THEOREM 1 Net Change as the Integral of a Rate
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Transcript of THEOREM 1 Net Change as the Integral of a Rate
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2 1't
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s t dt s t s t
THEOREM 1 Net Change as the Integral of a Rate The net change in s(t) over an interval [t1, t2] is given by the integral
Integral of a Rate of Change
1 2 the Net Change of over r ,os s t t
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4 2
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34 liters
2 5 2 5
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t dt t dt
tt
Water leaks from a tank at a rate of 2 + 5t liters/hour, where t is the number of hours after 7 AM. How much water is lost between 9 and 11 AM?
Traffic Flow The number of cars per hour passing an observation point along a highway is called the traffic flow rate q(t) (in cars per hour).
(a) Which quantity is represented by the integral 2
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.t
t
q t dt
1 2
The total number of cars passing through the observation point
during the interval , .t t
rate
Traffic Flow The number of cars per hour passing an observation point along a highway is called the traffic flow rate q (t) (in cars per hour).
(b) The flow rate is recorded at 15-min intervals between 7:00 and 9:00 AM. Estimate the number of cars using the highway during this 2-hour period.
8 0.25 1044 1297 1478 1844 1451 1378 1155 802 2612.25L
8 0.25 1297 1478 1844 1451 1378 1155 802 542 2486.75R
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Equivalent to a Trap Rule (Chapter 8)
2550 cars
Let s(t) be the position at time t of an object in linear motion. Then the object’s velocity is v(t) = s (t), and the integral of v(t) is equal to the net change in position or displacement over a time interval [t1, t2]:
We must distinguish between displacement and distance traveled. If you travel 10 km and return to your starting point, your displacement is zero but your distance traveled is 20 km. To compute distance traveled rather than displacement, integrate the speed |v (t)|.
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The Integral of Velocity
2 2
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2 1't t
t t
v t dt s t dt s t s t Displacement or net
change in position
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THEOREM 2 The Integral of Velocity For an object in linear motion with velocity v(t),
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1 2Displacement during ,t
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t t v t dt
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1 2Distance traveled during ,t
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t t v t dt
A particle has velocity v (t) = t3 −10t2 + 24t m/s. Compute:(a) Displacement over [0, 6]
6
3 2
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64 32
0
Displacement 10 2
36
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3m
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t t t dt
t tt
4 63 2 3 2
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4 43 2 3 2
0 6
4 44 3 4 32 2
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Total Distance 10 24 10 24
10 24 10 24
10 1012 12
4 3 4 3
128 12836
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t t t dt t t t dt
t t t dt t t t dt
t t t tt t
A particle has velocity v (t) = t3 −10t2 + 24t m/s. Compute:
(b) Total distance traveled over [0, 6]
63 2
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Total Distance 10 24
10 24 10 24 6 4
t t t dt
t t t t t t t t t
0 4 6
Use a sign analysis to break up the integral.
Consider the cost function C(x) of a manufacturer (the dollar cost of producing x units of a particular product or commodity). The derivative C (x) is called the marginal cost. The cost of increasing production from a to b is the net change C(b) − C(a), which is equal to the integral of the marginal cost:
Total versus Marginal Cost
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The marginal cost of producing x computer chips (in units of 1000) is
2' 300 4000 40,000 (dollars per thousand chips).C x x x
(a) Find the cost of increasing production from 10,000 to 15,000 chips.
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15 10 300 4000 40,000
100 2000 40,000
487,500 300,000 $187,500
C C x x dx
x x x
Cost of increasing production from to 'b
a
a b C x dx
Total versus Marginal Cost
The marginal cost of producing x computer chips (in units of 1000) is
2' 300 4000 40,000 (dollars per thousand chips).C x x x
15 0 487,5 $517,50000C C
(b) Determine the total cost of producing 15,000 chips, assuming that it costs $30,000 to set up the manufacturing run [that is, C(0) = 30,000].
Consider the cost function C(x) of a manufacturer (the dollar cost of producing x units of a particular product or commodity). The derivative C (x) is called the marginal cost. The cost of increasing production from a to b is the net change C(b) − C(a), which is equal to the integral of the marginal cost:
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Cost of increasing production from to 'b
a
a b C x dx
15 10 487,500 300,000 $187,500C C