/*! 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 2 A commuter regularly drives 70 m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A commuter regularly drives 70 miles from home to work, and the amount of time required for the trip varies widely as a result of road and traffic conditions. The average speed for such a trip is a function of the time required. For example, if the trip takes 2 hours, then the average speed is \(\frac{70}{2}=35\) miles per hour. a. What is the average speed if the trip takes an hour and a half? b. Find a formula for the average speed as a function of the time required for the trip. (You need to choose variable and function names. Be sure to state the units.) c. Make a graph of the average speed as a function of the time required. Include trips from 1 hour to 3 hours in length. d. Is the graph concave up or concave down? Explain in practical terms what this means.

Short Answer

Expert verified
a. 46.67 mph b. \( S(t) = \frac{70}{t} \) c. Graph: Hyperbola, \( t \) from 1 to 3 d. Concave down; diminishing rate of speed decrease with time.

Step by step solution

01

Calculate Average Speed for 1.5 Hours

To find the average speed when the trip takes 1.5 hours, use the formula for average speed: \( \text{Average Speed} = \frac{\text{Total Distance}}{\text{Time}} \). Here, the distance is 70 miles and time is 1.5 hours. Thus, \( \text{Average Speed} = \frac{70}{1.5} = 46.67 \) miles per hour.
02

Define the Average Speed Function

Let the total time for the trip be represented as \( t \) (in hours). The average speed function, \( S(t) \), can be defined as \( S(t) = \frac{70}{t} \), where \( S(t) \) is in miles per hour.
03

Make the Graph

We plot \( S(t) = \frac{70}{t} \) for \( t \) ranging from 1 to 3 hours. The x-axis represents the time \( t \) in hours, and the y-axis represents the average speed \( S(t) \) in miles per hour. The curve will show a hyperbolic graph, starting from \( S(1) = 70 \) and approaching 23.33 miles per hour as \( t \) approaches 3 hours.
04

Analyze the Graph's Concavity

The graph of \( S(t) = \frac{70}{t} \) is concave down because the second derivative \( S''(t) = \frac{140}{t^3} \) is negative for \( t > 0 \). In practical terms, this means that as travel time increases, there is a diminishing rate of decrease in average speed.

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 91Ó°ÊÓ!

Key Concepts

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

Concavity of Functions
Understanding the concavity of a function can help you comprehend how the rate of change is behaving over time. In mathematical terms, a function is concave down when its second derivative is negative. For instance, in the function for the average speed of a trip, \( S(t) = \frac{70}{t} \), the second derivative \( S''(t) = \frac{140}{t^3} \) is negative when \( t > 0 \). This implies that as time \( t \) increases, the rate of change of the average speed slows down.

In practical terms, this means that if a trip takes longer, each additional hour doesn't result in as significant a decrease in speed as the previous hour did. The curve of the graph flattens out as time increases, showing that while your speed is still decreasing, it is doing so at a slower rate.

Key Points to Remember:
  • The function is concave down if the second derivative is negative.
  • Concavity provides insights into the rate of change's acceleration or deceleration.
  • In this problem, it reflects the diminishing reduction in speed as time increases.
Distance-Time Relationship
The relationship between distance and time is fundamental when calculating speed. Speed measures how quickly distance is covered over a particular period, and it's commonly expressed in units like miles per hour (mph).

For this exercise, the total distance of the commute is 70 miles. If we let \( t \) represent the time in hours, the average speed \( S \) can be described by the function \( S(t) = \frac{70}{t} \). This formula shows that if you know how long the journey takes, you can compute the average speed.

Important Aspects:
  • A shorter travel time \( t \) increases average speed \( S \).
  • A longer travel time \( t \) decreases average speed \( S \).
  • Speed is inversely proportional to time when distance is constant.
This reciprocal linkage allows us to draw conclusions about speed from the time taken, providing a clear glance at how variability in trip duration affects overall travel efficiency.
Graphing Functions
Graphing is a powerful tool when analyzing functions, as it transforms abstract equations into visual data. For the average speed function \( S(t) = \frac{70}{t} \), plotting on a graph helps illustrate how speed changes as time varies.

Steps for Graphing:
1. Choose a range of time \( t \), here from 1 to 3 hours.
2. Calculate \( S(t) \) for various \( t \)-values within the range.
3. Plot these points on a graph where the x-axis is \( t \) (time in hours) and the y-axis is \( S(t) \) (average speed in mph).

The Resulting Graph:
  • Shows a hyperbolic curve, starting high at short durations and flattens over longer periods.
  • Illustrates the inverse relationship: as time increases, average speed decreases.
  • Visually demonstrates the concave down nature by the decreasing slope as \( t \) increases.
This visual representation aids in grasping how speed is sensitive to changes in travel time, facilitating a more intuitive understanding of the relationship between these variables.

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

The quantity \(S\) of barley, in billions of bushels, that barley suppliers in a certain country are willing to produce in a year and offer for sale at a price \(P\), in dollars per bushel, is determined by the relation $$ P=1.9 S-0.7 $$ The quantity \(D\) of barley, in billions of bushels, that barley consumers are willing to purchase in a year at price \(P\) is determined by the relation $$ P=2.8-0.6 D . $$ The equilibrium price is the price at which the quantity supplied is the same as the quantity demanded. Find the equilibrium price for barley.

The economic order quantity model tells a company how many items at a time to order so that inventory costs will be minimized. The number \(Q=Q(N, c, h)\) of items that should be included in a single order depends on the demand \(N\) per year for the product, the fixed cost \(c\) in dollars associated with placing a single order (not the price of the item), and the carrying cost \(h\) in dollars. (This is the cost of keeping an unsold item in stock.) The relationship is given by $$ Q=\sqrt{\frac{2 N c}{h}} $$ a. Assume that the demand for a certain item is 400 units per year and that the carrying cost is \(\$ 24\) per unit per year. That is, \(N=400\) and \(h=24\). i. Find a formula for \(Q\) as a function of the fixed ordering \(\operatorname{cost} c\), and plot its graph. For this particular item, we do not expect the fixed ordering costs ever to exceed \(\$ 25\). ii. Use the graph to find the number of items to order at a time if the fixed ordering cost is \$6 per order. iii. How should increasing fixed ordering cost affect the number of items you order at a time? b. Assume that the demand for a certain item is 400 units per year and that the fixed ordering cost is \(\$ 14\) per order. i. Find a formula for \(Q\) as a function of the carrying cost \(h\) and make its graph. We do not expect the carrying cost for this particular item ever to exceed \(\$ 25\). ii. Use the graph to find the optimal order size if the carrying cost is \(\$ 15\) per unit per year. iii. How should an increase in carrying cost affect the optimal order size? iv. What is the average rate of change per dollar in optimal order size if the carrying cost increases from \(\$ 15\) to \(\$ 18 ?\) v. Is this graph concave up or concave down? Explain what that tells you about how optimal order size depends on carrying costs.

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?

Recall that the APR (the annual percentage rate) is the percentage rate on a loan that the Truth in Lending Act requires lending institutions to report on loan agreements. It does not tell directly what the interest rate really is. If you borrow money for 1 year and make no payments, then in order to calculate how much you owe at the end of the year, you must use another interest rate, the EAR (the effective annual rate), which is not normally reported on loan agreements. The calculation is made by adding the interest indicated by the EAR to the amount borrowed. The relationship between the APR and the EAR depends on how often interest is compounded. If you borrow money at an annual percentage rate APR (as a decimal), and if interest is compounded \(n\) times per year, then the effective annual rate EAR (as a decimal) is given by $$ \mathrm{EAR}=\left(1+\frac{\mathrm{APR}}{n}\right)^{n}-1 $$ For the remainder of this problem, we will assume an APR of \(10 \%\). Thus in the formula above, we would use \(0.1\) in place of APR. a. Would you expect a larger or a smaller EAR if interest is compounded more often? Explain your reasoning. b. Make a table that shows how the EAR depends on the number of compounding periods. Use your table to report the EAR if interest is compounded once each year, monthly, and daily. (Note: The formula will give the EAR as a decimal. You should report your answer as a percent with three decimal places.) c. If you borrow \(\$ 5000\) and make no payments for 1 year, how much will you owe at the end of a year if interest is compounded monthly? If interest is compounded daily? d. If interest is compounded as often as possiblethat is, continuously - then the relationship between APR and EAR is given by $$ \mathrm{EAR}=e^{\mathrm{APR}}-1 \text {. } $$ Again using an APR of 10\%, compare the EAR when the interest is compounded monthly with the EAR when the interest is compounded continuously.

The amount of growth of plants in an ungrazed pasture is a function of the amount of plant biomass already present and the amount of rainfall. \({ }^{9}\) For a pasture in the arid zone of Australia, the formula \(Y=-55.12-0.01535 N-0.00056 N^{2}+3.946 R\) gives an approximation of the growth. Here \(R\) is the amount of rainfall, in millimeters, over a 3 -month period; \(N\) is the plant biomass, in kilograms per hectare, at the beginning of that period; and \(Y\) is the growth, in kilograms per hectare, of the biomass over that period. (For comparison, 100 millimeters is about \(3.9\) inches, and 100 kilograms per hectare is about 89 pounds per acre.) For this exercise, assume that the amount of plant biomass initially present is 400 kilograms per hectare, so \(N=400\). a. Find a formula for the growth \(Y\) as a function of the amount \(R\) of rainfall. b. Make a graph of \(Y\) versus \(R\). Include values of \(R\) from 40 to 160 millimeters. c. What happens to \(Y\) as \(R\) increases? Explain your answer in practical terms. d. How much growth will there be over a 3 -month period if initially there are 400 kilograms per hectare of plant biomass and the amount of rainfall is 100 millimeters?

See all solutions

Recommended explanations on Math Textbooks

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.