Linear Approximations and Differentials

We have just seen how derivatives allow us to compare related quantities that are changing over time. In this section, we examine another application of derivatives: the ability to approximate functions locally by linear functions. Linear functions are the easiest functions with which to work, so they provide a useful tool for approximating function values. In addition, the ideas presented in this section are generalized later in the text when we study how to approximate functions by higher-degree polynomials Introduction to Power Series and Functions.

Linear Approximation of a Function at a Point

far from 2, the equation of the tangent line does not give us a good approximation. For example, if

x = 10,

To show how useful the linear approximation can be, we look at how to find the linear approximation for

f (x) = \sqrt{x}

Linear Approximation of

\sqrt{x}

Find the linear approximation of $f (x) = \sqrt{x}$

at $x = 9$

and use the approximation to estimate $\sqrt{9.1} .$

Since we are looking for the linear approximation at $x = 9,$

using [link] we know the linear approximation is given by

L (x) = f (9) + f' (9) (x - 9) .

We need to find $f (9)$

and $f' (9) .$

\begin{matrix} f (x) = \sqrt{x} & \Rightarrow & f (9) = \sqrt{9} = 3 \\ f' (x) = \frac{1}{2 \sqrt{x}} & \Rightarrow & f' (9) = \frac{1}{2 \sqrt{9}} = \frac{1}{6} \end{matrix}

Therefore, the linear approximation is given by [link].

L (x) = 3 + \frac{1}{6} (x - 9)

Using the linear approximation, we can estimate $\sqrt{9.1}$

by writing

\sqrt{9.1} = f (9.1) \approx L (9.1) = 3 + \frac{1}{6} (9.1 - 9) \approx 3.0167 .

The function f(x) = the square root of x is shown with its tangent at (9, 3). The tangent appears to be a very good approximation from x = 6 to x = 12.

Analysis

Using a calculator, the value of $\sqrt{9.1}$

to four decimal places is 3.0166. The value given by the linear approximation, 3.0167, is very close to the value obtained with a calculator, so it appears that using this linear approximation is a good way to estimate $\sqrt{x},$

at least for $x$

near $9 .$

At the same time, it may seem odd to use a linear approximation when we can just push a few buttons on a calculator to evaluate $\sqrt{9.1} .$

However, how does the calculator evaluate $\sqrt{9.1} ?$

The calculator uses an approximation! In fact, calculators and computers use approximations all the time to evaluate mathematical expressions; they just use higher-degree approximations.

Linear Approximation of

sin x

Find the linear approximation of $f (x) = sin x$

at $x = \frac{π}{3}$

and use it to approximate $sin (62 °) .$

First we note that since $\frac{π}{3}$

rad is equivalent to $60 °,$

using the linear approximation at $x = π / 3$

seems reasonable. The linear approximation is given by

L (x) = f (\frac{π}{3}) + f' (\frac{π}{3}) (x - \frac{π}{3}) .

We see that

\begin{matrix} f (x) = sin x & \Rightarrow & f (\frac{π}{3}) = sin (\frac{π}{3}) = \frac{\sqrt{3}}{2} \\ f' (x) = cos x & \Rightarrow & f' (\frac{π}{3}) = cos (\frac{π}{3}) = \frac{1}{2} \end{matrix}

Therefore, the linear approximation of $f$

at $x = π / 3$

is given by [link].

L (x) = \frac{\sqrt{3}}{2} + \frac{1}{2} (x - \frac{π}{3})

To estimate $sin (62 °)$

using $L,$

we must first convert $62 °$

to radians. We have $62 ° = \frac{62 π}{180}$

radians, so the estimate for $sin (62 °)$

is given by

sin (62 °) = f (\frac{62 π}{180}) \approx L (\frac{62 π}{180}) = \frac{\sqrt{3}}{2} + \frac{1}{2} (\frac{62 π}{180} - \frac{π}{3}) = \frac{\sqrt{3}}{2} + \frac{1}{2} (\frac{2 π}{180}) = \frac{\sqrt{3}}{2} + \frac{π}{180} \approx 0.88348 .

The function f(x) = sin x is shown with its tangent at (π/3, square root of 3 / 2). The tangent appears to be a very good approximation for x near π / 3.

Linear approximations may be used in estimating roots and powers. In the next example, we find the linear approximation for

f (x) = {(1 + x)}^{n}

which can be used to estimate roots and powers for real numbers near 1. The same idea can be extended to a function of the form

f (x) = {(m + x)}^{n}

Differentials

We have seen that linear approximations can be used to estimate function values. They can also be used to estimate the amount a function value changes as a result of a small change in the input. To discuss this more formally, we define a related concept: differentials. Differentials provide us with a way of estimating the amount a function changes as a result of a small change in input values.

Although we used the expressions dy and dx in this notation, they did not have meaning on their own. Here we see a meaning to the expressions dy and dx. Suppose

y = f (x)

is a differentiable function. Let dx be an independent variable that can be assigned any nonzero real number, and define the dependent variable

d y

The expressions dy and dx are called differentials. We can divide both sides of [link] by

d x,

This is the familiar expression we have used to denote a derivative. [link] is known as the differential form of [link].

We now connect differentials to linear approximations. Differentials can be used to estimate the change in the value of a function resulting from a small change in input values. Consider a function

f

We now take a look at how to use differentials to approximate the change in the value of the function that results from a small change in the value of the input. Note the calculation with differentials is much simpler than calculating actual values of functions and the result is very close to what we would obtain with the more exact calculation.

Calculating the Amount of Error

Any type of measurement is prone to a certain amount of error. In many applications, certain quantities are calculated based on measurements. For example, the area of a circle is calculated by measuring the radius of the circle. An error in the measurement of the radius leads to an error in the computed value of the area. Here we examine this type of error and study how differentials can be used to estimate the error.

with an input that is a measured quantity. Suppose the exact value of the measured quantity is

a,

Since all measurements are prone to some degree of error, we do not know the exact value of a measured quantity, so we cannot calculate the propagated error exactly. However, given an estimate of the accuracy of a measurement, we can use differentials to approximate the propagated error

Δ y .

In the next example, we look at how differentials can be used to estimate the error in calculating the volume of a box if we assume the measurement of the side length is made with a certain amount of accuracy.

Volume of a Cube

Suppose the side length of a cube is measured to be 5 cm with an accuracy of 0.1 cm.

Use differentials to estimate the error in the computed volume of the cube.
Compute the volume of the cube if the side length is (i) 4.9 cm and (ii) 5.1 cm to compare the estimated error with the actual potential error.

The measurement of the side length is accurate to within $± 0.1$
cm. Therefore,

$−0.1 \leq d x \leq 0.1 .$

The volume of a cube is given by
$V = x^{3},$
which leads to

$d V = 3 x^{2} d x .$

Using the measured side length of 5 cm, we can estimate that

$−3 {(5)}^{2} (0.1) \leq d V \leq 3 {(5)}^{2} (0.1) .$

Therefore,

$−7.5 \leq d V \leq 7.5 .$
If the side length is actually 4.9 cm, then the volume of the cube is

$V (4.9) = {(4.9)}^{3} = 117.649 {cm}^{3} .$

If the side length is actually 5.1 cm, then the volume of the cube is

$V (5.1) = {(5.1)}^{3} = 132.651 {cm}^{3} .$

Therefore, the actual volume of the cube is between 117.649 and 132.651. Since the side length is measured to be 5 cm, the computed volume is
$V (5) = 5^{3} = 125 .$
Therefore, the error in the computed volume is

$117.649 - 125 \leq Δ V \leq 132.651 - 125 .$

That is,

$−7.351 \leq Δ V \leq 7.651 .$

We see the estimated error
$d V$
is relatively close to the actual potential error in the computed volume.

are absolute errors. We are typically interested in the size of an error relative to the size of the quantity being measured or calculated. Given an absolute error

Δ q

for a particular quantity, we define the relative error as

\frac{Δ q}{q},

is the actual value of the quantity. The percentage error is the relative error expressed as a percentage. For example, if we measure the height of a ladder to be 63 in. when the actual height is 62 in., the absolute error is 1 in. but the relative error is

\frac{1}{62} = 0.016,

By comparison, if we measure the width of a piece of cardboard to be 8.25 in. when the actual width is 8 in., our absolute error is

\frac{1}{4}

Therefore, the percentage error in the measurement of the cardboard is larger, even though 0.25 in. is less than 1 in.

Relative and Percentage Error

An astronaut using a camera measures the radius of Earth as 4000 mi with an error of $± 80$

mi. Let’s use differentials to estimate the relative and percentage error of using this radius measurement to calculate the volume of Earth, assuming the planet is a perfect sphere.

If the measurement of the radius is accurate to within $± 80,$

we have

−80 \leq d r \leq 80 .

Since the volume of a sphere is given by $V = (\frac{4}{3}) π r^{3},$

we have

d V = 4 π r^{2} d r .

Using the measured radius of 4000 mi, we can estimate

−4 π {(4000)}^{2} (80) \leq d V \leq 4 π {(4000)}^{2} (80) .

To estimate the relative error, consider $\frac{d V}{V} .$

Since we do not know the exact value of the volume $V,$

use the measured radius $r = 4000 mi$

to estimate $V .$

We obtain $V \approx (\frac{4}{3}) π {(4000)}^{3} .$

Therefore the relative error satisfies

\frac{−4 π {(4000)}^{2} (80)}{4 π {(4000)}^{3} / 3} \leq \frac{d V}{V} \leq \frac{4 π {(4000)}^{2} (80)}{4 π {(4000)}^{3} / 3},

which simplifies to

−0.06 \leq \frac{d V}{V} \leq 0.06 .

The relative error is 0.06 and the percentage error is $6 % .$

Key Concepts

Key Equations

What is the linear approximation for any generic linear function $y = m x + b ?$

Determine the necessary conditions such that the linear approximation function is constant. Use a graph to prove your result.

f^{'} (a) = 0

Explain why the linear approximation becomes less accurate as you increase the distance between $x$

and $a .$

Use a graph to prove your argument.

When is the linear approximation exact?

The linear approximation exact when $y = f (x)$

is linear or constant.

For the following exercises, find the linear approximation $L (x)$

to $y = f (x)$

near $x = a$

for the function.

[T] $f (x) = x + x^{4}, a = 0$

[T] $f (x) = \frac{1}{x}, a = 2$

L (x) = \frac{1}{2} - \frac{1}{4} (x - 2)

[T] $f (x) = tan x, a = \frac{π}{4}$

[T] $f (x) = sin x, a = \frac{π}{2}$

L (x) = 1

[T] $f (x) = x sin x, a = 2 π$

[T] $f (x) = {sin}^{2} x, a = 0$

L (x) = 0

For the following exercises, compute the values given within 0.01 by deciding on the appropriate $f (x)$

and $a,$

and evaluating $L (x) = f (a) + f^{'} (a) (x - a) .$

Check your answer using a calculator.

[T] ${(2.001)}^{6}$

[T] $sin (0.02)$

0.02

[T] $cos (0.03)$

[T] ${(15.99)}^{1 / 4}$

1.9996875

[T] $\frac{1}{0.98}$

[T] $sin (3.14)$

0.001593

For the following exercises, determine the appropriate $f (x)$

and $a,$

and evaluate $L (x) = f (a) + f^{'} (a) (x - a) .$

Calculate the numerical error in the linear approximations that follow.

{(1.01)}^{3}

cos (0.01)

1;

error, $~0.00005$

{(sin (0.01))}^{2}

{(1.01)}^{−3}

0.97;

error, $~0.0006$

{(1 + \frac{1}{10})}^{10}

\sqrt{8.99}

3 - \frac{1}{600};

error, $~4.632 \times 10^{−7}$

For the following exercises, find the differential of the function.

y = 3 x^{4} + x^{2} - 2 x + 1

y = x cos x

d y = (cos x - x sin x) d x

y = \sqrt{1 + x}

y = \frac{x^{2} + 2}{x - 1}

d y = (\frac{x^{2} - 2 x - 2}{{(x - 1)}^{2}}) d x

For the following exercises, find the differential and evaluate for the given $x$

and $d x .$

y = 3 x^{2} - x + 6,

x = 2,

d x = 0.1

y = \frac{1}{x + 1},

x = 1,

d x = 0.25

d y = - \frac{1}{{(x + 1)}^{2}} d x,

- \frac{1}{16}

y = tan x,

x = 0,

d x = \frac{π}{10}

y = \frac{3 x^{2} + 2}{\sqrt{x + 1}},

x = 0,

d x = 0.1

d y = \frac{9 x^{2} + 12 x - 2}{2 {(x + 1)}^{3 / 2}} d x,

−0.1

y = \frac{sin (2 x)}{x},

x = π,

d x = 0.25

y = x^{3} + 2 x + \frac{1}{x},

x = 1,

d x = 0.05

d y = (3 x^{2} + 2 - \frac{1}{x^{2}}) d x,

0.2

For the following exercises, find the change in volume $d V$

or in surface area $d A .$

d V

if the sides of a cube change from 10 to 10.1.

d A

if the sides of a cube change from $x$

to $x + d x .$

12 x d x

d A

if the radius of a sphere changes from $r$

by $d r .$

d V

if the radius of a sphere changes from $r$

by $d r .$

4 π r^{2} d r

d V

if a circular cylinder with $r = 2$

changes height from 3 cm to $3.05 cm .$

d V

if a circular cylinder of height 3 changes from $r = 2$

to $r = 1.9 cm .$

−1.2 π {cm}^{3}

For the following exercises, use differentials to estimate the maximum and relative error when computing the surface area or volume.

A spherical golf ball is measured to have a radius of $5 mm,$

with a possible measurement error of $0.1 mm .$

What is the possible change in volume?

A pool has a rectangular base of 10 ft by 20 ft and a depth of 6 ft. What is the change in volume if you only fill it up to 5.5 ft?

−100

ft³

An ice cream cone has height 4 in. and radius 1 in. If the cone is 0.1 in. thick, what is the difference between the volume of the cone, including the shell, and the volume of the ice cream you can fit inside the shell?

For the following exercises, confirm the approximations by using the linear approximation at $x = 0 .$

\sqrt{1 - x} \approx 1 - \frac{1}{2} x

\frac{1}{\sqrt{1 - x^{2}}} \approx 1

\sqrt{c^{2} + x^{2}} \approx c

Glossary

differential

the differential

d x

is an independent variable that can be assigned any nonzero real number; the differential

d y

is defined to be

d y = f' (x) d x

differential form

given a differentiable function

y = f' (x),

the equation

d y = f' (x) d x

is the differential form of the derivative of

y

with respect to

x

linear approximation

the linear function

L (x) = f (a) + f' (a) (x - a)

is the linear approximation of

f

x = a

percentage error: the relative error expressed as a percentage

propagated error: the error that results in a calculated quantity $f (x)$
resulting from a measurement error dx

relative error

given an absolute error

Δ q

for a particular quantity,

\frac{Δ q}{q}

is the relative error.

tangent line approximation (linearization)

since the linear approximation of

f

x = a

is defined using the equation of the tangent line, the linear approximation of

f

x = a

is also known as the tangent line approximation to

f

x = a