Calculus 04-Extrem Value

7/30/2019 Calculus 04-Extrem Value

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Absolute extreme values are either maximum or

minimum points on a curve.

They are sometimes called global extremes.

They are also sometimes called absolute extrema.

(Extrema is the plural of the Latin extremum.)

4.1 Extreme Values of

Functions


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FunctionsDefinition Absolute Extreme Values

Let f be a function with domain D. Thenf(c) is the

a. absolute minimum value on D if and only

iff(x) f(c) for allx in D.


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Extreme values can be in the interior or the end

points of a function.

0

1

2

3

4

-2 -1 1 2

2

y x

,D AbsoluteMinimum

No Absolute

Maximum


Functions


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0

1

2

3

4

-2 -1 1 2

2

y x 0,2D

Absolute Minimum

Absolute

Maximum

4.1 Extreme Values of Functions


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0

1

2

3

4

-2 -1 1 2

2y x

0,2D

No Minimum

AbsoluteMaximum


Functions


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0

1

2

3

4

-2 -1 1 2

2y x

0,2D No Minimum

NoMaximum


Functions


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Extreme Value Theorem:

Iff is continuous over a closed interval, [a,b] then f has a

maximum and minimum value over that interval.

Maximum &

minimum

at interior points

Maximum &

minimum

at endpoints

Maximum at

interior point,

minimum atendpoint



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Local Extreme Values:

A local maximum is the maximum value within some

open interval.

A local minimum is the minimum value within some

open interval.



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Absolute minimum

(also local minimum)

Local maximum

Local minimum

Absolute maximum

(also local maximum)

Local minimum

Local extremes

are also called

relative extremes.


Functions


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Local maximum

Local minimum

Notice that local extremes in the interior of the function

occur where is zero or is undefined.f f

Absolute maximum

(also local maximum)


Functions


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Local Extreme Values:

If a function f has a local maximum value or a

local minimum value at an interior point c of its

domain, and if exists at c, then

0f c

f


Functions


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Critical Point:

A point in the domain of a function f at which

or does not exist is a critical point off.

0f f

Note:Maximum and minimum points in the interior of a function

always occur at critical points, but critical points are not

always maximum or minimum values.


Functions


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EXAMPLE 3 FINDING ABSOLUTE EXTREMA

Find the absolute maximum and minimum values of

on the interval . 2/3f x x 2,3

2/3f x x

1

323

f x x

3

2

3

f x

x

There are no values of x that will make

the first derivative equal to zero.

The first derivative is undefined at x=0,

so (0,0) is a critical point.

Because the function is defined over a

closed interval, we also must check theendpoints.


Functions


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0 0

f

To determine if this critical point is

actually a maximum or minimum, we

try points on either side, without

passing other critical points.

2/3f x x

1 1f 1 1f

Since 0


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0 0f

2/3f x x 2,3D

At: 2x 232 2 1.5874f

At: 3x

Absolute

minimum:

Absolute

maximum:

0,0

3,2.08

2

33 3 2.08008f


Functions


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Functionsy =x2/3


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Finding Maximums and Minimums Analytically:

1 Find the derivative of the function, and determine

where the derivative is zero or undefined. Theseare the critical points.

2 Find the value of the function at each critical point.

3 Find values or slopes for points between thecritical points to determine if the critical points are

maximums or minimums.

4 For closed intervals, check the end points as

well.


Functions


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Functions

Find the absolute maximum and minimum of the function

]2,1[,2452)(23 onxxxxf

4106)('2 xxxf

410602 xx

Find the critical numbers2530

2 xx

)1)(23(0 xx 13

2 xx or


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Functions


]2,1[,2452)( 23 onxxxxf

Check endpoints and critical numbers

The absolute maximum is 2 whenx = -2The absolute minimum is -13 whenx = -1

22

11

27

26

3

2

131

xfx


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Functions


]3,0[,1

3)(

2

on

x

xxf 2

2

)1(

)1)(3()2)(1()('

x

xxxxf

3202 xx

Find the critical numbers

)1)(3(0 xx 13 xx or

2

2

)1(

32)('

x

xxxf


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Functions


]3,0[,1

3)(

2

on

x

xxf

3321

30

xfx

Check endpoints and critical numbers

The absolute maximum is 3 whenx = 0, 3

The absolute minimum is 2 whenx = 1


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2,0,sinsin)( 2 onxxxf

xxxxf cossin2cos)('

Find the critical numbers

xxx cossin2cos0

)sin21(cos0 xx

0cos x 0sin21 x

2

3,

2

x

6

5,

6

x


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Functions

Find the absolute maximum and

minimum of the function

2,0,sinsin)( 2 onxxxf

02

22

3

4

1

6

5

02

4

1

6

00

xfx

The absolute maximum is 1/4 whenx =

/6, 5

/6The absolute minimum is2 whenx =3/2


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Critical points are not always extremes!

-2

-1

0

1

2

-2 -1 1 2

3y x

0f (not an extreme)


Functions


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-2

-1

0

1

2

-2 -1 1 2

1/3y x

is undefined.f

(not an extreme)


Functions


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Iff(x) is a differentiable function over [a,b],

then at some point between a and b:

f b f a

f cb a

Mean Value Theorem for Derivatives

4.2 Mean Value Theorem


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Iff(x) is a differentiable function over [a,b],then at some point between a and b:

f b f af c

b a


Differentiable implies that the function is also continuous.



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f b f af c

b a


Differentiable implies that the function is also continuous.

The Mean Value Theorem only applies over a closed interval.



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f b f af c

b a


The Mean Value Theorem says that at some point

in the closed interval, the actual slope equals

the average slope.



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y

x0

A

B

a b

Slope of chord:

f b f a

b a

Slope of tangent:

f c

y f x

Tangent parallel

to chord.

c



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Iff(x) is a differentiable function over [a,b],and iff(a) =f(b) = 0, then there is at least one

point c between a and b such thatf (c)=0:

Rolles Theorem


(a,0) (b,0)


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4.2 Mean Value TheoremShow the function

satisfies the hypothesis of

the Mean Value Theorem

3,0oncos)(

xxf

The function is continuous on [0,/3] and differentiable on

(0,/3). Sincef(0) = 1 andf(/3) = 1/2, the Mean Value

Theorem guarantees a point c in the interval (0,/3) forwhich

f b f af c

b a

csin

03/

12/1

c = .498


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4.2 Mean Value Theorem(0,1)

(/3,1/2)

atx = .498, the slope

of the tangent line is

equal to the slope of

the chord.


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Definitions Increasing Functions, Decreasing FunctionsLet f be a function defined on an interval I and letx1 andx2be any two points in I.

1. f increases on I ifx1 f(x2).


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A function is increasing over an interval if thederivative is always positive.

A function is decreasing over an interval if the

derivative is always negative.

A couple of somewhat obvious definitions:

4.2 Mean Value TheoremCorollary Increasing Functions, Decreasing Functions

Let f be continuous on [a,b] and differentiable on (a,b).1. If f > 0 at each point of (a,b), then f increases on [a,b].

2. If f < 0 at each point of (a,b), then f decreases on [a,b].


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4.2 Mean Value TheoremFind where the function

is increasing and decreasing and find the local

extrema.

xxxxf 249)( 23

xxxxf 249)(23

24183)('2 xxxf

)86(302

xx

)86(02 xx

)2)(4(0 xx

2 4

0 0f(x)

+-+

),4()2,( inc)4,2(dec

x = 2, local maximum

x = 4, local minimum


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y

x0

y f x

y g x

These two functions have the

same slope at any value ofx.

Functions with the same

derivative differ by a constant.

C4.2 Mean Value Theorem


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Find the function whose derivative is and whose

graph passes through f x sin x

0,2

cos sind

x xdx

cos sind

x xdx

so:

cosf x x C

2 cos 0 C



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Find the functionf(x) whose derivative is sin(x) and

whose graph passes through (0,2).

cos sind

x xdx

cos sind

x xdx so:

cosf x x C

2 cos 0 C

2 1 C 3 C

cos 3f x x Notice that we had to have

initial values to determine

the value ofC.



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The process of finding the original function from the

derivative is so important that it has a name:

Antiderivative

A function is an antiderivative of a function

if for all x in the domain off. The process

of finding an antiderivative is antidifferentiation.

F x f x

F x f x

You will hear much more about antiderivatives in the future.

This section is just an introduction.



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Since acceleration is thederivative of velocity,

velocity must be the

antiderivative of

acceleration.

Example 7b: Find the velocity and position equations

for a downward acceleration of 9.8 m/sec2 and an

initial velocity of 1 m/sec downward.

9.8a t

9.8 1v t t

1 9.8 0 C 1 C

9.8v t t C (We let down be positive.)



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Since velocity is the derivative of position,

position must be the antiderivative of velocity.

9.8a t

9.8 1v t t

1 9.8 0 C

1 C

9.8v t t C

29.8

2s t t t C

The power rule in reverse:

Increase the exponent by one and

multiply by the reciprocal of the

new exponent.



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9.8a t

9.8 1v t t

1 9.8 0 C

1 C

9.8v t t C

29.8

2

s t t t C

24.9s t t t C The initial position is zero at time zero.

20 4.9 0 0 C 0 C

2

4.9s t t t



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In the past, one of the important uses of derivatives was

as an aid in curve sketching. We usually use a calculator

of computer to draw complicated graphs, it is still

important to understand the relationships between

derivatives and graphs.

4.3 Connectingf andf with the

Graph off


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First Derivative Test for Local Extrema at a critical point c


Graph off

1. If f changes sign from positive to

negative at c, then f has a localmaximum at c.

local max

f>0 f0 f>0


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First derivative:

y is positive Curve is rising.

y is negative Curve is falling.

y is zero Possible local maximum orminimum.

4.3 Connectingf andf with

the Graph off


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Graph offDefinition Concavity

The graph of a differentiablefunctiony =f(x) is

a. concave up on an open interval

I ify is increasing on I. (y>0)b. concave down on an open interval

I ify is decreasing on I. (y


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Graph off

Second Derivative Test for Local Extrema at a critical point c

1. If f(c) = 0 andf(c) < 0, then f has a local maximum atx = c.2. If f(c) = 0 andf(c) > 0, then f has a local minimum atx = c.

+ +


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Second derivative:

y is positive Curve is concave up.

y is negative Curve is concave down.

y is zero Possible inflection point(where concavity changes).


Graph off


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Graph off

Definition Point of Inflection

A point where the graph of a function has a tangent line and

where the concavity changes is called a point of inflection.

inflection point


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23 2

3 4 1 2y x x x x

23 6y x x

0y Set2

0 3 6x x 2

0 2x x

0 2x x

0, 2x

First derivative test:

y

0 2

0 0

2

1 3 1 6 1 3y

negative

2

1 3 1 6 1 9y positive

23 3 3 6 3 9

y positive

Possible extreme at .0, 2x


Graph offSketch the graph

zeros atx = -1,x = 2


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23 6y x x

0y Set

20 3 6x x

20 2x x

0 2x x

0, 2x

First derivative test:

y

0 2

0 0

maximum at 0x

minimum at 2x



Graph off


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23 6y x x

0y Set

20 3 6x x

20 2x x

0 2x x

0, 2x


Or you could use the second derivative test:

maximum at 0x minimum at 2x

6 6y x

0 6 0 6 6y negativeconcave down

local maximum

2 6 2 6 6y positiveconcave up

local minimum


Graph off


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6 6y x

We then look for inflection points by setting the second

derivative equal to zero.

0 6 6x

6 6x

1 x

Possible inflection point at .1x

y

1

0

0 6 0 6 6y negative

2 6 2 6 6y positive

inflection point at 1x


Graph off


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43210-1-2

5

4

3

2

1

0

-1

43210-1-2

5

4

3

2

1

0

-1

Make a summary table: x y y y

1 0 9 12 rising, concave down

0 4 0 6 local max

1 2 3 0 falling, inflection point

2 0 0 6 local min

3 4 9 12 rising, concave up

4.3 Connecting f and f with

the Graph of f


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A Classic Problem

You have 40 feet of fence to enclose a rectangular garden along

the side of a barn. What is the maximum area that you can

enclose?

4.4 Modeling and Optimization


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x x

40 2x

40 2A x x

240 2A x x

40 4A x

0 40 4x

4 40x

10x

10 40 2 10A

10 20A

2200 ftA

40 2l x

w x 10 ftw

20 ftl



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To find the maximum (or minimum) value of a function:


1. Understand the Problem.2. Develop a Mathematical Model.

3. Graph the Function.

4. Identify Critical Points and Endpoints.

5. Solve the Mathematical Model.

6. Interpret the Solution.


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What dimensions for a one liter cylindrical can will use

the least amount of material?

We can minimize the material by minimizing the area.

22 2A r rh

area of

ends

lateral

area

We need another

equation that relates

rand h:

2V r h

31 L 1000 cm

21000 r h

2

1000h

r

2

2

10 02

02A r r

r

2 20002A rr

2

20004A r

r



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22 2A r rh

area of

ends

lateral

area

2V r h

31 L 1000 cm2

1000 r h

2

1000h

r

22

10 02

02A r r

r

2 20002A r

r

2

20004A r

r

2

2000

0 4 r r

2

20004 r

r

32000 4 r

3500r

3 500r

5.42 cmr

2

1000

5.42h

10.83 cmh



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Find the radius and height of

the right-circular cylinder of

largest volume that can beinscribed in a right-circular

cone with radius 6 in. and

height 10 in. h

r

10 in

6 in


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h

r

10 in

6 in

The formula for the volume of

the cylinder is hrV 2

To eliminate one variable, we

need to find a relationship

between rand h.

61010 r

h

rh3

510

6

h

10-h

r10


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h

r

10 in

6 in

hrV2

322

3

5103

510 rrrrV

2520 rr

dr

dV

)4(50 rr

4,0 rr


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h

r

10 in

6 in

Check critical points and endpoints.

r= 0, V= 0

r= 4 V= 160/3

r= 6 V= 0

The cylinder will have amaximum volume when

r= 4 in. and h = 10/3 in.


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Determine the point on the

curvey =x2 that is closest to

the point (18, 0).


22)18( yxd

42)18( xxd

Substitute forx

42)32436( xxxd

)3624()32436(2

1 321

24

xxxxx

dx

ds


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the point (18, 0).


)3624()32436(

2

1 321

24

xxxxx

dx

ds

0dx

dsset 36240

3 xx 1820 3 xx

2x 4y


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the point (18, 0).


18203 xx

2x 4y

)942)(2(02 xxx

2

- 0 +


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If the end points could be the maximum or

minimum, you have to check.

Notes:

If the function that you want to optimize has morethan one variable, use substitution to rewrite the

function.

If you are not sure that the extreme youve found is a

maximum or a minimum, you have to check.


4 5 Li i i d


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For any functionf(x), the tangent is aclose approximation of the function for

some small distance from the tangent

point.

y

x

0 x a

f x f aWe call the equation of the

tangent the linearization of

the function.

4.5 Linearization and

Newtons Method

4 5 Li i ti d


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The linearization is the equation of the tangent line, and you

can use the old formulas if you like.

Start with the point/slope equation:

1 1y y m x x 1x a 1y f a m f a

y f a f a x a y f a f a x a

L x f a f a x a linearization offat a

f x L x is the standard linear approximation offat a.


Newtons Method

4 5 Li i ti d


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Find the linearization off(x) =x4 + 2x atx = 2

L x f a f a x a


Newtons Method

f (x) = 4x3 + 2

L (x) =f(3) +f(3)(x - 3)

L (x) = 87 + 110(x - 3)

L (x) = 110x - 243

4 5 Li i ti d


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Important linearizations forx near zero:

1k

x 1 kx

sinx

cosx

tanx

x

1

x

1

21

1 1 1

2

x x x

13 4 4 3

4 4

1 5 1 5

1 51 5 1

3 3

x x

x x

f x L x

This formula also leads to

non-linear approximations:


Newtons Method

4 5 Li i ti d


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Newtons Method

Estimate using local linearization.37

2

1

2

1)('

)(

xxf

xxf L x f a f a x a

)3637)(36(')36()37( ffL

)1(1216)37( L

0833.6)37( L

4 5 Li i ti d


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Newtons Method

Estimate sin 31 using local linearization.

xxf

xxf

cos)('

sin)(

L x f a f a x a

180)30(')30()31(

ffL

1802

321)31( L

360

3180)31(

L

Need to

be in radians

4 5 Li i ti d


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Differentials:

When we first started to talk about derivatives, we said

that becomes when the change in x and

change in y become very small.

y

x

dy

dx

dy can be considered a very small change iny.

dx can be considered a very small change inx.


Newtons Method

4 5 Li i ti d


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Let y =f(x) be a differentiable function.The differential dx is an independent

variable.

The differential dy is: dy =f(x)dx


Newtons Method

4 5 Li i ti d


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Example: Consider a circle of radius 10. If the radius increases by

0.1, approximately how much will the area change?

2

A r2dA r dr

2dA dr rdx dx

very small change in A

very small change in r

2 10 0.1dA

2dA

(approximate change in area)


Newtons Method

4 5 Linearization and


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Compare to actual change:

New area:

Old area:

2

10.1 102.01

2

10 100.00


Newtons Method

01.2A

2dAAbsoluteerror

%2

100

2

A

dA

%01.2100

01.2

A

Apercenterror



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Newtons Method

True Estimated

Absolute Change

Relative Change

Percent Change

)()( afdxaff dxafdf )('

)(af

f

)(af

df

%100)(

xaf

df%100)(

xaf

f



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Newtons Method

Newtons Method

0 x

y

y =f(x)

Root

sought

x1

First

(x1,f(x1))

x2

Second

x3

Third

(x2,f(x2))

(x3,f(x3))

)( 1212 xxmyy ))((')(0 121 xxxfxf

))((')(0 121 xxxfxf

)(')(')( 1121 xfxxfxxf

)('

)(

1

112

xf

xfxx



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This isNewtons Method of finding roots. It is an

example of an algorithm (a specific set of

computational steps.)

Newtons Method:

1n

n n

n

f xx x

f x

This is a recursive algorithm because a set of steps are

repeated with the previous answer put in the next

repetition. Each repetition is called an iteration.


Newtons Method



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Newtons Method

21

32f x x

Finding a root for:

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 1 2 3 4

We will use

Newtons Method to

find the rootbetween 2 and 3.


Newtons Method



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Newtons Method 2

13

2f x x

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 1 2 3 4


Newtons Method

xxf )('

Guessx1 = 2

)('

)(

1

112

xf

xfxx

5.22

122

x



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Newtons Method 2

13

2f x x

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 1 2 3 4


Newtons Method

xxf )('

Guessx2 = 2.5

)('

)(

2

223

xf

xfxx

45.25.2

125.5.23 x



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Find where crosses .3

y x x 1y

3

1 x x 3

0 1x x

3 1f x x x

23 1f x x


Newtons Method



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nx nf xn nf x 1

n

n n

n

f xx x

f x

0 1 1 2 11 1.52

1 1.5 .875 5.75.875

1.5 1.34782615.75

2 1.3478261 .1006822 4.4499055 1.3252004

3

1.3252004 1.3252004 1.0020584 1


Newtons Method



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There are some limitations to Newtons Method:

Wrong root found

Looking for this root.

Bad guess.

Failure to converge


Newtons Method


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First, a review problem:

Consider a sphere of radius 10 cm.

If the radius changes 0.1 cm (a very small amount)

how much does the volume change?

343

V r24dV r dr

2

4 10cm 0.1cmdV 3

40 cmdV

The volume would change by approximately 40 cm3

4.6 Related Rates


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Now, suppose that the radius is

changing at an instantaneous rate

of 0.1 cm/sec.

34

3V r 24

dV dr r

dt dt

2 cm

4 10cm 0.1

sec

dV

dt

3cm

40

sec

dV

dt

The sphere is growing at a rate of 40 cm3/sec .

Note: This is an exact answer, not an approximation like

we got with the differential problems.

4.6 Related Rates


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Water is draining from a cylindrical tank

at 3 liters/second. How fast is the surface

dropping?

L3

sec

dV

dt

3cm3000

sec

Finddh

dt

2V r h

2dV dhr

dt dt

(ris a constant.)

32cm

3000

sec

dhr

dt

3

2

cm3000

secdh

dt r

(We need a formula to

relate Vand h. )

4.6 Related Rates


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Steps for Related Rates Problems:

1. Draw a picture (sketch).

2. Write down known information.

3. Write down what you are looking for.

4. Write an equation to relate the variables.

5. Differentiate both sides with respect to t.

6. Evaluate.

4.6 Related Rates


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Hot Air Balloon Problem:

Given:

4

rad0.14

min

d

dt

How fast is the balloon rising?

Finddh

dttan

500

h

2 1sec

500

d dh

dt dt

2

1sec 0.14

4 500

dh

dt

h

500ft

4.6 Related Rates


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Hot Air Balloon Problem:

Given:

4

rad0.14

min

d

dt

How fast is the balloon rising?

Find

dh

dt tan 500

h

2 1sec

500

d dh

dt dt

21

sec 0.144 500

dh

dt

h

500ft

2

2 0.14 500dh

dt

1

1

2

4

sec 24

ft140

min

dh

dt

4.6 Related Rates


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4x

3y

B

A

5z

TruckProblem:

Truck A travels east at 40 mi/hr.

Truck B travels north at 30 mi/hr.

How fast is the distance between the

trucks changing 6 minutes later?

r t d

140 410

130 310

2 2 23 4 z

2

9 16 z

2

25 z 5 z

4.6 Related Rates


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4x

3y

30dy

dt

40dx

dt

B

A

5z

Truck Problem:



r t d

140 410

130 310

2 2 23 4 z

29 16 z

2 2 2x y z

2 2 2dx dy dzx y zdt dt dt

4 40 3 30 5dz

dt



4.6 Related Rates


97/97

250 5dz

dt 50

dz

d

miles50

hour

4.6 Related Rates

TruckProblem:





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Calculus 04-Extrem Value