AP CALCULUS AB Chapter 4: Applications of Derivatives Section 4.5: Linearization and Newton’s Method.

AP CALCULUS ABChapter 4:

Applications of DerivativesSection 4.5:

Linearization and Newton’s Method

What you’ll learn about Linear Approximation Newton’s Method Differentials Estimating Change with Differentials Absolute, Relative, and Percent Change Sensitivity to Change

…and whyEngineering and science depend on approximation

in most practical applications; it is important to understand how approximation techniques work.

Linear ApproximationAny differentiable curve is “Locally Linear”

if you zoom in enough times.

Do Exploration 1: Appreciating Local Linearity (p 233)

A fancy name for the equation of the tangent line at a is “The linearization of f at a”

y – f(a) = f’(a)(x – a)

Definition of Linearization

If is differentiable at , then the equation of the tangent line,

( ) ( ) '( )( - ), defines the . The

approximation ( ) ( ) is the

f x a

L x f a f a x a

f x L x

linearization of at

standard linear approximation

of at

f a

f . The point is the of the approximation.x a centera

Just Math TutoringYou Tube

What is Linearization?

Just math tutoringFinding the Linearization at a point

Followed by25) Linear Approximation

10 minutes total time needed – Watch if you miss class this day or do not understand!

Example 1 Finding a Linearization

Find the linearization of at x = 0 (center of approximation) and use it to approximate without a calculator.

Then use a calculator to determine the accuracy of the approximation.

Point of tangency f ‘(0) = L(x) = Equation of the tangent line:

Evaluate L(.02)

Calculator approximation?

Approximation error:

xxf 1)(02.1

You try: Find linearization L(x) of f(x) at x = a when and a = 2. How accurate is the approximation L(a + 0.1) ≈ f(a + 0.1)

Point of tangency f(2) = f ’(2)Tangent Line equation: L(x)

Evaluate |L(2.1) – f(2.1)|

Approximation error:

32)( 3 xxxf

Example 2: Find the linearization of f(x) = cos x at x = π/2 and use it to approximate cos 1.75 without a calculator. Then use a calculator to determine the accuracy of the approximation.

Point of tangency f (π/2) f ’(π/2)

Tangent Line equation: L(x)

Evaluate |L(1.75) – cos 1.75 by calculator |

Approximation error:

Example Finding a Linearization

Find the linearization of ( ) cos at / 2 and use it to approximate

cos 1.75 without a calculator.

f x x x

Since ( / 2) cos( / 2) 0, the point of tangency is ( / 2,0). The slope of the

tangent line is '( / 2) sin( / 2) 1. Thus ( ) 0 ( 1) .2 2

To approximate cos 1.75 (1.75) (1.75) 1.75 .2

f

f L x x x

f L

Summary

Every function is “locally linear” about a point x = a. If you evaluate the tangent line at x = a for points close to a, you will have a close approximation to the function’s actual value.

Steps1) Using f(x), find the equation of a tangent line at

some point (a, f(a)).

Find f(a) by plugging a into f(x).Find the slope from f’(a).L(x) = f(a) + f’(a) (x - a).

2) Evaluate L(x) for any x near a to get a close approximation of f(x) for points near a.

Example 3: Approximating Binomial Powers using the general formula

Use the formula to find polynomials that will approximate the following functions for values of x close to zero.

a) b) c) d)

How? Rewrite expression as (1 + x) k, Identify coefficients of x and k. Find L(x) = 1 + kx for each expression.

3 1 x

kxx k 1)1(

451 x 21

1

xx1

1

Example 4: Use linearizations to approximate roots. Find a) and b)

Identify function: f(x) = Let a be the perfect square closest to 123. Find L(x) at x = a. Use L(x) to estimate Error?

You try b.

3 123

x

123

123

Differentials

Let ( ) be a differentiable function. The is an

independent variable. The is '( ) .

y f x

dy f x dx

differential

differential

dx

dy

(With dx as in independent variable and dy a dependent variable that depends on both x and dx.)

Although Liebniz did most of his calculus using dy and dx as separable entities, he never quite settled the issue of what they were. To him, they were “infinitesimals” – nonzero numbers, but infinitesimally small. There was much debate about whether such things could exist in mathematics, but luckily for the early development of calculus it did not matter: thanks to the Chain Rule, dy/dx behaved like a quotient whether it was one or not.

Example Finding the Differential dy

5

Find the differential and evaluate for the given value of and .

2 , 1, 0.01

dy dy x dx

y x x x dx

45 2

5 2 0.01

0.07

dy x dx

dy

Example 6 Find the differential dy and evaluate dy for the given values of x and dx.

How? Find f ’(x), multiply both sides by dx, evaluate for given values.

a) y = x5 + 37x b) y = sin 3x c) x + y = xy x=1, dx = 0.01 x=π, dx = -0.02 x=2, dx = 0.05

You try:

1.0,2,1

22

dxxx

xy

More Notation…

)(')('

xfdx

dxxf

dx

dy

dxxfdf )('

)(' xfdx

df

Example 7 Finding Differentials of functions. Find dy/dx and multiply both sides by dx.

a) d (tan (2x)) b)

You try: d(e5x + x5)

)1

(xx

d

Estimating Change with DifferentialsSuppose we know the value of a differentiable

function f(x) at a point a and we want to predict how much this value will change if we move to a nearby point (a + dx).

If dx is small, f and its linearization L at “a” will change by nearly the same amount.

Since the values of L are simple to calculate, calculating the change in L offers a practical way to estimate the change in f.

Differential Estimate of Change Let ( ) be differentiable at . The approximate change in the value of

when changes from to is '( ) .

f x x a

f x a a dx df f a dx

Estimating Change with Differentials

Example Estimating Change with Differentials The radius of a circle increases from 5 m to 5.1 m. Use to estimate

the increase in the circle's area .

a dA

A

2

2

Since , the estimated increase is

2 2 5 0.1 m

A r

dA rdr

Example 8 The radius r of a circle increases from a = 10 to 10.1 m. Use dA to estimate the increase in the circle’s area A. Compare this estimate with the true change ∆A, and find the approximation error.

Area formula for a circle: A =

True change: f(10.1) – f(10) =

Estimated change: dA/dr = dA =

Approximation error: |∆A – dA| =

You try: f(x) = x3 - x, a = 1, dx = 0.1

In Review: The linear approximation of a differentiable

function at c is

because, from the slope of the tangent line

cxcfcfy

xf

cxcfcfy

cfcx

cfy

or

In Review Definition of Differentials:

is a differentiable function in an open interval containing x.The differential of x is any non-zero real number.The differential of y is

xfy

dx dy

dxxfdy

SummaryLinearization: The equation of a tangent line to f at a point a

will give a good approximation of the value of a function f at a.

The Linearization of (1 + x)k = 1 + kx

Newton’s Method is used to find the roots of a function by using successive tangent line approximations, moving closer and closer to the roots of f if you start with a reasonable value of a.

Differentials: Differentials simply estimate the change in y as it relates to the change in x for given values of x. We learned how to estimate with linearizations, differentials are simply a more efficient method of finding change.

FYI – not testedNewton’s Method for approximating a zero of a function

Approximate the zero of a function by finding the zeros of linearizations converging to an accurate approximation.

Just Math Tutoring – Newton’s Method (7:29 minutes)

Procedure for Newton’s Method 1. Guess a first approximation to a solution of the equation ( ) 0.

A graph of ( ) may help.

2. Use the first approximation to get a second, the second to get a third,

and so on, using the formula

f x

y f x

1

( )

'( )n

n n

n

f xx x

f x

Procedure for Newton’s Method

Using Newton’s Method 3Use Newton's method to solve 3 1 0.x x

3 2

3

1 1 2

1

1

2

3

4

Let ( ) 3 1, then '( ) 3 3 and

( ) 3 1

'( ) 3 3

A graph suggests that 0.3 is a good first approximation. Then,

0.3

0.322324159

0.3221853603

0.322185

n n n

n n n n

n n

f x x x f x x

f x x xx x x x

f x x

x

x

x

x

x

4

3

3546 The for 5 all appear to equal on the calculator.

The solution to 3 1 0 is about 0.3221853546.nx n x

x x