/*! 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 16 Astronauts looking at Earth from... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 16

Astronauts looking at Earth from a spacecraft can see only a portion of the surface. 10 See Figure \(2.59\) on the next page. The fraction \(F\) of the surface of Earth that is visible at a height \(h\), in kilometers, above the surface is given by the formula $$ F=\frac{0.5 h}{R+h} . $$ Here \(R\) is the radius of Earth, about 6380 kilometers. (For comparison, 1 kilometer is about \(0.62\) mile, and the moon is about 380,000 kilometers from Earth.) a. Make a graph of \(F\) versus \(h\) covering heights up to 100,000 kilometers. b. A value of \(F\) equal to \(0.25\) means that \(25 \%\), or one-quarter, of Earth's surface is visible. At what height is this fraction visible? c. During one flight of a space shuttle, astronauts performed an extravehicular activity at a height of 280 kilometers. What fraction of the surface of Earth is visible at that height? d. Is the graph of \(F\) concave up or concave down? Explain your answer in practical terms. e. Determine the limiting value for \(F\) as the height \(h\) gets larger. Explain your answer in practical terms.

Short Answer

Expert verified
25% visibility occurs at 6380 km; at 280 km, about 2.1% is visible; the graph is concave down with a limit of 0.5 as height increases.

Step by step solution

01

Graph the Function for Part a

To graph the function, we will use the equation \(F = \frac{0.5h}{R+h}\). We define \(R = 6380\) kilometers. We'll plot \(F\) on the y-axis and \(h\) on the x-axis, with \(h\) ranging from 0 to 100,000 km. This allows us to visualize how the fraction of Earth's surface visible from space changes with altitude.
02

Find Height for Part b

To find the height \(h\) where \(F = 0.25\), set up the equation \(0.25 = \frac{0.5h}{6380+h}\). Solving this equation involves multiplying both sides by \(6380+h\) to isolate \(h\), leading to \(0.25(6380 + h) = 0.5h\). Simplifying, we get \(1595 + 0.25h = 0.5h\). Thus, \(h = \frac{1595}{0.25} = 6380\) km. So, 25% of Earth's surface is visible from a height of 6380 km.
03

Calculate Fraction for Part c

Substitute \(h = 280\) into the formula to find \(F\). Compute \(F = \frac{0.5 \times 280}{6380 + 280} = \frac{140}{6660} \approx 0.021\). Therefore, at a height of 280 km, about 2.1% of Earth's surface is visible.
04

Determine Concavity for Part d

The graph of \(F\) is concave down because the function \(F = \frac{0.5h}{R+h}\) is a rational function where the degree of the numerator and denominator are equal, leading to a slow increase in \(F\) as \(h\) increases. Practically, this represents diminishing visibility gains with increasing height.
05

Find Limiting Value for Part e

As \(h\to \infty\), the \(+R\) becomes negligible, so \(F \approx \frac{0.5h}{h} = 0.5\). Thus, the limiting fraction is 0.5, meaning at very large altitudes, at most 50% of Earth's surface can be seen regardless of further increase in height.

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 a function helps us visualize how it behaves across different inputs. For this exercise, the function is given as \( F = \frac{0.5h}{R+h} \), where \( R \) is the Earth's radius. To graph this function, we plot \( F \) on the y-axis and \( h \) on the x-axis, showing how much of Earth's surface is visible from a spacecraft at various altitudes.
  • The x-axis represents the height above the Earth's surface in kilometers.
  • The y-axis represents the fraction of Earth's surface visible from this height.
  • We want the graph to cover a range from 0 to 100,000 kilometers, allowing us to see wide variations at different altitudes.
The graph will show a curve starting at \( F = 0 \) when \( h = 0 \), which means nothing is visible from the Earth's surface; as \( h \) increases, the value of \( F \) approaches 0.5, indicating that more of Earth's surface becomes visible.
Rational Functions
Rational functions consist of ratios of polynomials. In this problem, the function \( F(h) = \frac{0.5h}{R+h} \) is a rational function. Understanding rational functions is crucial because they can exhibit interesting behaviors such as horizontal asymptotes, which are particularly relevant in this context.
  • The numerator of the function is \( 0.5h \), and the denominator is \( R+h \).
  • As \( h \) increases, especially when \( h \) is much larger than \( R \), the function simplifies to \( 0.5 \), leading to a horizontal asymptote.
  • This simplification shows how a rational function can stabilize to a constant value as one variable becomes much larger than others.
Recognizing these forms helps predict the function's behavior without complex calculations. Ultimately, it illustrates how astronauts see more of Earth from higher altitudes, but there is a limit to this increase.
Concavity
Concavity describes the direction in which a graph curves. The graph of the function \( F(h) = \frac{0.5h}{R+h} \) is concave down. This can be understood through a simple analysis of rational functions: when the degrees of the numerator and denominator are equal, the function generally shows decreasing marginal returns.
  • A function is concave down when it "bows" downwards.
  • This implies that the rate of increase in the visible fraction of Earth's surface decreases as height increases.
  • This has a practical implication: from increasing altitudes, each additional kilometer of distance results in progressively less increase in visible surface area.
Understanding concavity helps us to grasp why the rewards of visibility lessen at higher heights and why space observation behaves in such a dramatically different manner compared to close proximity to Earth.
Limiting Values
Limiting values are outcomes a function approaches as the input becomes very large or very small. In the case of \( F(h) = \frac{0.5h}{R+h} \), the limiting behavior as \( h \) approaches infinity is crucial.
  • As \( h \) increases, \( R \) becomes negligible compared to \( h \).
  • This simplifies the function to \( F(h) = \frac{0.5h}{h} = 0.5 \).
  • This means that no matter how high the spacecraft goes, it will never see more than 50% of Earth's surface.
The concept of limiting values in rational functions helps us understand the boundaries of function behavior. Here, it describes the maximum visibility astronauts can expect from any altitude, emphasizing that even technological advancements can't overcome certain natural limits.

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

In 1933, Riatt found that the length \(L\) of haddock in centimeters as a function of the age \(t\) in years is given approximately by the formula $$ L=53-42.82 \times 0.82^{t} . $$ a. Calculate \(L(4)\) and explain what it means. b. Compare the average yearly rate of growth in length from age 5 to 10 years with the average yearly rate of growth from age 15 to 20 years. Explain in practical terms what this tells you about the way haddock grow. c. What is the longest haddock you would expect to find anywhere?

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.

Radium 223 is a radioactive substance that itself is a product of the radioactive decay of thorium 227. For one experiment, the amount \(A\) of radium 223 present, as a function of the time \(t\) since the experiment began, is given by the formula \(A=3\left(e^{-0.038 t}-e^{-0.059 t}\right)\), where \(A\) is measured in grams and \(t\) in days. a. Make a graph of \(A\) versus \(t\) covering the first 60 days of the experiment. b. What was the largest amount of radium 223 present over the first 60 days of the experiment? c. What was the largest amount of radium 223 present over the first 10 days of the experiment? d. What was the smallest amount of radium 223 present over the first 60 days of the experiment?

For retailers who buy from a distributor or manufacturer and sell to the public, a major concern is the cost of maintaining unsold inventory. You must have appropriate stock to do business, but if you order too much at a time, your profits may be eaten up by storage costs. One of the simplest tools for analysis of inventory costs is the basic or der quantity model. It gives the yearly inventory expense \(E=E(c, N, Q, f)\) when the following inventory and restocking cost factors are taken into account: \- The carrying cost \(c\), which is the cost in dollars per year of keeping a single unsold item in your warehouse. \- The number \(N\) of this item that you expect to sell in 1 year. \- The number \(Q\) of items you order at a time. \- The fixed costs \(f\) in dollars of processing a restocking order to the manufacturer. (Note: This is not the cost of the order; the price of an item does not play a role here. Rather, \(f\) is the cost you would incur with any order of any size. It might include the cost of processing the paperwork, fixed costs you pay the manufacturer for each order, shipping charges that do not depend on the size of the order, the cost of counting your inventory, or the cost of cleaning and rearranging your warehouse in preparation for delivery.) The relationship is given by $$ E=\left(\frac{Q}{2}\right) c+\left(\frac{N}{Q}\right) f \text { dollars per year. } $$ A new-car dealer expects to sell 36 of a particular model car in the next year. It costs \(\$ 850\) per year to keep an unsold car on the lot. Fixed costs associated with preparing, processing, and receiving a single order from Detroit total \(\$ 230\) per order. a. Using the information provided, express the yearly inventory expense \(E=E(Q)\) as a function of \(Q\), the number of automobiles included in a single order. b. What is the yearly inventory expense if 3 cars at a time are ordered? c. How many cars at a time should be ordered to make yearly inventory expenses a minimum? d. Using the value of \(Q\) you found in part c, determine how many orders to Detroit will be placed this year. e. What is the average rate of increase in yearly inventory expense from the number you found in part \(\mathrm{c}\) to an order of 2 cars more?

The background for this exercise can be found in Exercises 11, 12, 13, and \(14 \mathrm{in} \mathrm{Sec}\) tion 1.4. A manufacturer of widgets has fixed costs of \(\$ 700\) per month, and the variable cost is \(\$ 65\) per thousand widgets (so it costs \(\$ 65\) to produce 1 thousand widgets). Let \(N\) be the number, in thousands, of widgets produced in a month. a. Find a formula for the manufacturer's total cost \(C\) as a function of \(N\). b. The highest price \(p\), in dollars per thousand widgets, at which \(N\) can be sold is given by the formula \(p=75-0.02 \mathrm{~N}\). Using this, find a formula for the total revenue \(R\) as a function of \(N\). c. Use your answers to parts a and \(b\) to find \(a\) formula for the profit \(P\) of this manufacturer as a function of \(N\). d. Use your formula from part c to determine the two break-even points for this manufacturer. Assume that the manufacturer can produce at most 500 thousand widgets in a month.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

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.