# Linear Approximations and Differentials with examples

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In this section, we will learn about: Linear approximations and differentials and their applications.
1. 3
DIFFERENTIATION RULES
2. DIFFERENTIATION RULES
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.
4. DIFFERENTIATION RULES
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.
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)).
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)
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.
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.
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
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.
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
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
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.
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.
16. LINEAR APPROXIMATIONS
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.
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.
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?
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 ⎠
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.
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.
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
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.
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
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.
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.
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
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
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.
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.
32. APPLICATIONS TO PHYSICS
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.
34. The ideas behind linear approximations
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.
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.
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.
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.
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.
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
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
42. DIFFERENTIALS Example 3 a
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
44. Notice that:
 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.
46. In the notation of differentials,
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
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.
49. Our final example illustrates the use
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?
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.
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.
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
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.
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.