G\), in thousands of dollars per year, in sales of a product as a function of the sales level \(s\), in thousands of dollars. \({ }^{30}\) The model assumes that there is a limit to the total amount of sales that can be attained. In this situation we use the term unattained sales for the difference between this limit and the current sales level. For example, if we expect sales to grow to 3 thousand dollars in the long run, then \(3-s\) gives the unattained sales. The model states that the growth rate \(G\) is proportional to the product of the sales level \(s\) and the unattained sales. Assume that the constant of proportionality is \(0.3\) and that the sales grow to 2 thousand dollars in the long run. a. Find a formula for unattained sales. b. Write an equation that shows the proportionality relation for \(G\). c. On the basis of the equation from part b, make a graph of \(G\) as a function of \(s\). d. At what sales level is the growth rate as large as possible? e. What is the largest possible growth rate?

The rate of growth \(G\) in the weight of a fish is a function of the weight \(w\) of the fish. For the North Sea cod, the relationship is given by $$ G=2.1 w^{2 / 3}-0.6 w $$ Here \(w\) is measured in pounds and \(G\) in pounds per year. The maximum size for a North Sea cod is about 40 pounds. a. Make a graph of \(G\) against \(w\). b. Find the greatest rate of growth among all cod weighing at least 5 pounds. c. Find the greatest rate of growth among all cod weighing at least 25 pounds.

Competition between populations: In this exercise we consider the question of competition between two populations that vie for resources but do not prey on each other. Let \(m\) be the size of the first population and \(n\) the size of the second (both measured in thousands of animals), and assume that the populations coexist eventually. Here is an example of one common model for the interaction: Per capita growth rate for \(m=5(1-m-n)\), Per capita growth rate for \(n=6(1-0.7 m-1.2 n)\). a. An isocline is formed by the points at which the per capita growth rate for \(m\) is zero. These are the solutions of the equation \(5(1-m-n)=0\). Find a formula for \(n\) in terms of \(m\) that describes this isocline. b. The points at which the per capita growth rate for \(n\) is zero form another isocline. Find a formula for \(n\) in terms of \(m\) that describes this isocline. c. At an equilibrium point the per capita growth rates for \(m\) and for \(n\) are both zero. If the populations reach such a point, they will remain there indefinitely. Use your answers to parts \(a\) and \(b\) to find the equilibrium point.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.