Applications from Science and Statistics · Copyright © 2007 Pearson Education, Inc. Publishing as...

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

8.5 Applications from

Science and Statistics

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall Slide 7- 2

What you’ll learn about… n  Work Revisited n  Fluid Force and Fluid Pressure n  Normal Probabilities

…and why It is important to see

applications of integrals as various accumulation

functions

EQ: What are some of the applications for using integrals?


Finding the Work Done by a Force Work is defined as a force applied through a distance.

W = Fd

When the force is constant, this is an easy problem. The force needed

to raise a 5-lb weight 10 feet is 5 x 10 = 50 ft-lb.



Finding the Work Done by a Force

The problem arises when the force is variable. Consider a 10-ft chain that weighs 5 lb/ft and lies coiled on the ground. To take an end of the chain and lift it so that it is fully extended (in other words, to a height of 10 feet), the weight is constantly changing. The weight of the chain can be expressed as 5y, where y is the height off the ground of the end that is being raised. The work done to raise the chain 10 feet can be calculated by:

€

5y dy0

10

∫ =5y2

20

10

= 250− 0 = 250 ft - lb



Example Finding the Work Done by a Force ( )Find the work done by the force ( ) sin newtons along the

1-axis from 0 meters to meter.2

F x x

x x x

π=

= =


( )

( )

( )

12

0

12

0

1 cos

sin

1 0 1

1

x

W x dx

ππ

π

π

π

∫

#− %&

=

=

= − −

=

€

sin(πx)dx01/ 2∫

N-m


Example Work Done Lifting A heavy rope, 50 ft. long, weighs 0.5 lb/ft and hangs over the edge of a building 120 ft. high. How much work is done pulling the rope to the top of the building? (Stewart, © 1997, 1998)

Weight of rope: 0.5(50 - x) lbs, varies as rope rises. Work done:

€

0.5(50 − x)[ ]0

50∫ dx

= 25x − x2

4

$

% &

'

( )

0

50

= 1250 − 625( ) − 0 − 0( )

= 625 ft − lb


Alternate solution:

€

0.5x0

50

∫ dx

=x2

4

#

$ %

&

' (

0

50

= 625− 0( ) = 625 ft − lb


Another Example Work Done Lifting A cable that weighs 2 lb/ft is used to lift 800 lbs of coal up a mineshaft 500 ft deep. Find the work done. (Stewart, © 1997, 1998)

Weight of coal: 800 lbs, constant. Weight of cable: 2(500 - x) lbs, varies as coal rises. Work done:

€

800 + 2(500 − x)[ ]0

500∫ dx

= 1800x − x2[ ]0

500= 900000 − 250000( ) − 0 − 0( )

= 650,000 ft − lb



p.429 #1-4

Slide 7- 8


Yet Another Example Work Done Lifting

A leaky bucket weighs 15 newtons (N) empty. It is lifted fromthe ground at a constant rate to a point 10 m above the ground by a rope weighing 0.5 N/m. The bucket starts with 70 N of water,but it leaks at a constant rate and just finishes draining as the bucket reaches the top. Find the amount of work done

lifting the bucket alone; lifting the water alone; lifting the rope alone; li

(a)(b)(c)(d) fting the bucket, water, and rope together.

( ) ( )

Since the bucket's weight is constant, you must exert a forceof 15 N through the entire 10-meter interval.W= 15 N 10 m 150 N m=150 J× = ⋅

(a)




(b) The force needed to lift the water is equal to the water's weight, which decreases steadily from 70 N to 0 N over the10-m lift. The work done is:

€

W = 70− 7x( )dx0

10∫ = 70x − 7x2

2$

% &

'

( )

0

10

= 350J.


W = 70− 70x10

"

#$

%

&'

0

10

∫ dx



The total work is the sum: 25+350+150=525 J(d)

The force needed to lift the rope also varies, starting at(0.5)(10)=5 N when the bucket is on the ground and 0 Nwhen the bucket is is at the top. The rate of decrease is constant. At elevation

(c)

( )( )10

0

10

0

25 -0.25

meters, the (10 - ) meters of ropestill there to lift weigh ( ) 0.5 10 - N.

0.5 10 -

=25 J.

x x

x xF x x

W x dx∫

" #$ %

=

=

=

€

W = .5 10− x( )dx0

10∫

= 5x − x2

4$

% &

'

( )

0

10

= 25J.



Another Yet Another Example Work Done Lifting A bucket of sand weighing 100 lbs is hoisted 10 ft by a cable of negligible

weight. As the bucket rises, sand pours out of a hole in the bottom at a constant rate and only 80 lbs of sand is left when it reaches the top. How much work is done? (Simon ©1982) Weight of bucket: 100-2x lbs. Work done =

€

100− 2x[ ]0

10

∫ dx = 100x − x2[ ]0

10

= 1000−100( )− 0− 0( ) = 900 ft - lb

(follow-up) Same problem, but now take into account that the cable weighs 5lbs/ft. How much work is done? (assume 10 ft of cable) (Simon ©1982) Weight of cable: 5(10 - x) lbs. Work done =

€

50 − 5x[ ]0

10∫ dx = 50x − 5x2

2

$

% &

'

( )

0

10

= 500 − 250( ) − 0 − 0( ) = 250 ft - lb

So, the total work done to lift the sand and cable would be 1150 ft-lb.



Assignment: worksheet

Slide 7- 13


Finding the Work Done by a Force: Springs

A spring is compressed 3 cm from its natural state by a force of 9 N. How much work is done in stretching it 5 cm from its natural state?


3x dx0

5∫ =

3x2

2 0

5

=752− 0 = 75

2 N-cm

the force it takes to compress or stretch a spring x units from its natural (unstressed) length is a constant times x. In symbols: F = kx.

€

9 = 3k ⇒ k = 3

F = kx


Another Example Work Done: Springs

It takes a force of 10 N to stretch a spring 2 cm beyond its natural state. How much work is done in stretching the spring 4 cm from its natural state?

F = kx 10 = k(2) k = 5

€

W = Fd = 5x dx0

4

∫=

5x2

20

4

= 40− 0 = 40 N ⋅cm



Yet Another Example Work Done: Springs A force of 750 pounds compresses a spring 3 in beyond its natural length of

15 inches. Find the work done in compressing the spring an additional 3 in.

F = kx 750 = k(3) k = 250

€

W = Fd = 250x dx3

6

∫= 125x2

3

6= 4500−1125 = 3375 inch− pounds



Another Yet Another Example Work Done: Springs

A force of 250 newtons stretches a spring 30 cm. How much work is done in stretching the spring from 20 cm to 50 cm?

F = kx 250 = k(30) k = 25/3

€

W = Fd =25x

3dx

20

50

∫=

25x2

620

50

=62500

6−

100006

=52500

6 N - cm = 87.5 N - m

Source: Larson/Hostetler 6th ed.



Assignment: p. 429 #5-10

Slide 7- 18


Example Work Done Pumping

A rectangular tank with a base 4 feet by 5 feet and a height of 4 feet is full of water. The water weighs 62.4 pounds per cubic foot. How much work is done in pumping the water out over the top edge in order to empty all of the tank? (Larson, © 1998)

Weight of water: 62.4(20 dy) lbs, constant. Work done: 62.4(20 dy) y, where y is the height lifted.

62.4 20( )(4− y)⎡⎣ ⎤⎦0

4∫ dy

= 4992−1248 y2

2⎡

⎣⎢

⎤

⎦⎥0

4

= 9984( )− 0( )

= 9984 ft − lb


4 ft

5 ft

4 ft



A rectangular tank with a base 4 feet by 5 feet and a height of 4 feet is full of water. The water weighs 62.4 pounds per cubic foot. How much work is done in pumping the water out over the top edge in order to empty all of the tank? Variation: There is only 3 feet of water in the tank. What changes?

Weight of water: 62.4(20 dy) lbs, constant. Work done: 62.4(20 dy) (4 - y), where y is the height of the water.

€

62.4 20( ) 4 − y( )[ ]0

3

∫ dy

= 9360 ft − lb


4 ft

5 ft

4 ft



A cylindrical tank with a radius of 4 feet and a height of 10 feet is full of oil that weighs 50 pounds per cubic foot. How much work is done in pumping the water out over the top edge in order to empty all of the tank?

Weight of oil: 50π(42)dy lbs, constant. Work done: 50π(42)(10 - y)dy , where (10 - y) is the height lifted.

€

50π 16( ) 10 − y( )[ ]0

10

∫ dy ≈125,663.7 ft - lb


4 ft

10 ft



A cylindrical tank with a radius of 4 feet and a height of 10 feet is full of oil that weighs 50 pounds per cubic foot. How much work is done in pumping the water out over the top edge in order to empty all of the tank? Variation: The tank is only half-full. What changes?

Weight of oil: 50π(42)dy lbs, constant. Work done: 50π(42)(10 - y)dy , where (10 - y) is the height lifted.

€

50π 16( ) 10 − y( )[ ]0

5

∫ dy ≈ 94,247.8 ft - lb


4 ft

10 ft



A tank in the shape of a right circular cone is full of water that weighs 62.4 pounds per cubic foot. The vertex of the tank is at the bottom.The tank is 8 feet across the top and 6 feet high. How much work is done in pumping the water out over the top edge in order to empty all of the tank?

€

62.4π 23

y#

$ %

&

' (

2

6 − y( )*

+ , ,

-

. / / 0

6

∫ dy ≈ 9,409.7 ft - lb


€

rh

=46⇒ r =

23

h

Area of each slice : π 23

h$

% &

'

( )

2



A tank in the shape of a right circular cone is full of water that weighs 62.4 pounds per cubic foot. The vertex of the tank is at the bottom.The tank is 8 feet across the top and 6 feet high. How much work is done in pumping the water out over the top edge in order to empty all of the tank? Variation: The water in the tank is only 2 feet deep. What changes?

€

62.4π 23

y#

$ %

&

' (

2

6 − y( )*

+ , ,

-

. / / 0

2

∫ dy ≈1,045.5 ft - lb


€

rh

=46⇒ r =

23

h

Area of each slice : π 23

h$

% &

'

( )

2


p.430 #17, 18, 20-22

Slide 7- 25


Fluid Force and Fluid Pressure

In any liquid, the (force per unit area) atdepth is , where is the weight-density (weightper unit volume) of the liquid.

ph p wh w=

fluid pressure


Example Fluid Force on a Vertical Surface

A vertical side of a tank is in the shape of a rectangle that is 4 feet long and 3 feet high. Assuming that the tank is full of water, what is the fluid force on the vertical side?

4 feet

3 feet

62.4 y( )4dy0

3∫ =1123.2



A vertical side of a tank is in the shape of a trapezoid that is 4 feet long on one base and 2 feet long on the other, and is 3 feet high.. Assuming that the tank is full of water, what is the fluid force on the vertical side?

€

62.4 3− y( )(2) y + 33

#

$ %

&

' ( dy

0

3

∫ = 41.6 9 − y2( ) dy0

3

∫

= 41.6 9y + −y3

3

*

+ ,

-

. /

0

3

= 41.6 27 − 9 − 0 − 0( )[ ]= 748.8

4 feet

3 feet

2 feet

€

y = 3 x −1( ) = 3x − 3

x =y + 3

3



A vertical gate in a dam has the shape of an isosceles trapezoid 8 feet across the top and 6 feet across the bottom, with a height of 5 feet, as shown below. What is the fluid force on the gate if the top of the gate is 3 feet below the surface of the water?

€

62.4 2( ) y + 235

"

# $

%

& ' 0 − y( ) dy

−8

−3

∫ =11752

€

y + 8 = 5 x − 3( ) = 5x − 23

x =y + 23

58 feet

5 feet

6 feet

3 feet


p.431 #25 & 26

Slide 7- 30


Normal Probability Density Function (pdf)

( ) ( )2 22

The fora population with mean and standard deviation is

1( ) .2

xf x e µ σ

µ σ

σ π− −=

normal probability density function (Gaussian curve)

€

f (x) =1

σ 2πe− χ−µ( )2 /(2σ 2 )

p. 423


The 68-95-99.7 Rule for Normal Distributions

Given a normal curve, 68% of the area will lie within of the mean , 95% of the area will lie within 2 of the mean , 99.7% of the area will lie within 3 of the mean .

σ µ

σ µ

σ µ

•

•

•

p. 423


Example A Telephone Help Line

Suppose a telephone help line takes a mean of 1.5 minutes to answer a call. If the standard deviation is 0.2, then 68% of the calls are answered in the range of 1.3 to 1.7 minutes and 99.7% of the calls are answered in the range of 0.9 to 2.1 minutes.


Example A Tribe of Pygmies

The average height of a certain tribe of pygmies is 3.2 feet, with a standard deviation of 0.2 feet. What is the probability that a member of the tribe chosen at random is between 3.5 and 4 feet tall, given that the height of the tribe is normally distributed?

10.2 2π

e−(x−3.2)2 /0.08

Find the area under the curve of f (x) from 3.5 to 4:

Assume a normal pdf will model these probabilities.

10.2 2π

e− x−3.2( )2 /0.08

3.5

4

∫ dx ≈ 0.067 = 6.7%


Example An IQ Test

Repeated measurements of a student's IQ yield a mean of 135, with a standard deviation of 5. What is the probability that the student has an IQ between 132 and 138?

€

f ( x) =1

5 2πe−(x−135)2 / 50

Find the area under the curve of f (x) from 132 to 138:


15 2π

e− x−135( )2 /50

132

138

∫ dx ≈ 0.45149 = 45.1%


Example A Psychology Test

It is known that subjects score an average of 100 points on a new personality test. If the standard deviation is 10 points, what percentage of all subjects will score between 75 and 80?

€

f ( x) =1

10 2πe−(x−100)2 / 200

Find the area under the curve of f (x) from 75 to 80:


110 2π

e− x−100( )2 /200

75

80

∫ dx ≈ 0.01654 =1.7%


Example An IQ Test

Professor Easy's students earned an average grade of 3.5, with a standard deviation of 0.2. What percentage of his students earned between 3.5 and 3.9?

€

f ( x) =1

0.2 2πe−(x−3.5)2 / 0.08

Find the area under the curve of f (x) from 3.5 to 3.9:


10.2 2π

e− x−3.5( )2 /0.08

3.5

3.9

∫ dx ≈ 0.47725= 47.7%


Assignment: p. 431 #27-31

Slide 7- 38


Example Weights of Coffee Cans

Suppose that coffee cans marked as "8 ounces" of coffee havea mean weight of 8.2 ounces and a standard deviation of 0.3 ounces.

What percentage of all such cans can be expected to weigh between 8 a(a)

nd 9 ounces? What percentage would we expect to weigh less than 8 ounces? What is the probability that a can weighs exactly 8 ounces?

(b)(c)

( )28.2 / 0.18

Assume a normal pdf will model these probabilities.1( )

0.3 2 Find the area under the curve of ( ) from 8 to 9:

NINT( ( ), ,8,9) 0.744 About 74.4% of the cans will have between 8 and

xf x e

f xf x x

π− −=

≈

(a)

9 ounces.

€

f (x) =1

0.3 2πe−( x−8.2)2 / 0.18


Example Weights of Coffee Cans

The graph of ( ) approaches the -axis as an asymptote. When 5, the graph of ( ) is very close to the -axis.NINT( ( ), ,5,8) 0.252. Expect about 25.2% of the boxes to weigh less than 8 ounc

f x xx f x xf x x=

≈

(b)

es.

This would be the integral from 8 to 8, which is zero.(c)



A trough with an end in the shape of a semicircle is full of water that weighs 62.4 pounds per cubic foot.The trough is 50 feet long and the radius of the ends is 7 feet. The water weighs 62.4 pounds per cubic foot. How much work is done in pumping the water out over the top edge in order to empty all of the trough?

€

62.4 100( ) 49 − y2( ) 0 − y( )# $ %

& ' ( −7

0

∫ dy ≈ 713,440ft - lb


€

Area of each slice : 50 2 49 - y2( )



A trough with an end in the shape of a semicircle is full of water that weighs 62.4 pounds per cubic foot.The trough is 50 feet long and the radius of the ends is 7 feet. The water weighs 62.4 pounds per cubic foot. How much work is done in pumping the water out over the top edge in order to empty all of the trough?

€

62.4 100( ) 14y − y2( ) 7 − y( )# $ %

& ' ( 0

7

∫ dy ≈ 713,440ft - lb


€

Area of each slice : 50 2 14y - y2( )


Example Work Done: Emptying tanks of liquid

A spherical tank of radius 8 feet is half full of oil that weighs 50 pounds per cubic foot. Find the work required to pump oil out through a hole in the top of the tank.

Source: Larson/Hostetler 6th ed.

€

x2 + 8 − y( )2

= 82

x2 + 64 −16y + y2 = 64

x = 16y − y2

8

x

y

16

8

We need to find a way to express the radius at any point as the oil level drops.

So, the weight of each "slice" is given by:

€

50 π 16y − y2( )2$

% &

'

( ) dy


Example Work Done: Emptying tanks of liquid

A spherical tank of radius 8 feet is half full of oil that weighs 50 pounds per cubic foot. Find the work required to pump oil out through a hole in the top of the tank.

€

W = Fd = 50 π 16 y − y2( )2$

% &

'

( ) 16 − y( )dy

0

8

∫= 50π 16y − y2( ) 16 − y( )dy

0

8

∫= 50π y3 − 32y2 + 256y( )dy

0

8

∫= 50π y 4

4−

32y3

3+128y2

+

, -

.

/ 0

0

8

≈ 589,782 foot - poundsSource: Larson/Hostetler 6th ed. Source: Larson/Hostetler 6th ed.

8

x

y

16

8

We will integrate from y = 0 to y = 8.

Distance lifted will be 16 - y


Example The Great Molasses Flood of 1919

At 1:00 pm on January 15, 1919, a 90-ft-high, 90-foot-diameter cylindrical metal tank in which the Puritan Distilling Company stored molasses at the corner of Foster and Commercial streets in Boston’s North End exploded. Molasses flooded the streets 30 feet deep, trapping pedestrians and horses, knocking down buildings and oozing into homes. It was eventually tracked all over town and even made its way into the suburbs via trolley cars and people’s shoes. It took weeks to clean up. (a) Given that the tank was full of molasses weighing 100 lb/ft3, what was the

total force exerted by the molasses on the bottom of the tank at the time it ruptured?

(b) What was the total force against the bottom foot-wide band of the tank wall?



( )

( )

3 2

2

2

At the bottom of the tank, the molasses exerted a constant lb lbpressure of 100 90 ft 9000 .ft ft

The area of the base was 45 and the total force on the base

lbwas 9000 2025 ftft

p wh

π

π

" #= = =$ %& '

" #$ %& '

(a)

( )2 57,225,526 lb.≈



2

Partition the band from depth 89 ft to depth 90 ft into narrowerbands of width and choose a depth in each one. The pressureat this depth is 100 lb/ft . The force against each na

k

k k

y yy p wh yΔ

= =

(b)

( )( )rrow

band is approximately pressure area 100 90

9000 lb.Add the forces against all bands in the partition and take the limit

k

k

y yy y

π

π

× = Δ

= Δ

90 90

89 89

as the norms go to zero. The force against the bottom foot of tank wall is:9000 9000 2,530,553 lb.F ydy ydyπ π∫ ∫= = ≈


Probability Density Function (pdf)

[ ]

-

A is a function ( ) with domainall reals such that ( ) 0 for all and ( ) 1.

Then the probability associated with an interval , is

( ) .b

a

f xf x x f x dx

a b

f x dx

∞

∞∫

∫

≥ =

probability density function

€

f (x)−∞

∞

∫ dx =1.

€

f (x)a

b∫ dx.

p. 422

Applications from Science and Statistics · Copyright © 2007 Pearson Education, Inc. Publishing as...

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