Calculus: Single and Multivariable

Calculus: Single and Multivariable

Title: Calculus: Single and Multivariable

Author: Deborah Hughes-Hallett, Andrew M. Gleason, William G. McCallum, Daniel E. Flath, Patti Frazer Lock, Thomas W. Tucker, David O. Lomen, David Lovelock, David Mumford, Brad G. Osgood, Douglas Quinney, Karen Rhea, Jeff Tecosky-Feldman

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Week #10 - The Integral Section 5.3 From “Calculus, Single Variable” by Hughes-Hallett, Gleason, McCallum et. al. Copyright 2005 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc. SUGGESTED PROBLEMS 1. If f (t) is measured in dollars per year and t is measured in years, what are the units of b f (t)dt? a The units are the product of the the function, f (t), and dt...

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#10 Week - The Integral Section 5.3 From “Calculus, Single Variable” by Hughes-Hallett, Gleason, McCallum et. al. Copyright 2005 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc. SUGGESTED PROBLEMS 1. If f (t) is measured in dollars per year and t is measured in years, what are the units of b f (t)dt? a The units are the product of the the function, f (t), and dt (or ∆t): (dollars / year) · (years) = dollars. In Exercises 4-7, explain in words what the integral represents and give units. 6 5. 0 a(t)dt, where a(t) is acceleration in km/hr2 and t is time in hours. As seen in class, the integral of velocity is the net change in position, so we can infer that the integral of acceleration is the net change in velocity. In particular, the net change in velocity under the acceleration a(t) during the time t = 0 to t = 6. The units bear this out: (km/hr2 ) · (hr) = km/hr. 3 25. (a) Using Figure 5.42, estimate −3 f (x)dx. (b) Which of the following average values of f (x) is larger? (i). Between x = −3 and x = 3 (ii). Between x = 0 and x = 3 Figure 5.42 (a) The integral is the area above the x-axis minus the area below the x−axis. By 3 counting squares, we can estimate that −3 f (x)dx is about −6 + 2 = −4 (the negative of the area from t = −3 to t = 1 plus the area from t = 1 to t = 3.) (b) Since the integral in part (a) is negative, the average value of f (x) between x = −3 and x = 3 is negative. From the graph, however, it appears that the average value of f (x) from x = 0 to x = 3 is positive. Hence (ii) is the larger quantity. 1 34. A bicyclist pedals along a straight road with velocity, v, given in Figure 5.45. She starts 5 miles from a lake; positive velocities take her away from the lake and negative velocities take her toward the lake. When is the cyclist farthest from the lake, and how far away is she then? Figure 5.45 First, it helps to interpret the graph. Each t-axis square represents 1/6th of an hour, or 10 minutes. Each square then represents a distance of (10 miles per hour) ·(1/6 hour = 5 3 miles. For the first ten minutes, the graph is below the t axis, so her velocity is negative, meaning she is moving towards the lake. At exactly the 20 minute mark, her velocity is zero, so she has stopped. From 20-60 minutes, the graph is above the t-axis, so her velocity is positive, or she is moving away from the lake. At the 60 minute mark, her velocity drops to zero, so she is stopped again. Remember that the area between the graph above the t-axis represents distance traveled away from the lake. Since the area below the t-axis represents distance traveled towards the lake, while the area above represents distance away from the lake, it is clear from the larger area above the axis that the cyclist is much further away from the lake at the end of the trip than when she started. Thus during her trip, she is furthest away from the lake at t = 1 hour. Computing the actual distance from the lake combines the fact that she starts 5 miles away, cycles towards the lake for the first twenty minutes, and then cycles away. We can estimate each distance by counting the squares between the graph and the t-axis: • First 20 minutes: ≈ 1.5 squares = 1.5 · 53 = 2.5 miles towards the lake • Last 40 minutes: ≈ 7.5 squares, = 7.5 · 53 = 12.5 miles away from the lake In total, she will have gone from 5 miles away to ≈ 5 − 2.5 + 12.5 = 15 miles away from the lake when she stops at the end of the hour. We could write this more mathematically as 1/3 1 D =5+ 0 v dt + 1/3 v dt ≈ 15 Note that the limits are in units of hours (1/3 hour instead 20 minutes), and that the first integral would have a negative value because the velocity is negative from t = 0 to t = 1/3. 35. A car speeds up at a constant rate from 10 to 70 mph over a period of half an hour. Its fuel efficiency (in miles per gallon) at various speeds is shown in the table. Make lower and upper estimates of the quantity of fuel used during the half hour. 2 Speed (mph) Fuel efficiency (mpg) 10 15 20 18 30 21 40 23 50 24 60 25 70 26 This question is complicated by the fact that the table gives efficiency as miles per gallon as a function of velocity. We want to know the total consumption, in gallons. If the velocity were constant, we could compute distance (miles) efficiency (miles per gallon) consumption (gallons) = Unfortunately, the velocity is changing with time, so the best we can do is approximate the consumption over small intervals of time. Since the acceleration is from 10 mph to 70 mph over half an hour, we are gaining 2 mph per minute, or 10 mph every five minutes. Since the efficiency ratings are given every 10 mph, we will choose ∆t = 5 minutes = 1/12 hour. On the first five minutes, we can approximate the consumption by saying v(0) = 10, so efficiency(v) = 15 mpg and the distance traveled = v ∗ ∆t = 10(1/12) = 5/6 miles. Thus the consumption in the first five minutes could be approximated as (5/6 miles) / (15 mpg) = 1/18 gallons. Repeating this over next the few five minute intervals we can build the following table: dist(v) consumption 10 10 ≈ 0.0556 0 10 15 6 6 · 18 10 10 5 20 18 ≈ 0.0926 6 6 · 18 15 15 ≈ 0.1190 10 30 21 6 6 · 21 20 20 15 40 23 ≈ 0.1449 6 6 · 23 25 25 ≈ 0.1736 20 50 24 6 6 · 24 30 30 25 60 25 ≈ 0.2000 6 6 · 25 35 35 ≈ 0.2244 30 70 26 6 6 · 26 Using left-hand sums, or the values of t = 5, 10, . . . , 25, we would compute the approximate consumption as 0.79 gallons. Using right-hand sums, we would estimate 0.95 gallons were consumed. There are several other ways to obtain estimates of the consumption, depending on how you mix + match the velocity and efficiencies at each end of the 5 minute intervals. This is merely one example. To get better approximations, you would ideally have more detailed efficiency tables. 36. Height velocity graphs are used by endocrinologists to follow the progress of children with growth deficiencies. Figure 5.46 shows the height velocity curves of an average boy and an average girl between age 3 and 18. (a) Which curve is for girls and which is for boys? Explain how you can tell. 3 t v(t) eff(v) (b) About how much does the average boy grow between ages 3 and 10? (c) The growth spurt associated with adolescence and the onset of puberty occurs between ages 12 and 15 for the average boy and between ages 10 and 12.5 for the average girl. Estimate the height gained by each average child during this growth spurt. (d) When fully grown, about how much taller is the average man than the average woman? (The average boy and girl are about the same height at age 3.) Figure 5.46 (a) The area under each curve represents the change in height in centimeters. Since men are generally taller than women, the curve with the larger area under it is the height velocity of boys. Thus the black curve is for boys, the blue one is for girls. (b) Each square below the height velocity curve has area 1 cm/yr · 1 yr = 1 cm. Counting squares lying below the black curve gives about 43 cm. Thus, on average, boys grow about 43 cm between ages 3 and 10. (c) Counting squares lying below the black curve gives about 23 cm growth for boys during their growth spurt. Counting squares lying below the blue curves gives about 18 cm for girls during their growth spurt. (d) We can measure the difference in growth by counting squares that lie between the two curves. Between age 2 and 12.5, the average girl grows faster than the average boy. Counting squares yields about 5 cm between the blue and black curves for 2 ≤ x ≤ 12.5. Counting squares between the curves for 12.5 ≤ x ≤ 18 gives about 18 squares. Thus there is a net increase of boys over girls by about 18 - 5 = 13 cm. QUIZ PREPARATION PROBLEMS 17. The following table gives the emissions, E, of nitrogen oxides in millions of metric tons per year in the US.4 Let t be the number of years since 1940 and E = f (t). 50 (a) What are the units and meaning of 0 50 f (t)dt? (b) Estimate 0 f (t)dt. 1950 9.4 1960 13.0 1970 18.5 4 1980 20.9 1990 19.6 Year E 1940 6.9 50 (a) The integral 0 f (t)dt represents the total emission of nitrogen oxides, in millions of metric tons, during the period 1940 to 1990. (b) We estimate the integral using left- and right-hand sums: Left sum = (6.9)(10) + (9.4)(10) + (13.0)(10) + (18.5)(10) + (20.9)(10) = 687 Right sum = (9.4)(10) + (13.0)(10) + (18.5)(10) + (20.9)(10) + (19.6)(10) = 814 We average the left-a dn right-hand sums to find the best estimate of the integral: 50 687 + 814 f (t)dt ≈ = 750.5 million metric tons. 2 0 20. A warehouse charges its customers $5 per day for every 10 cubic feet of space used for storage. Figure 5.37 records the storage used by one company over a month. How much will the company have to pay? Figure 5.37 The area under the curve represents the number of cubic feet of storage times the number of days the storage was used. This area is given by Area under graph = Area of rectangle + Area of triangle 1 = 30 · 10, 000 + · 30(30, 000 − 10, 000) 2 = 600, 000. Since the warehouse charges $5 for every 10 cubic feet of storage used for a day, the (5)(600, 000) company will have to pay =$300,000. 10 23. (a) Using Figures 5.39 and 5.40, find the average value on 0 ≤ x ≤ 2 of (i). f (x) (ii). g(x) (iii). f (x) · g(x) (b) Is the following statement true? Explain your answer. Average(f )· Average(g) = Average(f · g) Figure 5.39 5 Figure 5.40 (a) (i) Since the triangular region under the graph of f (x) has area 1/2, we have 1 2−0 2 Average(f ) = (ii) Similarly, Average(g) = f (x)dx = 0 1 11 ·= 22 4 1 2−0 2 g(x)dx = 0 1 11 ·= 22 4 (iii) Since f (x) is nonzero only for 0 ≤ x < 1 and g(x) is nonzero only for 1 < x ≤ 2, the product f (x)g(x) = 0 for all x. Thus, 1 2−0 2 Average(f · g) = f (x)g(x)dx = 0 1 2 2 0 dx = 0 0 (b) Since the average values of f (x) and g(x) are nonzero, their product is nonzero. Thus the left side of the statement is nonzero. However, the average of the product f (x)g(x) is zero. Thus, the right side of the statement is zero, so the statement is not true. 6

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