How to Draw a Tangent Line in Latex

4. Applications of Derivatives

4.2 Linear Approximations and Differentials

Learning Objectives

Describe the linear approximation to a function at a point.
Write the linearization of a given function.
Draw a graph that illustrates the use of differentials to approximate the change in a quantity.
Calculate the relative error and percentage error in using a differential approximation.

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 in the second volume of this text, when we studied how to approximate functions by higher-degree polynomials in the Introduction to Power Series and Functions.

Linear Approximation of a Function at a Point

Consider a function that is differentiable at a point . Recall that the tangent line to the graph of at is given by the equation

$y=f(a)+f^{\prime}(a)(x-a)$ .

For example, consider the function $f(x)=\frac{1}{x}$ at . Since is differentiable at and $f^{\prime}(x)=-\frac{1}{x^2}$ , we see that $f^{\prime}(2)=-\frac{1}{4}$ . Therefore, the tangent line to the graph of at is given by the equation

$y=\frac{1}{2}-\frac{1}{4}(x-2)$ .

(Figure)(a) shows a graph of $f(x)=\frac{1}{x}$ along with the tangent line to at . Note that for near 2, the graph of the tangent line is close to the graph of . As a result, we can use the equation of the tangent line to approximate for near 2. For example, if , the value of the corresponding point on the tangent line is

$y=\frac{1}{2}-\frac{1}{4}(2.1-2)=0.475$ .

The actual value of is given by

$f(2.1)=\frac{1}{2.1}\approx 0.47619$ .

Therefore, the tangent line gives us a fairly good approximation of ((Figure)(b)). However, note that for values of far from 2, the equation of the tangent line does not give us a good approximation. For example, if , the -value of the corresponding point on the tangent line is

$y=\frac{1}{2}-\frac{1}{4}(10-2)=\frac{1}{2}-2=-1.5$ ,

whereas the value of the function at is .

In general, for a differentiable function , the equation of the tangent line to at can be used to approximate for near . Therefore, we can write

$f(x)\approx f(a)+f^{\prime}(a)(x-a)$ for .

We call the linear function

$L(x)=f(a)+f^{\prime}(a)(x-a)$

the linear approximation, or tangent line approximation, of at . This function is also known as the linearization of at .

To show how useful the linear approximation can be, we look at how to find the linear approximation for $f(x)=\sqrt{x}$ at .

Linear Approximation of $\sqrt{x}$

Find the linear approximation of $f(x)=\sqrt{x}$ at and use the approximation to estimate $\sqrt{9.1}$ .

Solution

Since we are looking for the linear approximation at , using (Figure) we know the linear approximation is given by

$L(x)=f(9)+f^{\prime}(9)(x-9)$ .

We need to find and $f^{\prime}(9)$ .

$\begin{array}{lll} f(x)=\sqrt{x}& \Rightarrow & f(9)=\sqrt{9}=3 \\ f^{\prime}(x)=\frac{1}{2\sqrt{x}}& \Rightarrow & f^{\prime}(9)=\frac{1}{2\sqrt{9}}=\frac{1}{6} \end{array}$

Therefore, the linear approximation is given by (Figure).

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

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

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

Find the local linear approximation to $f(x)=\sqrt[3]{x}$ at Use it to approximate $\sqrt[3]{8.1}$ to five decimal places.

Solution

$L(x)=2+\frac{1}{12}(x-8);$ 2.00833

Linear Approximation of $\sin x$

Find the linear approximation of $f(x)= \sin x$ at $x=\frac{\pi }{3}$ and use it to approximate $\sin (62\text{°}).$

Solution

First we note that since $\frac{\pi }{3}$ rad is equivalent to $60\text{°},$ using the linear approximation at $x=\pi \text{/}3$ seems reasonable. The linear approximation is given by

$L(x)=f(\frac{\pi }{3})+f\prime (\frac{\pi }{3})(x-\frac{\pi }{3}).$

We see that

$\begin{array}{ccc}\hfill f(x)= \sin x& ⇒\hfill & f(\frac{\pi }{3})= \sin (\frac{\pi }{3})=\frac{\sqrt{3}}{2}\hfill \\ \hfill f\prime (x)= \cos x& ⇒\hfill & f\prime (\frac{\pi }{3})= \cos (\frac{\pi }{3})=\frac{1}{2}\hfill \end{array}$

Therefore, the linear approximation of at $x=\pi \text{/}3$ is given by (Figure).

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

To estimate $\sin (62\text{°})$ using we must first convert 62° to radians. We have $62\text{°}=\frac{62\pi }{180}$ radians, so the estimate for $\sin (62\text{°})$ is given by

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

Find the linear approximation for $f(x)= \cos x$ at $x=\frac{\pi }{2}.$

$L(x)=\text{−}x+\frac{\pi }{2}$

Linear approximations may be used in estimating roots and powers. In the next example, we find the linear approximation for $f(x)={(1+x)}^{n}$ at 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 $f(x)={(m+x)}^{n}$ to estimate roots and powers near a different number

Approximating Roots and Powers

Find the linear approximation of $f(x)={(1+x)}^{n}$ at Use this approximation to estimate ${(1.01)}^{3}.$

Solution

The linear approximation at is given by

$L(x)=f(0)+f\prime (0)(x-0).$

Because

$\begin{array}{ccc}\hfill f(x)={(1+x)}^{n}& ⇒\hfill & f(0)=1\hfill \\ \hfill f\prime (x)=n{(1+x)}^{n-1}& ⇒\hfill & f\prime (0)=n,\hfill \end{array}$

the linear approximation is given by (Figure)(a).

We can approximate ${(1.01)}^{3}$ by evaluating when We conclude that

${(1.01)}^{3}=f(1.01)\approx L(1.01)=1+3(0.01)=1.03.$

Find the linear approximation of $f(x)={(1+x)}^{4}$ at without using the result from the preceding example.

Solution

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.

When we first looked at derivatives, we used the Leibniz notation $dy\text{/}dx$ to represent the derivative of with respect to 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 is a differentiable function. Let dx be an independent variable that can be assigned any nonzero real number, and define the dependent variable by

$dy=f\prime (x)dx.$

It is important to notice that is a function of both and The expressions dy and dx are called differentials. We can divide both sides of (Figure) by which yields

$\frac{dy}{dx}=f\prime (x).$

This is the familiar expression we have used to denote a derivative. (Figure) is known as the differential form of (Figure).

Computing differentials

Solution

The key step is calculating the derivative. When we have that, we can obtain dy directly.

Since $f(x)={x}^{2}+2x,$ we know $f\prime (x)=2x+2,$ and therefore

When and
Since $f(x)= \cos x,$ $f\prime (x)=\text{−} \sin (x).$ This gives us
$dy=\text{−} \sin xdx.$

When and

$dy=\text{−} \sin (3)(0.1)=-0.1 \sin (3).$

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 that is differentiable at point Suppose the input changes by a small amount. We are interested in how much the output changes. If changes from to then the change in is (also denoted $\text{Δ}x),$ and the change in is given by

$\text{Δ}y=f(a+dx)-f(a).$

Instead of calculating the exact change in however, it is often easier to approximate the change in by using a linear approximation. For near can be approximated by the linear approximation

$L(x)=f(a)+f\prime (a)(x-a).$

Therefore, if is small,

$f(a+dx)\approx L(a+dx)=f(a)+f\prime (a)(a+dx-a).$

That is,

$f(a+dx)-f(a)\approx L(a+dx)-f(a)=f\prime (a)dx.$

In other words, the actual change in the function if increases from to is approximately the difference between and where is the linear approximation of at By definition of this difference is equal to $f\prime (a)dx.$ In summary,

$\text{Δ}y=f(a+dx)-f(a)\approx L(a+dx)-f(a)=f\prime (a)dx=dy.$

Therefore, we can use the differential $dy=f\prime (a)dx$ to approximate the change in if increases from to We can see this in the following graph.

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.

Approximating Change with Differentials

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.

Consider a function with an input that is a measured quantity. Suppose the exact value of the measured quantity is but the measured value is We say the measurement error is dx (or $\text{Δ}x).$ As a result, an error occurs in the calculated quantity This type of error is known as a propagated error and is given by

$\text{Δ}y=f(a+dx)-f(a).$

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 $\text{Δ}y.$ Specifically, if is a differentiable function at the propagated error is

$\text{Δ}y\approx dy=f\prime (a)dx.$

Unfortunately, we do not know the exact value However, we can use the measured value and estimate

$\text{Δ}y\approx dy\approx f\prime (a+dx)dx.$

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.

Solution

The measurement of the side length is accurate to within $\text{±}0.1$ cm. Therefore,
$-0.1\le dx\le 0.1.$

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

$dV=3{x}^{2}dx.$

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

$-3{(5)}^{2}(0.1)\le dV\le 3{(5)}^{2}(0.1).$

Therefore,

$-7.5\le dV\le 7.5.$
If the side length is actually 4.9 cm, then the volume of the cube is
$V(4.9)={(4.9)}^{3}=117.649{\text{cm}}^{3}.$

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

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

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

$117.649-125\le \text{Δ}V\le 132.651-125.$

That is,

$-7.351\le \text{Δ}V\le 7.651.$

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

Estimate the error in the computed volume of a cube if the side length is measured to be 6 cm with an accuracy of 0.2 cm.

Solution

The volume measurement is accurate to within $21.6{\text{cm}}^{3}.$

The measurement error dx $(\text{=Δ}x)$ and the propagated error $\text{Δ}y$ 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 $\text{Δ}q$ for a particular quantity, we define the relative error as $\frac{\text{Δ}q}{q},$ where 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 $\frac{1}{62}=0.016,$ or $1.6\text{%}.$ 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 $\frac{1}{4}$ in., whereas the relative error is $\frac{0.25}{8}=\frac{1}{32},$ or $3.1\text{%}.$ 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 $\text{±}80$ 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.

Solution

If the measurement of the radius is accurate to within $\text{±}80,$ we have

$-80\le dr\le 80.$

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

$dV=4\pi {r}^{2}dr.$

Using the measured radius of 4000 mi, we can estimate

$-4\pi {(4000)}^{2}(80)\le dV\le 4\pi {(4000)}^{2}(80).$

To estimate the relative error, consider $\frac{dV}{V}.$ Since we do not know the exact value of the volume use the measured radius $r=4000\text{mi}$ to estimate We obtain $V\approx (\frac{4}{3})\pi {(4000)}^{3}.$ Therefore the relative error satisfies

$\frac{-4\pi {(4000)}^{2}(80)}{4\pi {(4000)}^{3}\text{/}3}\le \frac{dV}{V}\le \frac{4\pi {(4000)}^{2}(80)}{4\pi {(4000)}^{3}\text{/}3},$

which simplifies to

$-0.06\le \frac{dV}{V}\le 0.06.$

The relative error is 0.06 and the percentage error is $6\text{%}.$

Determine the percentage error if the radius of Earth is measured to be 3950 mi with an error of $\text{±}100$ mi.

Key Concepts

Key Equations

1. What is the linear approximation for any generic linear function

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

Solution

${f}^{\prime }(a)=0$

3. Explain why the linear approximation becomes less accurate as you increase the distance between and Use a graph to prove your argument.

4. When is the linear approximation exact?

Solution

The linear approximation exact when is linear or constant.

For the following exercises, find the linear approximation to near for the function.

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

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

Solution

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

7. [T] $f(x)= \tan x,a=\frac{\pi }{4}$

8. [T] $f(x)= \sin x,a=\frac{\pi }{2}$

Solution

9. [T] $f(x)=x \sin x,a=2\pi$

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

Solution

For the following exercises, compute the values given within 0.01 by deciding on the appropriate and and evaluating $L(x)=f(a)+{f}^{\prime }(a)(x-a).$ Check your answer using a calculator.

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

12. [T] $\sin (0.02)$

13. [T] $\cos (0.03)$

14. [T] ${(15.99)}^{1\text{/}4}$

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

16. [T] $\sin (3.14)$

For the following exercises, determine the appropriate and and evaluate $L(x)=f(a)+{f}^{\prime }(a)(x-a).$ Calculate the numerical error in the linear approximations that follow.

17. ${(1.01)}^{3}$

19. ${( \sin (0.01))}^{2}$

21. ${(1+\frac{1}{10})}^{10}$

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

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

24. $y=x \cos x$

Solution

$dy=( \cos x-x \sin x)dx$

25. $y=\sqrt{1+x}$

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

Solution

$dy=(\frac{{x}^{2}-2x-2}{{(x-1)}^{2}})dx$

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

27. $y=3{x}^{2}-x+6,$ [latex]x=2,[/latex]

28. $y=\frac{1}{x+1},$ [latex]x=1,[/latex]

Solution

$dy=-\frac{1}{{(x+1)}^{2}}dx,$ [latex]-\frac{1}{16}[/latex]

29. $y= \tan x,$ [latex]x=0,[/latex] $dx=\frac{\pi }{10}$

30. $y=\frac{3{x}^{2}+2}{\sqrt{x+1}},$ [latex]x=0,[/latex]

Solution

$dy=\frac{9{x}^{2}+12x-2}{2{(x+1)}^{3\text{/}2}}dx,$ -0.1

31. $y=\frac{ \sin (2x)}{x},$ [latex]x=\pi ,[/latex]

32. $y={x}^{3}+2x+\frac{1}{x},$ [latex]x=1,[/latex]

Solution

$dy=(3{x}^{2}+2-\frac{1}{{x}^{2}})dx,$ 0.2

For the following exercises, find the change in volume or in surface area

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

34. if the sides of a cube change from to

Solution

36. if the radius of a sphere changes from by

Solution

$4\pi {r}^{2}dr$

38. if a circular cylinder of height 3 changes from to $r=1.9\text{cm}.$

Solution

$-1.2\pi {\text{cm}}^{3}$

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

39. A spherical golf ball is measured to have a radius of $5\text{mm},$ with a possible measurement error of $0.1\text{mm}.$ What is the possible change in volume?

40. 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?

41. 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

42. $\sqrt{1-x}\approx 1-\frac{1}{2}x$

43. $\frac{1}{\sqrt{1-{x}^{2}}}\approx 1$

44. $\sqrt{{c}^{2}+{x}^{2}}\approx c$

How to Draw a Tangent Line in Latex

Source: https://opentextbc.ca/calculusv1openstax/chapter/linear-approximations-and-differentials/

How to Draw a Tangent Line in Latex

4.2 Linear Approximations and Differentials

Learning Objectives

Linear Approximation of a Function at a Point

Linear Approximation of

Solution

Solution

Linear Approximation of

Solution

Approximating Roots and Powers

Solution

Solution

Differentials

Computing differentials

Solution

Approximating Change with Differentials

Calculating the Amount of Error

Volume of a Cube

Solution

Solution

Relative and Percentage Error

Solution

Key Concepts

Key Equations

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

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Linear Approximation of $\sqrt{x}$

Linear Approximation of $\sin x$