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In this section, we will learn about: Linear approximations and differentials and their applications.

1.
3

DIFFERENTIATION RULES

DIFFERENTIATION RULES

2.
DIFFERENTIATION RULES

We have seen that a curve lies

very close to its tangent line near

the point of tangency.

We have seen that a curve lies

very close to its tangent line near

the point of tangency.

3.
DIFFERENTIATION RULES

In fact, by zooming in

toward a point on the graph

of a differentiable function,

we noticed that the graph

looks more and more like

its tangent line.

In fact, by zooming in

toward a point on the graph

of a differentiable function,

we noticed that the graph

looks more and more like

its tangent line.

4.
DIFFERENTIATION RULES

This observation is

the basis for a method

of finding approximate

values of functions.

This observation is

the basis for a method

of finding approximate

values of functions.

5.
DIFFERENTIATION RULES

3.10

Linear Approximations

and Differentials

In this section, we will learn about:

Linear approximations and differentials

and their applications.

3.10

Linear Approximations

and Differentials

In this section, we will learn about:

Linear approximations and differentials

and their applications.

6.
LINEAR APPROXIMATIONS

The idea is that it might be easy to calculate

a value f(a) of a function, but difficult (or even

impossible) to compute nearby values of f.

So, we settle for

the easily computed

values of the linear

function L whose graph

is the tangent line

of f at (a, f(a)).

The idea is that it might be easy to calculate

a value f(a) of a function, but difficult (or even

impossible) to compute nearby values of f.

So, we settle for

the easily computed

values of the linear

function L whose graph

is the tangent line

of f at (a, f(a)).

7.
LINEAR APPROXIMATIONS

In other words, we use the tangent line

at (a, f(a)) as an approximation to the curve

y = f(x) when x is near a.

An equation of

this tangent line is

y = f(a) + f’(a)(x - a)

In other words, we use the tangent line

at (a, f(a)) as an approximation to the curve

y = f(x) when x is near a.

An equation of

this tangent line is

y = f(a) + f’(a)(x - a)

8.
LINEAR APPROXIMATION Equation 1

The approximation

f(x) ≈ f(a) + f’(a)(x – a)

is called the linear approximation

or tangent line approximation of f at a.

The approximation

f(x) ≈ f(a) + f’(a)(x – a)

is called the linear approximation

or tangent line approximation of f at a.

9.
LINEARIZATION Equation 2

The linear function whose graph is

this tangent line, that is,

L(x) = f(a) + f’(a)(x – a)

is called the linearization of f at a.

The linear function whose graph is

this tangent line, that is,

L(x) = f(a) + f’(a)(x – a)

is called the linearization of f at a.

10.
LINEAR APPROXIMATIONS Example 1

Find the linearization of the function

f ( x) = x + 3 at a = 1 and use it to

approximate the numbers 3.98 and 4.05

Are these approximations overestimates or

Find the linearization of the function

f ( x) = x + 3 at a = 1 and use it to

approximate the numbers 3.98 and 4.05

Are these approximations overestimates or

11.
LINEAR APPROXIMATIONS Example 1

The derivative of f(x) = (x + 3)1/2 is:

−1/ 2 1

f '( x) = ( x + 3)

1

2 =

2 x+3

So, we have f(1) = 2 and f’(1) = ¼.

The derivative of f(x) = (x + 3)1/2 is:

−1/ 2 1

f '( x) = ( x + 3)

1

2 =

2 x+3

So, we have f(1) = 2 and f’(1) = ¼.

12.
LINEAR APPROXIMATIONS Example 1

Putting these values into Equation 2,

we see that the linearization is:

L( x) = f (1) + f '(1)( x −1)

1

= 2 + ( x −1)

4

7 x

= +

4 4

Putting these values into Equation 2,

we see that the linearization is:

L( x) = f (1) + f '(1)( x −1)

1

= 2 + ( x −1)

4

7 x

= +

4 4

13.
LINEAR APPROXIMATIONS Example 1

The corresponding linear approximation is:

7 x (when x is near 1)

x+3 ≈ +

4 4

In particular, we have:

7 0.98

3.98 ≈ + = 1.995

4 4

7 1.05

and 4.05 ≈ + = 2.0125

4 4

The corresponding linear approximation is:

7 x (when x is near 1)

x+3 ≈ +

4 4

In particular, we have:

7 0.98

3.98 ≈ + = 1.995

4 4

7 1.05

and 4.05 ≈ + = 2.0125

4 4

14.
LINEAR APPROXIMATIONS Example 1

The linear approximation is illustrated here.

We see that:

The tangent line approximation is a good approximation

to the given function when x is near 1.

Our approximations are overestimates, because

the tangent line

lies above

the curve.

The linear approximation is illustrated here.

We see that:

The tangent line approximation is a good approximation

to the given function when x is near 1.

Our approximations are overestimates, because

the tangent line

lies above

the curve.

15.
LINEAR APPROXIMATIONS Example 1

Of course, a calculator could give us

approximations for 3.98 and 4.05

The linear approximation, though, gives

an approximation over an entire interval.

Of course, a calculator could give us

approximations for 3.98 and 4.05

The linear approximation, though, gives

an approximation over an entire interval.

16.
LINEAR APPROXIMATIONS

In the table, we compare the estimates

from the linear approximation in

Example 1 with the true values.

In the table, we compare the estimates

from the linear approximation in

Example 1 with the true values.

17.
LINEAR APPROXIMATIONS

Look at the table

and the figure.

The tangent line

approximation

gives good

estimates if x

is close to 1.

However,

the accuracy

decreases

when x is farther

away from 1.

Look at the table

and the figure.

The tangent line

approximation

gives good

estimates if x

is close to 1.

However,

the accuracy

decreases

when x is farther

away from 1.

18.
LINEAR APPROXIMATIONS

How good is the approximation that

we obtained in Example 1?

The next example shows that, by using a graphing

calculator or computer, we can determine an interval

throughout which a linear approximation provides

a specified accuracy.

How good is the approximation that

we obtained in Example 1?

The next example shows that, by using a graphing

calculator or computer, we can determine an interval

throughout which a linear approximation provides

a specified accuracy.

19.
LINEAR APPROXIMATIONS Example 2

For what values of x is the linear

7 x

approximation x+3 ≈ + accurate

4 4

to within 0.5?

What about accuracy to within 0.1?

For what values of x is the linear

7 x

approximation x+3 ≈ + accurate

4 4

to within 0.5?

What about accuracy to within 0.1?

20.
LINEAR APPROXIMATIONS Example 2

Accuracy to within 0.5 means that

the functions should differ by less than 0.5:

⎛7 x ⎞

x + 3 −⎜ + ⎟ < 0.5

⎝4 4 ⎠

Accuracy to within 0.5 means that

the functions should differ by less than 0.5:

⎛7 x ⎞

x + 3 −⎜ + ⎟ < 0.5

⎝4 4 ⎠

21.
LINEAR APPROXIMATIONS Example 2

Equivalently, we could write:

7 x

x + 3 −0.5 < + < x + 3 + 0.5

4 4

This says that the linear approximation should lie

between the curves obtained by shifting the curve

y = x+3 upward and downward by an amount

0.5.

Equivalently, we could write:

7 x

x + 3 −0.5 < + < x + 3 + 0.5

4 4

This says that the linear approximation should lie

between the curves obtained by shifting the curve

y = x+3 upward and downward by an amount

0.5.

22.
LINEAR APPROXIMATIONS Example 2

The figure shows the tangent line

y = (7 + x) / 4 intersecting the upper

y = x + 3 + 0.5

curve at P

and Q.

The figure shows the tangent line

y = (7 + x) / 4 intersecting the upper

y = x + 3 + 0.5

curve at P

and Q.

23.
LINEAR APPROXIMATIONS Example 2

Zooming in and using the cursor,

we estimate that the x-coordinate of P

is about -2.66 and the x-coordinate of Q

is about 8.66

Zooming in and using the cursor,

we estimate that the x-coordinate of P

is about -2.66 and the x-coordinate of Q

is about 8.66

24.
LINEAR APPROXIMATIONS Example 2

Thus, we see from the graph

7 x

that the approximation x+3 ≈ +

4 4

is accurate to within 0.5 when

-2.6 < x < 8.6

We have rounded

to be safe.

Thus, we see from the graph

7 x

that the approximation x+3 ≈ +

4 4

is accurate to within 0.5 when

-2.6 < x < 8.6

We have rounded

to be safe.

25.
LINEAR APPROXIMATIONS Example 2

Similarly, from this figure, we see that

the approximation is accurate to within 0.1

when -1.1 < x < 3.9

Similarly, from this figure, we see that

the approximation is accurate to within 0.1

when -1.1 < x < 3.9

26.
APPLICATIONS TO PHYSICS

Linear approximations are often used

in physics.

In analyzing the consequences of an equation,

a physicist sometimes needs to simplify a function

by replacing it with its linear approximation.

Linear approximations are often used

in physics.

In analyzing the consequences of an equation,

a physicist sometimes needs to simplify a function

by replacing it with its linear approximation.

27.
APPLICATIONS TO PHYSICS

For instance, in deriving a formula for

the period of a pendulum, physics textbooks

obtain the expression aT = -g sin θ for

tangential acceleration and then replace

sin θ by θ with the remark that sin θ

is very close to θ if θ is not too large.

For instance, in deriving a formula for

the period of a pendulum, physics textbooks

obtain the expression aT = -g sin θ for

tangential acceleration and then replace

sin θ by θ with the remark that sin θ

is very close to θ if θ is not too large.

28.
APPLICATIONS TO PHYSICS

You can verify that the linearization of

the function f(x) = sin x at a = 0 is L(x) = x.

So, the linear approximation at 0 is:

sin x ≈ x

You can verify that the linearization of

the function f(x) = sin x at a = 0 is L(x) = x.

So, the linear approximation at 0 is:

sin x ≈ x

29.
APPLICATIONS TO PHYSICS

So, in effect, the derivation of the formula

for the period of a pendulum uses the

tangent line approximation for the sine

So, in effect, the derivation of the formula

for the period of a pendulum uses the

tangent line approximation for the sine

30.
APPLICATIONS TO PHYSICS

Another example occurs in the theory

of optics, where light rays that arrive at

shallow angles relative to the optical axis

are called paraxial rays.

Another example occurs in the theory

of optics, where light rays that arrive at

shallow angles relative to the optical axis

are called paraxial rays.

31.
APPLICATIONS TO PHYSICS

In paraxial (or Gaussian) optics,

both sin θ and cos θ are replaced by

their linearizations.

In other words, the linear approximations

sin θ ≈ θ and cos θ ≈ 1

are used because θ is close to 0.

In paraxial (or Gaussian) optics,

both sin θ and cos θ are replaced by

their linearizations.

In other words, the linear approximations

sin θ ≈ θ and cos θ ≈ 1

are used because θ is close to 0.

32.
APPLICATIONS TO PHYSICS

The results of calculations made

with these approximations became

the basic theoretical tool used to

design lenses.

The results of calculations made

with these approximations became

the basic theoretical tool used to

design lenses.

33.
APPLICATIONS TO PHYSICS

In Section 11.11, we will present

several other applications of the idea

of linear approximations to physics.

In Section 11.11, we will present

several other applications of the idea

of linear approximations to physics.

34.
The ideas behind linear approximations

are sometimes formulated in the

terminology and notation of differentials.

are sometimes formulated in the

terminology and notation of differentials.

35.
If y = f(x), where f is a differentiable

function, then the differential dx

is an independent variable.

That is, dx can be given the value of any real

number.

function, then the differential dx

is an independent variable.

That is, dx can be given the value of any real

number.

36.
DIFFERENTIALS Equation 3

The differential dy is then defined in terms

of dx by the equation

dy = f’(x)dx

So, dy is a dependent variable—it depends on

the values of x and dx.

If dx is given a specific value and x is taken

to be some specific number in the domain of f,

then the numerical value of dy is determined.

The differential dy is then defined in terms

of dx by the equation

dy = f’(x)dx

So, dy is a dependent variable—it depends on

the values of x and dx.

If dx is given a specific value and x is taken

to be some specific number in the domain of f,

then the numerical value of dy is determined.

37.
The geometric meaning of differentials

is shown here.

Let P(x, f(x)) and

Q(x + ∆x, f(x + ∆x))

be points on

the graph of f.

Let dx = ∆x.

is shown here.

Let P(x, f(x)) and

Q(x + ∆x, f(x + ∆x))

be points on

the graph of f.

Let dx = ∆x.

38.
The corresponding change in y is:

∆y = f(x + ∆x) – f(x)

The slope of the

tangent line PR is

the derivative f’(x).

Thus, the directed

distance from S

to R is f’(x)dx = dy.

∆y = f(x + ∆x) – f(x)

The slope of the

tangent line PR is

the derivative f’(x).

Thus, the directed

distance from S

to R is f’(x)dx = dy.

39.
dy represents the amount that the tangent line

rises or falls (the change in the linearization).

∆y represents the amount that the curve y = f(x)

rises or falls when changes by an amount dx.

rises or falls (the change in the linearization).

∆y represents the amount that the curve y = f(x)

rises or falls when changes by an amount dx.

40.
DIFFERENTIALS Example 3

Compare the values of ∆y and dy

if y = f(x) = x3 + x2 – 2x + 1

and x changes from:

a. 2 to 2.05

b. 2 to 2.01

Compare the values of ∆y and dy

if y = f(x) = x3 + x2 – 2x + 1

and x changes from:

a. 2 to 2.05

b. 2 to 2.01

41.
DIFFERENTIALS Example 3 a

We have:

f(2) = 23 + 22 – 2(2) + 1 = 9

f(2.05) = (2.05)3 + (2.05)2 – 2(2.05) + 1

= 9.717625

∆y = f(2.05) – f(2) = 0.717625

In general,

dy = f’(x)dx = (3x2 + 2x – 2) dx

We have:

f(2) = 23 + 22 – 2(2) + 1 = 9

f(2.05) = (2.05)3 + (2.05)2 – 2(2.05) + 1

= 9.717625

∆y = f(2.05) – f(2) = 0.717625

In general,

dy = f’(x)dx = (3x2 + 2x – 2) dx

42.
DIFFERENTIALS Example 3 a

When x = 2 and dx = ∆x,

this becomes:

dy = [3(2)2 + 2(2) – 2]0.05

= 0.7

When x = 2 and dx = ∆x,

this becomes:

dy = [3(2)2 + 2(2) – 2]0.05

= 0.7

43.
DIFFERENTIALS Example 3 b

We have:

f(2.01) = (2.01)3 + (2.01)2 – 2(2.01) + 1

= 9.140701

∆y = f(2.01) – f(2) = 0.140701

When dx = ∆x = 0.01,

dy = [3(2)2 + 2(2) – 2]0.01 = 0.14

We have:

f(2.01) = (2.01)3 + (2.01)2 – 2(2.01) + 1

= 9.140701

∆y = f(2.01) – f(2) = 0.140701

When dx = ∆x = 0.01,

dy = [3(2)2 + 2(2) – 2]0.01 = 0.14

44.
Notice that:

The approximation ∆y ≈ dy becomes better

as ∆x becomes smaller in the example.

dy was easier to compute than ∆y.

The approximation ∆y ≈ dy becomes better

as ∆x becomes smaller in the example.

dy was easier to compute than ∆y.

45.
For more complicated functions, it may

be impossible to compute ∆y exactly.

In such cases, the approximation by differentials

is especially useful.

be impossible to compute ∆y exactly.

In such cases, the approximation by differentials

is especially useful.

46.
In the notation of differentials,

the linear approximation can be

written as:

f(a + dx) ≈ f(a) + dy

the linear approximation can be

written as:

f(a + dx) ≈ f(a) + dy

47.
For instance, for the function f ( x) = x+3

in Example 1, we have:

dy = f '( x)dx

dx

=

2 x+3

in Example 1, we have:

dy = f '( x)dx

dx

=

2 x+3

48.
If a = 1 and dx = ∆x = 0.05, then

0.05

dy = = 0.0125

2 1+ 3

and 4.05 = f (1.05) ≈ f (1) + dy = 2.0125

This is just as we found in Example 1.

0.05

dy = = 0.0125

2 1+ 3

and 4.05 = f (1.05) ≈ f (1) + dy = 2.0125

This is just as we found in Example 1.

49.
Our final example illustrates the use

of differentials in estimating the errors

that occur because of approximate

of differentials in estimating the errors

that occur because of approximate

50.
DIFFERENTIALS Example 4

The radius of a sphere was measured

and found to be 21 cm with a possible error

in measurement of at most 0.05 cm.

What is the maximum error in using this

value of the radius to compute the volume

of the sphere?

The radius of a sphere was measured

and found to be 21 cm with a possible error

in measurement of at most 0.05 cm.

What is the maximum error in using this

value of the radius to compute the volume

of the sphere?

51.
DIFFERENTIALS Example 4

If the radius of the sphere is r, then

its volume is V = 4/3π r3.

If the error in the measured value of r

is denoted by dr = ∆r, then the corresponding

error in the calculated value of V is ∆V.

If the radius of the sphere is r, then

its volume is V = 4/3π r3.

If the error in the measured value of r

is denoted by dr = ∆r, then the corresponding

error in the calculated value of V is ∆V.

52.
DIFFERENTIALS Example 4

This can be approximated by the differential

dV = 4π r2 dr

When r = 21 and dr = 0.05, this becomes:

dV = 4π(21)2 0.05 ≈ 277

The maximum error in the calculated volume

is about 277 cm3.

This can be approximated by the differential

dV = 4π r2 dr

When r = 21 and dr = 0.05, this becomes:

dV = 4π(21)2 0.05 ≈ 277

The maximum error in the calculated volume

is about 277 cm3.

53.
DIFFERENTIALS Note

Although the possible error in the example

may appear to be rather large, a better

picture of the error is given by the relative

Although the possible error in the example

may appear to be rather large, a better

picture of the error is given by the relative

54.
RELATIVE ERROR Note

Relative error is computed by dividing

the error by the total volume:

2

ΔV dV 4π r dr dr

≈ = 4 3 =3

V V 3πr r

Thus, the relative error in the volume is about

three times the relative error in the radius.

Relative error is computed by dividing

the error by the total volume:

2

ΔV dV 4π r dr dr

≈ = 4 3 =3

V V 3πr r

Thus, the relative error in the volume is about

three times the relative error in the radius.

55.
RELATIVE ERROR Note

In the example, the relative error in the radius

is approximately dr/r = 0.05/21 ≈ 0.0024

and it produces a relative error of about 0.007

in the volume.

The errors could also be expressed as

percentage errors of 0.24% in the radius

and 0.7% in the volume.

In the example, the relative error in the radius

is approximately dr/r = 0.05/21 ≈ 0.0024

and it produces a relative error of about 0.007

in the volume.

The errors could also be expressed as

percentage errors of 0.24% in the radius

and 0.7% in the volume.