Biocalculus: Calculus For Life Sciences
15th Edition
ISBN: 9781305177291
Author: James Stewart, Troy Day
Publisher: Cengage Learning
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Chapter 1.1, Problem 12E
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To Explain: the effect of crowing with the use of given graph.
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Chapter 1 Solutions
Biocalculus: Calculus For Life Sciences
Ch. 1.1 - Prob. 1ECh. 1.1 - Prob. 2ECh. 1.1 - Prob. 3ECh. 1.1 - Prob. 4ECh. 1.1 - Prob. 5ECh. 1.1 - Prob. 6ECh. 1.1 - Prob. 7ECh. 1.1 - Prob. 8ECh. 1.1 - Prob. 9ECh. 1.1 - Prob. 10E
Ch. 1.1 - Prob. 11ECh. 1.1 - Prob. 12ECh. 1.1 - Prob. 13ECh. 1.1 - Prob. 14ECh. 1.1 - Prob. 15ECh. 1.1 - Prob. 16ECh. 1.1 - Prob. 17ECh. 1.1 - Prob. 18ECh. 1.1 - Prob. 19ECh. 1.1 - Prob. 20ECh. 1.1 - Prob. 21ECh. 1.1 - Prob. 22ECh. 1.1 - Prob. 23ECh. 1.1 - Prob. 24ECh. 1.1 - Prob. 25ECh. 1.1 - Prob. 26ECh. 1.1 - Prob. 27ECh. 1.1 - Prob. 28ECh. 1.1 - Prob. 29ECh. 1.1 - Prob. 30ECh. 1.1 - Prob. 31ECh. 1.1 - Prob. 32ECh. 1.1 - Prob. 33ECh. 1.1 - Prob. 34ECh. 1.1 - Prob. 35ECh. 1.1 - Prob. 36ECh. 1.1 - Prob. 37ECh. 1.1 - Prob. 38ECh. 1.1 - Prob. 39ECh. 1.1 - Prob. 40ECh. 1.1 - Prob. 41ECh. 1.1 - Prob. 42ECh. 1.1 - Prob. 43ECh. 1.1 - Prob. 44ECh. 1.1 - Prob. 45ECh. 1.1 - Prob. 46ECh. 1.1 - Prob. 47ECh. 1.1 - Prob. 48ECh. 1.1 - Prob. 49ECh. 1.1 - Prob. 50ECh. 1.1 - Prob. 51ECh. 1.1 - Prob. 52ECh. 1.1 - Prob. 53ECh. 1.1 - Prob. 54ECh. 1.1 - Prob. 55ECh. 1.1 - Prob. 56ECh. 1.1 - Prob. 57ECh. 1.1 - Prob. 58ECh. 1.1 - Prob. 59ECh. 1.1 - Prob. 60ECh. 1.1 - Prob. 61ECh. 1.1 - Prob. 62ECh. 1.1 - Prob. 63ECh. 1.1 - Prob. 64ECh. 1.1 - Prob. 65ECh. 1.1 - Prob. 66ECh. 1.1 - Prob. 67ECh. 1.1 - Prob. 68ECh. 1.1 - Prob. 69ECh. 1.1 - Prob. 70ECh. 1.1 - Prob. 71ECh. 1.1 - Prob. 72ECh. 1.1 - Prob. 73ECh. 1.1 - Prob. 74ECh. 1.2 - Prob. 1ECh. 1.2 - Prob. 2ECh. 1.2 - Prob. 3ECh. 1.2 - Prob. 4ECh. 1.2 - Prob. 5ECh. 1.2 - Prob. 6ECh. 1.2 - Prob. 7ECh. 1.2 - Prob. 8ECh. 1.2 - Prob. 9ECh. 1.2 - Prob. 10ECh. 1.2 - Prob. 11ECh. 1.2 - Prob. 12ECh. 1.2 - Prob. 13ECh. 1.2 - Prob. 14ECh. 1.2 - Prob. 15ECh. 1.2 - Prob. 16ECh. 1.2 - Prob. 17ECh. 1.2 - Prob. 18ECh. 1.2 - Prob. 19ECh. 1.2 - Prob. 20ECh. 1.2 - Prob. 21ECh. 1.2 - Prob. 22ECh. 1.2 - Prob. 23ECh. 1.2 - Prob. 24ECh. 1.2 - Prob. 25ECh. 1.2 - Prob. 26ECh. 1.2 - Prob. 27ECh. 1.3 - Prob. 1ECh. 1.3 - Prob. 2ECh. 1.3 - Prob. 3ECh. 1.3 - Prob. 4ECh. 1.3 - Prob. 5ECh. 1.3 - Prob. 6ECh. 1.3 - Prob. 7ECh. 1.3 - Prob. 8ECh. 1.3 - Prob. 9ECh. 1.3 - Prob. 10ECh. 1.3 - Prob. 11ECh. 1.3 - Prob. 12ECh. 1.3 - Prob. 13ECh. 1.3 - Prob. 14ECh. 1.3 - Prob. 15ECh. 1.3 - Prob. 16ECh. 1.3 - Prob. 17ECh. 1.3 - Prob. 18ECh. 1.3 - Prob. 19ECh. 1.3 - Prob. 20ECh. 1.3 - Prob. 21ECh. 1.3 - Prob. 22ECh. 1.3 - Prob. 23ECh. 1.3 - Prob. 24ECh. 1.3 - Prob. 25ECh. 1.3 - Prob. 26ECh. 1.3 - Prob. 27ECh. 1.3 - Prob. 28ECh. 1.3 - Prob. 29ECh. 1.3 - Prob. 30ECh. 1.3 - Prob. 31ECh. 1.3 - Prob. 32ECh. 1.3 - Prob. 33ECh. 1.3 - Prob. 34ECh. 1.3 - Prob. 35ECh. 1.3 - Prob. 36ECh. 1.3 - Prob. 37ECh. 1.3 - Prob. 38ECh. 1.3 - Prob. 39ECh. 1.3 - Prob. 40ECh. 1.3 - Prob. 41ECh. 1.3 - Prob. 42ECh. 1.3 - Prob. 43ECh. 1.3 - Prob. 44ECh. 1.3 - Prob. 45ECh. 1.3 - Prob. 46ECh. 1.3 - Prob. 47ECh. 1.3 - Prob. 48ECh. 1.3 - Prob. 49ECh. 1.3 - Prob. 50ECh. 1.3 - Prob. 51ECh. 1.3 - Prob. 52ECh. 1.3 - Prob. 53ECh. 1.3 - Prob. 54ECh. 1.3 - Prob. 55ECh. 1.3 - Prob. 56ECh. 1.3 - Prob. 57ECh. 1.3 - Prob. 58ECh. 1.3 - Prob. 59ECh. 1.3 - Prob. 1PCh. 1.3 - Prob. 2PCh. 1.3 - Prob. 3PCh. 1.4 - Prob. 1ECh. 1.4 - Prob. 2ECh. 1.4 - Prob. 3ECh. 1.4 - Prob. 4ECh. 1.4 - Prob. 5ECh. 1.4 - Prob. 6ECh. 1.4 - Prob. 7ECh. 1.4 - Prob. 8ECh. 1.4 - Prob. 9ECh. 1.4 - Prob. 10ECh. 1.4 - Prob. 11ECh. 1.4 - Prob. 12ECh. 1.4 - Prob. 13ECh. 1.4 - Prob. 14ECh. 1.4 - Prob. 15ECh. 1.4 - Prob. 16ECh. 1.4 - Prob. 17ECh. 1.4 - Prob. 18ECh. 1.4 - Prob. 19ECh. 1.4 - Prob. 20ECh. 1.4 - Prob. 21ECh. 1.4 - Prob. 22ECh. 1.4 - Prob. 23ECh. 1.4 - Prob. 24ECh. 1.4 - Prob. 25ECh. 1.4 - Prob. 26ECh. 1.4 - Prob. 27ECh. 1.4 - Prob. 28ECh. 1.4 - Prob. 29ECh. 1.4 - Prob. 30ECh. 1.4 - Prob. 31ECh. 1.4 - Prob. 32ECh. 1.4 - Prob. 33ECh. 1.4 - Prob. 34ECh. 1.4 - Prob. 35ECh. 1.4 - Prob. 36ECh. 1.4 - Prob. 37ECh. 1.4 - Prob. 38ECh. 1.5 - Prob. 1ECh. 1.5 - Prob. 2ECh. 1.5 - Prob. 3ECh. 1.5 - Prob. 4ECh. 1.5 - Prob. 5ECh. 1.5 - Prob. 6ECh. 1.5 - Prob. 7ECh. 1.5 - Prob. 8ECh. 1.5 - Prob. 9ECh. 1.5 - Prob. 10ECh. 1.5 - Prob. 11ECh. 1.5 - Prob. 12ECh. 1.5 - Prob. 13ECh. 1.5 - Prob. 14ECh. 1.5 - Prob. 15ECh. 1.5 - Prob. 16ECh. 1.5 - Prob. 17ECh. 1.5 - Prob. 18ECh. 1.5 - Prob. 19ECh. 1.5 - Prob. 20ECh. 1.5 - Prob. 21ECh. 1.5 - Prob. 22ECh. 1.5 - Prob. 23ECh. 1.5 - Prob. 24ECh. 1.5 - Prob. 25ECh. 1.5 - Prob. 26ECh. 1.5 - Prob. 27ECh. 1.5 - Prob. 28ECh. 1.5 - Prob. 29ECh. 1.5 - Prob. 30ECh. 1.5 - Prob. 31ECh. 1.5 - Prob. 32ECh. 1.5 - Prob. 33ECh. 1.5 - Prob. 34ECh. 1.5 - Prob. 35ECh. 1.5 - Prob. 36ECh. 1.5 - Prob. 37ECh. 1.5 - Prob. 38ECh. 1.5 - Prob. 39ECh. 1.5 - Prob. 40ECh. 1.5 - Prob. 41ECh. 1.5 - Prob. 42ECh. 1.5 - Prob. 43ECh. 1.5 - Prob. 44ECh. 1.5 - Prob. 45ECh. 1.5 - Prob. 46ECh. 1.5 - Prob. 47ECh. 1.5 - Prob. 48ECh. 1.5 - Prob. 49ECh. 1.5 - Prob. 50ECh. 1.5 - Prob. 51ECh. 1.5 - Prob. 52ECh. 1.5 - Prob. 53ECh. 1.5 - Prob. 54ECh. 1.5 - Prob. 55ECh. 1.5 - Prob. 56ECh. 1.5 - Prob. 57ECh. 1.5 - Prob. 58ECh. 1.5 - Prob. 59ECh. 1.5 - Prob. 60ECh. 1.5 - Prob. 61ECh. 1.5 - Prob. 62ECh. 1.5 - Prob. 63ECh. 1.5 - Prob. 64ECh. 1.5 - Prob. 65ECh. 1.5 - Prob. 66ECh. 1.5 - Prob. 67ECh. 1.5 - Prob. 68ECh. 1.5 - Prob. 69ECh. 1.5 - Prob. 70ECh. 1.5 - Prob. 1PCh. 1.5 - Prob. 2PCh. 1.5 - Prob. 3PCh. 1.5 - Prob. 4PCh. 1.5 - Prob. 5PCh. 1.6 - Prob. 1ECh. 1.6 - Prob. 2ECh. 1.6 - Prob. 3ECh. 1.6 - Prob. 4ECh. 1.6 - Prob. 5ECh. 1.6 - Prob. 6ECh. 1.6 - Prob. 7ECh. 1.6 - Prob. 8ECh. 1.6 - Prob. 9ECh. 1.6 - Prob. 10ECh. 1.6 - Prob. 11ECh. 1.6 - Prob. 12ECh. 1.6 - Prob. 13ECh. 1.6 - Prob. 14ECh. 1.6 - Prob. 15ECh. 1.6 - Prob. 16ECh. 1.6 - Prob. 17ECh. 1.6 - Prob. 18ECh. 1.6 - Prob. 19ECh. 1.6 - Prob. 20ECh. 1.6 - Prob. 21ECh. 1.6 - Prob. 22ECh. 1.6 - Prob. 23ECh. 1.6 - Prob. 24ECh. 1.6 - Prob. 25ECh. 1.6 - Prob. 26ECh. 1.6 - Prob. 27ECh. 1.6 - Prob. 28ECh. 1.6 - Prob. 29ECh. 1.6 - Prob. 30ECh. 1.6 - Prob. 31ECh. 1.6 - Prob. 32ECh. 1.6 - Prob. 33ECh. 1.6 - Prob. 34ECh. 1.6 - Prob. 35ECh. 1.6 - Prob. 36ECh. 1.6 - Prob. 37ECh. 1.6 - Prob. 38ECh. 1.6 - Prob. 39ECh. 1.6 - Prob. 40ECh. 1.6 - Prob. 1PCh. 1.6 - Prob. 2PCh. 1.6 - Prob. 3PCh. 1.6 - Prob. 4PCh. 1 - Prob. 1CCCh. 1 - Prob. 2CCCh. 1 - Prob. 3CCCh. 1 - Prob. 4CCCh. 1 - Prob. 5CCCh. 1 - Prob. 6CCCh. 1 - Prob. 7CCCh. 1 - Prob. 8CCCh. 1 - Prob. 9CCCh. 1 - Prob. 10CCCh. 1 - Prob. 11CCCh. 1 - Prob. 12CCCh. 1 - Prob. 13CCCh. 1 - Prob. 14CCCh. 1 - Prob. 15CCCh. 1 - Prob. 16CCCh. 1 - Prob. 1TFQCh. 1 - Prob. 2TFQCh. 1 - Prob. 3TFQCh. 1 - Prob. 4TFQCh. 1 - Prob. 5TFQCh. 1 - Prob. 6TFQCh. 1 - Prob. 7TFQCh. 1 - Prob. 8TFQCh. 1 - Prob. 9TFQCh. 1 - Prob. 10TFQCh. 1 - Prob. 11TFQCh. 1 - Prob. 12TFQCh. 1 - Prob. 1ECh. 1 - Prob. 2ECh. 1 - Prob. 3ECh. 1 - Prob. 4ECh. 1 - Prob. 5ECh. 1 - Prob. 6ECh. 1 - Prob. 7ECh. 1 - Prob. 8ECh. 1 - Prob. 9ECh. 1 - Prob. 10ECh. 1 - Prob. 11ECh. 1 - Prob. 12ECh. 1 - Prob. 13ECh. 1 - Prob. 14ECh. 1 - Prob. 15ECh. 1 - Prob. 16ECh. 1 - Prob. 17ECh. 1 - Prob. 18ECh. 1 - Prob. 19ECh. 1 - Prob. 20ECh. 1 - Prob. 21ECh. 1 - Prob. 22ECh. 1 - Prob. 23ECh. 1 - Prob. 24ECh. 1 - Prob. 25ECh. 1 - Prob. 26ECh. 1 - Prob. 27ECh. 1 - Prob. 28ECh. 1 - Prob. 29ECh. 1 - Prob. 30ECh. 1 - Prob. 31ECh. 1 - Prob. 32ECh. 1 - Prob. 33ECh. 1 - Prob. 34ECh. 1 - Prob. 35ECh. 1 - Prob. 36ECh. 1 - Prob. 37ECh. 1 - Prob. 38ECh. 1 - Prob. 39ECh. 1 - Prob. 40ECh. 1 - Prob. 1CSCh. 1 - Prob. 2CSCh. 1 - Prob. 3CSCh. 1 - Prob. 4CSCh. 1 - Prob. 5CSCh. 1 - Prob. 6CSCh. 1 - Prob. 7CSCh. 1 - Prob. 8CSCh. 1 - Prob. 9CS
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- Modelling the species area relationship that table gives the area of several caves in Central Mexico and the number of bat species that lives in each cave.* (a) Find the power function that model that data. (b) Draw a graph of the function you found in part (a) and a scatter plot of the data on the same graph .does the model fit that that are well? (c) the cave called EI Saponear Puebla,mexico,has a surface area of A=205m2 .use the model to estimate the number of bed species you would expect to find in that cave.arrow_forwardSpawner-Recruit Model In fish management it is important to know the relationship between the abundance of the spawners also called the parent stock and the abundance of the recruitsthat is, those hatchlings surviving to maturity. According to the Ricker model, the number of recruits R as a function of the number of spawners P has the form R=APeBp for some positive constants A and B. This model describes well a phenomenon observed in some fisheries: A large spawning group can actually lead to a small group of recruits. In a study of the sockeye salmon, it was determined that A=4 and B=0.7. Here we measure P and R in thousands of salmon. a. Make a graph of R versus P for the sockeye salmon. Assume there are at most 3000 spawners. b. Find the maximum number of salmon recruits possible. c. If the number of recruits R is greater than the number of spawners P, then the difference R-P of the recruits can be removed by fishing, and next season there will once again be P spawners surviving to renew the cycle. What value of P gives the maximum value of R-P, the number of fish available for removal by fishing?arrow_forwardMaximum Sales Growth This is a continuation of Exercise 10. In this exercise, we determine how the sales level that gives the maximum growth rate is related to the limit on sales. Assume, as above, that the constant of proportionality is 0.3, but now suppose that sales grow to a level of 4 thousand dollars in the limit. a. Write an equation that shows the proportionality relation for G. b. On the basis of the equation from part a, make a graph of G as a function of s. c. At what sales level is the growth rate as large as possible? d. Replace the limit of 4 thousand dollars with another number, and find at what sales level the growth rate is as large as possible. What is the relationship between the limit and the sales level that gives the largest growth rate? Does this relationship change if the proportionality constant is changed? e. Use your answers in part d to explain how to determine the limit if we are given sales data showing the sales up to a point where the growth rate begins to decrease.arrow_forward
- Sales Growth In this exercise, we develop a model for the growth rate G, in thousands of dollars per year, in sales of the product as a function of the sales level s, in thousands of dollars. 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 difference this limit and the current sales level. For example, if we expect sales grow to 3 thousand dollars in the long run, then 3-s is 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 the formula for unattained sales. b.Write an equation that shows the proportionality relation for G. c.On the basis of the equation from the 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?arrow_forwardPopulation Growth The projected population of the United States for the years 2025 through 2055 can be modeled by P=307.58e0.0052t, where P is the population (in millions) and t is the time (in years), with t=25 corresponding to 2025. (a) Use a graphing utility to graph the function for the years 2025 through 2055. (b) Use the table feature of the graphing utility to create a table of values for the same time period as in part (a). (c) According to the model, during what year will the population of the United States exceed 430 million?arrow_forwardGrazing Kangaroos The amount of vegetation eaten in a day by a grazing animal V of food available measured as biomass, in units such as pounds per acre. This relationship is called the functional response. If there is little vegetation available, the daily intake will be small, since the animal will have difficulty finding and eating the food. As the amount of food biomass increases, so does the daily intake. Clearly, though, there is a limit to the amount the animal will eat, regardless of the amount of food available. This maximum amount eaten is the satiation level. a.For the western grey kangaroo of Australia, the functional response is G=2.54.8e0.004V, where G=G(V) is the daily intake measured in pounds and V is the vegetation biomass measured in pounds per acre. i. Draw a graph of G against V. Include vegetation biomass levels up to 2000 pounds per acre. ii. Is the graph you found in part i concave up or concave down? Explain in practical terms what your answer means about how this kangaroo feeds. iii. There is a minimal vegetation biomass level below which the western grey kangaroo will eat nothing. Another way of expressing this is to say that the animal cannot reduce the food biomass below this level. Find this minimal level. iv. Find the satiation level for the western grey kangaroo. b. For the red kangaroo of Australia, the functional response is R=1.91.9e0.033V, Where R is the daily intake measured in pounds and V is the vegetation biomass measured in pounds per acre. i. Add the graph of R against V to the graph of G you drew in part a. ii. A simple measure of the grazing efficiency of an animal involves the minimal vegetation biomass level described above: The lower the minimal level for an animal, the more efficient it is at grazing. Which is more efficient at grazing, the western grey kangaroo or the red kangaroo?arrow_forward
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