applidifferentiation_part2_iry14

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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Chapter 4Applications of Derivatives (Part 2)



4 - 2

Subtopics

Error Analysis

Indeterminate Forms and L’ Hopital’s Rule

Newton Raphson Method

Taylor/Maclaurin polynomial

© 2012, 2007 Pearson. All rights reserved.



Error Analysis



4 - 5

Error Analysis (cont)

If two quantities a and b are measured withabsolute error a and b respectively, thena)Error for p = a b The largest possible value is

p+ p = (a+b)+(a + b)

The smallest possible value is p- p = (a+b)-(a + b)

p = a + b



4 - 6

Cont’: Example ( p=a+b)

a) Two lines AB and BC, part of a straight

line AC are measured as

AB = 120 0.005m ; BC = 300 0.005m

AC = (120 + 300) (0.005 + 0.005)

= 420 0.01m

The true value of AC could be anywhere

from 419.99 to 420.01m

A B C


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Cont’ : Example ( p=a-b)

b) If a line PQ is measured 280.00 0.005m

and PR (part of the line PQ) is measured

as 114.57 0.005m,

RQ = PQ-PR

RQ = (280 – 114.57) (0.005+0.005)

= 165.43 0.01

P QR



Cont’ b) Error for p = ab

The largest possible value is p+ p = (a+a)(b+b)=ab+ab+ba+ab

The smallest possible value is

p- p = (a-a)(b-b)=ab-(ab+ba)+ab

b

b

a

a

p

p r p

Negligible p = ba + ab

abba

abab

p pr p

p=abRelative

error



c) Error for p=a/b

For relative error, it is the same as the case

where p = ab, thus

Copyright © 2007 Pearson Education, Inc. Publishing as

Pearson Addison-Wesley

2babba p

b

b

a

a

p

pr p



Cont’: Example ( p=ab)

c) A rectangular block of land is measured as 20 0.005m by 40 0.005m. Find p the area, therelative error in p, and the absolute error in p,ie. p .

Area, p =20x40=800m2

40

005.0

20

005.0

b

b

a

ar p

000375.0000125.000025.0 pr

p

pr p

2

3.0000375.0800 m pr p p

b

b

a

a

p

p

r p



Error analysis



Measurement,

p

Absolute error

, p

Measurement

p = a + b p =a +b

p = a - b p =a +b

p = ab

p = a/b

p = ba + ab

2b

abba p

b

b

a

a

p

p r p

b

b

a

a

p

pr p



cont’: Try This! (p=a/b)

The voltage drop, V , across a resistor is

measured as 12.0 0.05 volts, and the current I

is measured as 2.74 0.005 amps. Given that

resistance R = V/I ,find (i) R ; (ii) the relative error in R and (iii) R ,

the absolute error in R.

Ans: (i)4.3796ohms; (ii)0.0059914; (iii)0.0262ohms

b b

a a

p p r p

p

p r p



4 - 13

(i)

(ii)

(iii)



Solution:



4 - 14

Error for y = f ( x)

If x is measured with a maximum error of x and y is calculated from x using a

formula y = f ( x), then the maximum error y in y can be found by considering the graphof y = f ( x)



4 - 15

The gradient at the point P is d y/d x.

If x is small, then

and we have

dx

dp

x

p

x

dx

dp p



4 - 16

Example

A circular oval has its radius measured as 63.660.005m.Calculate its area and the maximum error in its area, andthe perimeter and its maximum error

Area, A =r 2= (63.66)2=12731.6sq.m.

r

dr

dA

r A

2

2

00.2)005.0)(66.63(22 r r r dr

dA A

..00.260.12731 m sq A

xdx

dp p ||

Ans:399.990.03m



4 - 17

Cont’

Perimeter, P :



m r 99.39966.6322

03142.099.399

03142.0)005.0(2

2

P

r dr

dP P

dr

dP



4 - 18

Example (Try This!)

The radius of a circle is increased from 2.00 to

2.20m.

a) Estimate the resulting change, A in area, A.

b) Express the estimate as a percentage of

the circle’s original area.

r

dr

dA A ||



4 - 19

Answer

a)

b)



5133.2)2.0)(2(22

2

2.022.2

2

r r A

r dr

dA

r A

r

%20%100

)2( 22

A

A

r A

r

dr

dA A




4.5Indeterminate Forms and

L’ Hopital’s Rule ^

-Using derivative to calculate limits of fraction



4 - 22

L’Hopital’s Rule

When gives an indeterminate

form (and the limit exists)

It is possible to find a limit by

Note: this only works when the originallimit gives an indeterminate form

( )lim

( ) x c

f x

g x

'( )lim'( ) x c

f x g x

001 0

0



4 - 23

L’Hopital’s Rule



4 - 24

Example: L’Hô pital’s Rule

20

cos1lim

x x

x

x

00

x

x

x x

x

x x 21

sinlim

cos1lim 020

01

0

21

sinlim

0

x

x

x

2

1

2

coslim0

x

x

Limit isfound!



4 - 25

Example:

0

0

0

0

2

0

211

lim

x

x x

x

x

x

x 2

2

1)1)(

2

1(

lim

2

1

0

2

)1)(4

1(

lim

2

3

0

x

x 8

1

Limit is found!



4 - 26

Other Indeterminate Forms /, 0 or-

L’Hopital’s Rule applies to the determinate

form 0/0 as well as to /.

If f ( x) and g ( x) as xa , then

)('

)('

lim)(

)(

lim x g

x f

x g

x f

a xa x



4 - 27

Example:

x

x

x

2

lnlim

x

x

x /1

/1lim

x x

1lim

0

Limit is found!



4 - 28

1

1 1lim

ln 1 x

x x

1

1 lnlim

1 ln x

x x

x x

0

0

1

11

lim1

ln x

x x

x

x

Use L’Hôpital’s rule.

1

1lim

1 ln x

x

x x x

0

0

Example:

1

1lim

1 1 ln x

x

Use L’Hôpital’s

rule again.1

2=

Limit is found!

Continue..



4 - 29

Indeterminate Power 1, 00, 0

x

x x /1

lim

0Limit leads to

First by taking logarithm of the function.

Second, use l’HÔpital Rule to find the Limit of thelogarithm expression.

Then, exponentiate the result to find the originalfunction limit.



4 - 30Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Indeterminate Power 1, 00, 0

L x f

a xa x ee x f

)(ln

lim)(lim

L x f a x

)(lnlim

Here a may be either finite or infinite.



4 - 31

Example: x

x x /1

lim

x

x x f

1

)( Let and find )(lnlim x f x

x

x x x f x

lnln)(ln

1

x

x x f

x x

lnlim)(lnlim

0

1

/1lim

x

x

Find

1limlim)(lim 0)(ln

1

e e x x f x f

x

x

x x

Exercise: x

x x

/1

lim

TryThisquestion



4 - 32

Example 2:


Find x

x x

/1

lim

x x x f 1

)( Let and find )(lnlim x f x

)(ln x f

)(lnlim x f x

)(ln1

limlim)(lim x f

x

x

x xe x x f



4 - 33 © 2012, 2007 Pearson. All rights reserved.

Extra example:



4 - 34

Exercise (Try These!)

x

x

x

11lim

0

2

1

30

sinlim

x

x x

x

6

1



4 - 35

Solutions:



4 - 36

Exercise (Try These!!)

a)

b)

c)

x

x x

x

sin3lim

0

2

0

2lim x

e x

x

x

x x 2

lnlim



4 - 37

Solutions:




4.7 Newton’s Method

-A technique to approximate the solution to anequation f ( x) = 0



4 - 41



4 - 42

The Intermediate Value Theorem


If f ( x) is a continuous function on [a, b],then for every d between f (a)and f (b), there exists a value c in between a

and b such that f (c) = d .Example:Show that the function f (x) = ln (x) -1 has a solution between 2 and 3.Solution:

Sub x = 2 and x= 3 into f ( x), f (2) = ln(2)-1 = -0.307 f (3) = ln(3) -1 = 0.099

By the Intermediate Value Theorem, f ( x) must have a solution between 2 and 3.

Different sign



4 - 43

Find the x-coordinate of the point where the curve y = x3 - x crosses the horizontal line y = 1

Solution: x3 - x = 1 f ( x) = x3 - x - 1 = 0By the intermediate Value

Theorem, there is a root

in (1,2)



4 - 44

cont’

Starting the value x0=1,

f ( x) = x3 - x - 1 = 0,



13)(' 2 x x f 13

1

)('

)(

2

3

1

1

n

nn

nn

n

nnn

x

x x x x

x f

x f x x

347826087.175.5

875.0

5.115.13

15.15.1

5.1

5.1113

1111

2

3

2

2

3

1

x

x

Continue to substitute the value till the value is consistent.

...

325200399.11)347826087.1(3

1347826087.1347826087.1347826087.1

4

2

3

3

x

x



4 - 45

cont’

Starting the value x0=1,

f ( x) = x3 - x - 1 = 0,



13)(' 2 x x f 13

1

)('

)(

2

3

1

1

n

nn

nn

n

nnn

x

x x x x

x f

x f x x

Root



4 - 46




E i



4 - 48

Newton’s method:

Exercise:

,...3,2,1 '

1 n x f

x f x x

n

n

nn

By using ,approximate the root of x4 − 4x + 1 = 0 and continuethe approximation procedure until two successive approximationsdiffer by less than 0.0001.

00 x



4 - 49

Solution:



4 - 50

Exercise:

Use the Newton Method to find a solutionfor the following equations

a) x – cos x = 0, let x0 = 1b) x5 – 5 x3 – 2 = 0; x > 0. Let x0 = 2

ANS: a) 0.7391 b) 2.2738 )(')(1

n

nnn

x f x f x x

(Hint: use Radian mode in calculatorto solve trigonometry function)



Taylor and Maclaurin Series

-A function f ( x) with derivatives of all ordersexists on an interval I, can be expressed as a power

series



4 - 52



4 - 53

Expand

Solution:

By using formula for the nth Taylor with a=0,

11

1)(

x x f around a=0, to get

3rd degree of approximations.

)0('''),0(''),0('),0( f f f f

nn

n a x

n

a f a x

a f a xa f a f x P )(

!

)(...)(

!2

)(''))((')()(

)(2

Calculate



4 - 54

11

1)(

x x f

0)0( f

3)1(

2)(''

x x f

2)1(

1)('

x x f

4)1(

6)('''

x x f

1)0(' f 2)0('' f 6)0(''' f

!3

))(('''

!2

))((''))((')()(

32

3

a xa f a xa f a xa f a f x P

3232

!3

6

!2

2))(1(01

1

1 x x x

x x x

x

3 rd degree approx. :



4 - 55

Exercise:

Find the Taylor series generated by f ( x)=e x

at x = 0up to n degrees.

621)(

32

3

x x x x P

1)0()(

1)0()(

1)0(')('

)0()(

33

22

0

f e x f

f e x f

f e x f

e f e x f

x

x

x

x

!3

))(('''

!2

))((''))((')()(

32

3

a xa f a xa f a xa f a f x P

Example:



4 - 56

Find the Taylor series and theTaylor polynomial generated

by f ( x) = e x at x = 0.

Example:



4 - 57



4 - 58

Exercise

Find the Taylor series generated by f ( x) =1/ x at a = 2.

Find the Maclaurin series generated by f ( x) = 1/(1+ x).

ANS:

13

2

22

21...

2

2

2

2

2

1

n

n

n x x x

n n x x x x )1(...1

32

nn

n a xn

a f a x

a f a xa f a f x P )(

!

)(...)(

!2

)(''))((')()(

)(2



~End Of Application of Derivative Part 2~

applidifferentiation_part2_iry14

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