Linear approximations and_differentials

Section 2.8 Linear Approximations and Differentials

2010 Kiryl Tsishchanka

Linear Approximations and Differentials
PROBLEM: Approximate the number

√
4

1.1.

IDEA: We have seen that a curve lies very close to its tangent
line near the point of tangency. This observation is the basis for a
method of finding approximate values of functions.
The idea is that it might be easy to calculate a value f (a) of a func√
tion (in our case 4 1), but difficult (or even impossible) to compute
√
nearby values of f (in our case 4 1.1). 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)).
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.
Solution: The point-slope equation of the tangent line is
y = f (a) + f ′ (a)(x − a)
therefore

f (x) ≈ f (a) + f ′ (a)(x − a) if x is close to a
√
In particular, if f (x) = 4 x, then
√
4

x≈

√
4
a+

1
(x − a)
4a3/4

1
. Plugging in x = 1.1 and a = 1, we get
4a3/4
√
√
1
1
4
4
1.1 ≈ 1 +
(1.1 − 1) = 1 + (0.1) = 1.025 (the true value is 1.024113689...)
3/4
4·1
4

since f ′ (a) =

The approximation
f (x) ≈ f (a) + f ′ (a)(x − a)

(1)

is called the linear approximation or tangent line approximation 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.

1

(2)



√
EXAMPLE: 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 underestimates?
√
Solution: The derivative of f (x) = x + 3 is
(
)′ 1
1
1
f ′ (x) = (x + 3)1/2 = (x + 3)1/2−1 = (x + 3)−1/2 = √
2
2
2 x+3
and so we have f (1) = 2 and f ′ (1) = 1 . Putting these values into (2), we see that the lineariza4
tion is
1
7 x
L(x) = f (1) + f ′ (1)(x − 1) = 2 + (x − 1) = +
4
4 4
The corresponding linear approximation (1) is
√
7 x
x+3≈ +
4 4

(when x is near 1)

In particular, we have
√
√
7 0.98
7 1.05
3.98 ≈ +
= 1.995 and
4.05 ≈ +
= 2.0125
4
4
4
4

(3)

The linear approximation is illustrated in the ﬁgure below.

We see that, indeed, the tangent line approximation is a good approximation to the given
function when x is near 1. We also see that our approximations are overestimates because the
tangent line lies above the curve.
√
√
Of course, a calculator could give us approximations for 3.98 and 4.05, but the linear
approximation gives an approximation over an entire interval.
In the table above we compare estimates from the obtained linear approximation with the
true values. Notice from this table, and also from the ﬁgure above, that the tangent line
approximation gives good estimates when x is close to 1 but the accuracy of the approximation
deteriorates when x is farther away from 1.
√
EXAMPLE: Find the linearization of the function f (x) = x at a = 4 and use it to approximate
√
√
the numbers 3.98 and 4.05. Are these approximations overestimates or underestimates?

2



√
EXAMPLE: Find the linearization of the function f (x) = x at a = 4 and use it to approximate
√
√
the numbers 3.98 and 4.05. Are these approximations overestimates or underestimates?
√
Solution: The derivative of f (x) = x is
(
)′ 1
1
1
f ′ (x) = x1/2 = x1/2−1 = x−1/2 = √
2
2
2 x
tion is
1
x
L(x) = f (4) + f ′ (4)(x − 4) = 2 + (x − 4) = 1 +
4
4
√
x
x≈1+
4

(when x is near 4)

√
√
3.98
4.05
3.98 ≈ 1 +
= 1.995 and
4.05 ≈ 1 +
= 2.0125
4
4
Note that we got the same results as in (3) since
1+

3.98
3 + 0.98
3 0.98
7 0.98
=1+
=1+ +
= +
4
4
4
4
4
4

1+

4.05
3 + 1.05
3 1.05
7 1.05
=1+
=1+ +
= +
4
4
4
4
4
4

and

or, in general,

x
3+x−3
3 x−3
7 x−3
=1+
=1+ +
= +
4
4
4
4
4
4
Our approximations are overestimates because the tangent line lies above the curve.
1+

√
EXAMPLE: Find the linearization of the function f (x) = 3 1 + x at a = 0 and use it to
√
√
approximate the numbers 3 0.95 and 3 1.1. Are these approximations overestimates or underestimates?

3



√
EXAMPLE: Find the linearization of the function f (x) = 3 1 + x at a = 0 and use it to
√
√
approximate the numbers 3 0.95 and 3 1.1. Are these approximations overestimates or underestimates?
√
Solution: The derivative of f (x) = 3 1 + x is
(
)′ 1
1
1
f ′ (x) = (1 + x)1/3 = (1 + x)1/3−1 = (1 + x)−2/3 = √
3
3
3 3 (1 + x)2
tion is
x
L(x) = f (0) + f ′ (0)(x − 0) = 1 +
3
√
x
3
1+x≈1+
(when x is near 0)
3
√
−0.05
3
0.95 ≈ 1 +
= 0.9833... (the true value is 0.9830475725...)
3
and
√
0.1
3
1.1 ≈ 1 +
= 1.0333... (the true value is 1.032280115...)
3
Our approximations are overestimates because the tangent line lies above the curve.
√
EXAMPLE: Find the linearization of the function f (x) = 3 x at a = 1 and use it to approximate
√
√
the numbers 3 0.95 and 3 1.1. Are these approximations overestimates or underestimates?
√
Solution: The derivative of f (x) = 3 x is
(
)′ 1
1
1
f ′ (x) = x1/3 = x1/3−1 = x−2/3 = √
3
3
3
3 x2
tion is
x−1
2 x
L(x) = f (1) + f ′ (1)(x − 1) = 1 +
= +
3
3 3
√
2 x
3
x≈ +
(when x is near 1)
3 3
√
2 0.95
3
= 0.9833... (the true value is 0.9830475725...)
0.95 ≈ +
3
3
and
√
2 1.1
3
= 1.0333... (the true value is 1.032280115...)
1.1 ≈ +
3
3
EXAMPLE: Find the linearization of the function f (x) = sin x at a = 0 and use it to approximate the numbers sin(−0.1) and sin(0.1). Are these approximations overestimates or underestimates?
4



EXAMPLE: Find the linearization of the function f (x) = sin x at a = 0 and use it to approximate the numbers sin(−0.1) and sin(0.1). Are these approximations overestimates or underestimates?
Solution: The derivative of f (x) = sin x is
f ′ (x) = (sin x)′ = cos x
and so we have f (0) = 0 and f ′ (0) = 1. Putting these values into (2), we see that the linearization is
L(x) = f (0) + f ′ (0)(x − 0) = 0 + 1 · (x − 0) = x
sin x ≈ x (when x is near 0)

sin(−0.1) ≈ −0.1 (the true value is -0.09983341665...)
and
sin(0.1) ≈ 0.1 (the true value is 0.09983341665...)
The ﬁrst approximation is an underestimate because the tangent line lies below the curve when
x is near 0 from the left. The second approximation is an overestimate because the tangent
line lies above the curve when x is near 0 from the right.
EXAMPLE: For what values of x is the linear approximation sin x ≈ x accurate to within 0.1?
Solution: Accuracy to within 0.1 means that the functions should diﬀer by less than 0.1:
| sin x − x| < 0.1

⇐⇒

−0.1 < sin x − x < 0.1

Using a graphing calculator we can conclude that the approximation sin x ≈ x is accurate to
within 0.1 when −0.86 < x < 0.86.

5



Differentials
The ideas behind linear approximations are sometimes formulated in the terminology and notation of differentials. 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. 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.
Let P (x, f (x)) and Q(x + ∆x, f (x + ∆x)) be points on the graph of f and let dx = ∆x. The
corresponding change in y is
∆y = f (x + ∆x) − f (x)
The slope of the tangent line P R is the derivative f ′ (x). Thus the directed distance from S to
R is f ′ (x)dx = dy. Therefore, dy represents the amount that the tangent line rises or falls (the
change in the linearization), whereas ∆y represents the amount that the curve y = f (x) rises
or falls when x changes by an amount dx. Notice that the approximation ∆y ≈ dy becomes
better as ∆x becomes smaller.

If we let dx = x − a, then x = a + dx and we can rewrite the linear approximation (1)
f (x) ≈ f (a) + f ′ (a)(x − a)
in the notation of differentials:
f (a + dx) ≈ f (a) + dy
√
For instance, for the function f (x) = x + 3 in Example 1, we have
dx
dy = f ′ (x)dx = √
2 x+3
If a = 1 and dx = ∆x = 0.05, then
0.05
= 0.0125
dy = √
2 1+3
and

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

just as we found in Example 1.
6

Linear approximations and_differentials

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