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7-1 Integral as change Using FTC
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Change in position π π₯ π£(π‘) ππ‘ 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) = __ ππ‘. (FTC) The integral is the change in position (displacement). Final position = Initial position + change (displacement)
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problem A carβs velocity in m/s is modeled by π£ π‘ = π π‘ cos t. At t = o the car is 25 m north of home traveling north. Give the carβs position at t = 10 s.
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Finding total distance traveled
0 10 |π£(π‘)| ππ‘ This is easy to do on a calculator. But, by hand, you must find the zeros and integrate each section individually. Then make βnegative areasβ positive and add up all parts. v(t) t cccccccccc β§Ύ β§Ώ
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problem A carβs velocity in m/s is modeled by π£ π‘ = π π‘ cos t. Give the total distance that the car travels from t = 0 to 10 sec.
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Change in velocity This is similar to the change-in-position problem.
0 8 π π‘ ππ‘ 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
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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. Thatβs 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 β73 as the beginning of 1974: 1974 β 1972 = 2. Do not subtract the original years; reason it out.
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Work Work = Force Β· distance π= π π πΉ(π₯) ππ₯
If πΉ=ππ₯, then W = π π ππ₯ ππ₯ . 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
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Discuss 385: 1, 2 phase line problems 386: 8 group, 23 group
10, 17, 27
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Homework 386: 2, 6, 8, 9, 11-16, 20-22, 30a, 32, 33, 35
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