AP CALCULUS AB
Chapter 4:
Applications of Derivatives
Section 4.5:
Linearization and Newton’s Method
What you’ll learn about
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Linear Approximation
Newton’s Method
Differentials
Estimating Change with Differentials
Absolute, Relative, and Percent Change
Sensitivity to Change
…and why
Engineering and science depend on approximation
in most practical applications; it is important to
understand how approximation techniques work.
Linear Approximation
Any 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 f is differentiable at x  a , then the equa tion of the tangent line,
L ( x )  f ( a )  f '( a )( x - a ), defines the lin earization of f at a . T he
approxim ation f ( x )  L ( x ) is the stan d ard lin ear ap p roxim ation
of f at a . T he point x  a is the cen ter of the approxim ation.
Just Math Tutoring
You Tube
What is Linearization?
Just math tutoring
Finding the Linearization at a point
Followed by
25) 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 f ( x )  1  x at x = 0 (center of
approximation) and use it to approximate 1 . 02 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:
You try: Find linearization L(x) of f(x) at x = a when
3
f ( x )  x  2 x  3 and a = 2.
How accurate is the approximation L(a + 0.1) ≈ f(a + 0.1)
Point of tangency f(2) =
Tangent Line equation: L(x)
Evaluate |L(2.1) – f(2.1)|
Approximation error:
f ’(2)
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)
Tangent Line equation:
L(x)
f ’(π/2)
Evaluate |L(1.75) – cos 1.75 by calculator |
Approximation error:
Example Finding a Linearization
Find the linearization of f ( x )  cos x at x   / 2 an d use it to approxim ate
cos 1.75 w ithout a calculator.
S ince f (  / 2)  cos(  / 2)  0, the point of tangenc y is ( / 2, 0). T he slope of the
 
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tangent line is f '(  / 2)   sin(  / 2)   1. T hus L ( x )  0  (  1)  x     x  .
2
2

T o approxim ate cos 1.75  f (1.75)  L (1.75)   1.75 

2
.
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.
Steps
1)
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
(1  x )
k
 1  kx
Use the formula to find polynomials that will approximate the following
functions for values of x close to zero.
a)
3
1 x
b)
1
1 x
How?
Rewrite expression as (1 + x) k,
Identify coefficients of x and k.
Find L(x) = 1 + kx for each expression.
c)
1 5x
4
d)
1
1 x
2
Example 4: Use linearizations to approximate roots. Find
a) 123 and b) 123
3
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Identify function: f(x) = x
Let a be the perfect square closest to 123. Find L(x) at x = a.
Use L(x) to estimate 123
Error?
You try b.
Differentials
L et y  f ( x ) be a differentiable function. T h e d ifferen tial d x is an
independent variable. T he d ifferen tial d y is dy  f '( x ) dx .
(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
Find the differential dy and evaluate dy for the given value of x and dx .
y  x  2 x , x  1, dx  0.01
5
dy   5 x  2  dx
4
dy   5  2   0.01 
 0.07
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
x=1, dx = 0.01
b) y = sin 3x
x=π, dx = -0.02
You try:
y
2x
1 x
2
, x   2 , dx  0 . 1
c) x + y = xy
x=2, dx = 0.05
More Notation…
dy

dx
df
f ' ( x ) dx
 f '( x)
dx

f '( x)
dx
df  f ' ( x ) dx
Example 7 Finding Differentials of functions. Find
dy/dx and multiply both sides by dx.
a)
d (tan (2x))
You try: d(e5x + x5)
b)
d(
x
x 1
)
Estimating Change with Differentials
Suppose 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
L et f ( x ) b e d ifferen tiab le at x  a . T h e ap p ro x im ate ch an g e in th e valu e o f
f w h en x ch an g es fro m a to a  d x is d f  f '( a ) d x .
Estimating Change with
Differentials
Example Estimating Change with
Differentials
T he radius of a circle increases from a  5 m to 5.1 m . U se dA to estim ate
the increase in the circle's area A.
S ince A   r , the estim ated increase is
2
dA  2  rdr  2   5   0.1    m
2
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:
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The linear approximation of a differentiable
function f  x  at c is
y  f c   f  c  x  c 
because, from the slope of the tangent line
y  f c 
xc
 f c 
or y  f c   f c  x  c 
In Review
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Definition of Differentials:
y  f  x  is a differentiable function in an open
interval containing x.
The differential of x  dx  is any non-zero real
number.
The differential of y  dy  is
dy  f   x dx
Summary
Linearization: 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 tested
Newton’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. G uess a first approxim ation to a solu tion of the equation f ( x )  0.
A graph of y  f ( x ) m ay help.
2. U se the first approxim ation to get a second, the second to get a third,
and so on, using the form ula
x
n 1
 x 
n
f (x )
n
f '( x )
n
Procedure for Newton’s Method
Using Newton’s Method
U se N ew ton's method to solve x  3 x  1  0.
3
Let f ( x )  x  3 x  1, then f '( x )  3 x  3 and
3
x
n 1
 x 
n
2
f (x )
n
x  3x  1
3
 x
f '( x )
n 1
 x 
n
n
3x  3
n
2
n
n
A graph suggests that x   0.3 is a good firs t approxim ation. T hen,
1
x   0.3
1
x   0.322324159
2
x   0.3221853603
3
x   0.322185 3546
4
T he x for n  5 all appear to equ al x on the calculator.
n
T he solution to x  3 x  1  0 is about  0.322185354 6.
3
4