theorem 1 net change as the integral of a rate

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THEOREM 1 Net Change as the Integral of a Rate The net change in s ( t ) over an interval [ t 1 , t 2 ] is given by the integral. Water leaks from a tank at a rate of 2 + 5 t liters/hour , where t is the number of hours after 7 AM. How much water is lost between 9 and 11 AM?. - PowerPoint PPT Presentation

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Page 1: THEOREM  1    Net  Change as the Integral of a Rate
Page 2: THEOREM  1    Net  Change as the Integral of a Rate

2

1

2 1't

t

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

FTC

Page 3: THEOREM  1    Net  Change as the Integral of a Rate

4 2

2 4

22

4

34 liters

2 5 2 5

52

2

4 10 8 40

14 48

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?

Page 4: THEOREM  1    Net  Change as the Integral of a 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).

(a) Which quantity is represented by the integral 2

1

.t

t

q t dt

1 2

The total number of cars passing through the observation point

during the interval , .t t

rate

Page 5: THEOREM  1    Net  Change as the Integral of a 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

2

1

1

22549.5

t

n n

t

q t dt R L

?t

Equivalent to a Trap Rule (Chapter 8)

2550 cars

Page 6: THEOREM  1    Net  Change as the Integral of a Rate

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)|.

'

The Integral of Velocity

2 2

1 1

2 1't t

t t

v t dt s t dt s t s t Displacement or net

change in position

FTC

Page 7: THEOREM  1    Net  Change as the Integral of a Rate

THEOREM 2 The Integral of Velocity For an object in linear motion with velocity v(t),

2

1

1 2Displacement during ,t

t

t t v t dt

2

1

1 2Distance traveled during ,t

t

t t v t dt

Page 8: THEOREM  1    Net  Change as the Integral of a Rate

A particle has velocity v (t) = t3 −10t2 + 24t m/s. Compute:(a) Displacement over [0, 6]

6

3 2

0

64 32

0

Displacement 10 2

36

4

01

3m

12

4

t t t dt

t tt

Page 9: THEOREM  1    Net  Change as the Integral of a Rate

4 63 2 3 2

0 4

4 43 2 3 2

0 6

4 44 3 4 32 2

0 6

Total Distance 10 24 10 24

10 24 10 24

10 1012 12

4 3 4 3

128 12836

148

33m

3

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

0

3 2 2

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.

Page 10: THEOREM  1    Net  Change as the Integral of a Rate

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

'

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.

152

10

153 2

10

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

Page 11: THEOREM  1    Net  Change as the Integral of a Rate

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:

'

Cost of increasing production from to 'b

a

a b C x dx

15 10 487,500 300,000 $187,500C C

Page 12: THEOREM  1    Net  Change as the Integral of a Rate
Page 13: THEOREM  1    Net  Change as the Integral of a Rate