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M 112 Short Course in Calculus. Chapter 5 – Accumulated Change: The Definite Integral Sections 5.1 – Distance and Accumulated Change V. J. Motto. Suppose a car is moving with increasing velocity and suppose we measure the car's velocity every two seconds, obtaining the data in Table 5.1:.
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M 112 Short Course in Calculus Chapter 5 – Accumulated Change: The Definite Integral Sections 5.1 – Distance and Accumulated Change V. J. Motto
Suppose a car is moving with increasing velocity and suppose we measure the car's velocity every two seconds, obtaining the data in Table 5.1: At most how far has the car traveled? At least how far has the car traveled? the data in Table 5.1:
Figure 5.2: Shaded area estimates distance traveled. Velocity measured every 2 seconds
Figure 5.3: Shaded area estimates distance traveled. Velocity measured every 2 seconds
If the velocity is positive, the total distance traveled is the area under the velocity curve. Figure 5.4: Velocity measured every ½ second Figure 5.5: Velocity measured every ¼ second Figure 5.6: Distance traveled is area under curve
Problem 10 • The rate of change of the world’s population, in millions of people per year, is given in the following table. • Use this data to estimate the total change in the world’s • population between 1950 and 2000. • The world population was 2555 million people in 1950 and 6085 million people in 2000. Calculate the true value of the total change in the population. How does this compare with your estimate in part (a)?
Solution to 10-a: • Based on the data, we will calculate the underestimate and the overestimate of the total change. A good estimate will be the average of both results. • Underestimate of total change = 37 · 10 + 41 · 10 + 77 · 10 + 77 · 10 + 79 · 10 = 3110. • 77 was considered twice since we needed to calculate the area under the graph. • Overestimate of total change = 41 · 10 + 78 · 10 + 78 · 10 + 86 · 10 + 86 · 10 = 3690. • 78 and 86 were considered twice since we needed to calculate the area over the graph. • The average is: (3110 + 3690)/2 = 3400 million people.
Solution 10-b: The total change in the world’s population between 1950 and 2000 is given by the difference between the populations in those two years. That is, the change in population equals 6085 (population in 2000) − 2555 (population in 1950) = 3530 million people. Our estimate of 3400 million people and the actual difference of 3530 million people are close to each other, suggesting our estimate was a good one.
Problem 15 • Two cars start at the same time and travel in the same direction along a straight road. Figure 5.11 gives the velocity, v, of each car as a function of time, t. Which car: • Attains the larger maximum velocity? • Stops first? • Travels farther? Figure 5.11
Solution to 15: Part a Car A has the largest maximum velocity because the peak of car A’s velocity curve is higher than the peak of B’s. Part b Car A stops first because the curve representing its velocity hits zero (on the t-axis) first. Part c. Car B travels farther because the area under car B’s velocity curve is the larger.
