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7-1 Integral as change Using FTC.

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1 7-1 Integral as change Using FTC

2 Change in position π‘Ž π‘₯ 𝑣(𝑑) 𝑑𝑑 gives the change in position from a to x sec. If the initial condition for the position is s 0 =4, then the final position is s(x) = __ 𝑑𝑑. (FTC) The integral is the change in position (displacement). Final position = Initial position + change (displacement)

3 problem A car’s velocity in m/s is modeled by 𝑣 𝑑 = 𝑒 𝑑 cos t. At t = o the car is 25 m north of home traveling north. Give the car’s position at t = 10 s.

4 Finding total distance traveled
0 10 |𝑣(𝑑)| 𝑑𝑑 This is easy to do on a calculator. But, by hand, you must find the zeros and integrate each section individually. Then make β€œnegative areas” positive and add up all parts. v(t) t cccccccccc β§Ύ β§Ώ

5 problem A car’s velocity in m/s is modeled by 𝑣 𝑑 = 𝑒 𝑑 cos t. Give the total distance that the car travels from t = 0 to 10 sec.

6 Change in velocity This is similar to the change-in-position problem.
0 8 π‘Ž 𝑑 𝑑𝑑 gives the change in velocity during the first 8 sec of acceleration. To get the actual final velocity, you have to add the initial velocity v(0) to the change (integral). FTC

7 Consumption p. 383: To get the total change in consumption,
integrate the rate. Usually the starting time is zero (t = 0) and often the starting amount consumed is also 0. That’s an initial condition. Note p. 383 #5. This a two-year period, the beginning of to the end of 1973. Think of the end of β€˜73 as the beginning of 1974: 1974 – 1972 = 2. Do not subtract the original years; reason it out.

8 Work Work = Force Β· distance π‘Š= π‘Ž 𝑏 𝐹(π‘₯) 𝑑π‘₯
If 𝐹=π‘˜π‘₯, then W = π‘Ž 𝑏 π‘˜π‘₯ 𝑑π‘₯ . a and b represent stretch Problem The spring constant k of a spring is 4 N/cm. Give the work done in stretching the spring out 5 cm from its original length. force distance

9 Discuss 385: 1, 2 phase line problems 386: 8 group, 23 group
10, 17, 27

10 Homework 386: 2, 6, 8, 9, 11-16, 20-22, 30a, 32, 33, 35


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