1. Linearization and Differentials
Chapter 5.5

2. Linear Approximation
A useful characteristic of the tangent line to a curve at a point is that, for 𝑥-values near the point, the curve is approximately linear
In fact, the function values of the curve are approximated by the derivative values near the point of tangency
We say that differentiable curves are always locally linear
In this section we will explore local linearity

3. Linear Approximation

4. Linear Approximation

5. Linear Approximation

6. Linearization DEFINITION:
If 𝑓 is differentiable at 𝑥=𝑎, then the equation of the tangent line,
𝐿 𝑥 = 𝑓 ′ 𝑎 𝑥−𝑎 +𝑓 𝑎
defines the linearization of 𝒇 at 𝒂. The approximation 𝑓 𝑥 ≈𝐿(𝑥) is the standard linear approximation of 𝒇 at 𝒂. The point 𝑥=𝑎 is the center of approximation.

7. Example 1: Finding a Linearization
Find the linearization of 𝑓 𝑥 = 1+𝑥 at 𝑥=0, and use it to approximate without a calculator. Then use a calculator to determine the accuracy of the approximation.

8. Example 1: Finding a Linearization
Find the linearization of 𝑓 𝑥 = 1+𝑥 at 𝑥=0, and use it to approximate without a calculator. Then use a calculator to determine the accuracy of the approximation.
Note that we are approximating 𝑓 0.02 = The center of the approximation is at 𝑥=0; that is, 𝑎=0 in
𝐿 𝑥 = 𝑓 ′ 𝑎 𝑥−𝑎 +𝑓(𝑎)
Find the derivative: 𝑓 ′ 𝑥 = 𝑥 ; find 𝑓 ′ 0 = 1 2 and 𝑓 0 =1. So our linear approximation is 𝑓 𝑥 ≈ 1 2 𝑥−0 +1= 1 2 𝑥+1≈ 1+𝑥
Therefore, 𝑓 0.2 ≈ =1.01

9. Example 1: Finding a Linearization
Find the linearization of 𝑓 𝑥 = 1+𝑥 at 𝑥=0, and use it to approximate without a calculator. Then use a calculator to determine the accuracy of the approximation.
The calculator value is ≈
The difference is 1.01− = < 10 −4
Note that as we choose values farther from zero, the approximation becomes less accurate. For example, the approximation for 3 = would be 1 2 ⋅2+1=2, which differs from the true value by ≈0.268 or about 16%. We could get a better approximation by choosing a center closer to 𝑥=2

10. Example 1: Finding a Linearization

11. Example 1: Finding a Linearization

12. Example 2: Finding a Linearization
Find the linearization of 𝑓 𝑥 = cos 𝑥 at 𝑥= 𝜋 2 and use it to approximate cos without a calculator. Then use a calculator to determine the accuracy of the approximation.

13. Example 2: Finding a Linearization
Find the linearization of 𝑓 𝑥 = cos 𝑥 at 𝑥= 𝜋 2 and use it to approximate cos without a calculator. Then use a calculator to determine the accuracy of the approximation.
Here, 𝑎= 𝜋 2 , so 𝐿 𝑥 =− sin 𝜋 2 𝑥− 𝜋 2 + cos 𝜋 2 =−𝑥+ 𝜋 2
The approximation is
𝑓 1.75 ≈−1.75+ 𝜋 2 ≈−0.179
The calculator value to 9 decimal places is − The difference is
− = < 10 −3
This amounts to a percent difference of about 0.5%

14. Example 3: Approximating Binomial Powers
Example 1 introduces a special case of a general linearization formula that applies to powers of 1+𝑥 for small values of 𝑥: 1+𝑥 𝑘 ≈1+𝑘𝑥 If 𝑘 is a positive integer this follows from the Binomial Theorem, but the formula actually holds for all real values of 𝑘. Use this formula to find polynomials that will approximate the following functions for values of 𝑥 close to zero: a) 3 1−𝑥 b) 1 1−𝑥 c) 1+5 𝑥 4 d) 1 1− 𝑥 2

15. Example 3: Approximating Binomial Powers
a) 3 1−𝑥 b) 1 1−𝑥 c) 𝑥 4 d) − 𝑥 2
The first example does not seem to follow the pattern. One technique to use when this happens is to substitute a variable so that the pattern is the same. Let 𝑦=−𝑥. The function is therefore 1+𝑦 ≈ 𝑦. Now back-substitute to get
3 1−𝑥 ≈1− 1 3 𝑥
This example shows that we need merely use the term in place of 𝑥 in the formula. So the rest of the examples are:
1 1−𝑥 ≈1− −𝑥 =1+𝑥, 𝑥 4 ≈ 𝑥 4 , 1 1− 𝑥 2 ≈ 𝑥 2

16. Example 4: Approximating Roots
Use linearization to approximate. State how accurate your approximation is.
123
3 123

17. Example 4: Approximating Roots
123
Use the function 𝑓 𝑥 = 𝑥 and center it around 𝑥=121. Then 𝑓 ′ 𝑥 = 1 2 𝑥 and 𝑓 121 =11, 𝑓 ′ 121 = So, 𝐿 𝑥 = 𝑥− Linearization means that 𝐿 𝑥 ≈𝑓(𝑥) at values of 𝑥 near the center. Therefore, 𝐿 123 ≈𝑓 123 = 𝐿 123 = ≈11.091
The difference between and our approximation is: − 1 11 −11 ≈
This number is less than 10 −4 , so 𝐿 123 −𝑓 123 < 10 −4

18. Example 4: Approximating Roots
3 123
The nearest perfect cube to 123 is 5 3 =125, so we will center around 𝑥=125. Our function is 𝑓 𝑥 = 3 𝑥 and we want an approximation to 𝑓 Linearization requires 𝑓 ′ 𝑥 = 1 3 𝑥 Now, 𝑓 125 =5 and 𝑓 ′ 125 = Our linearization formula is 𝐿 𝑥 = 𝑥− Use this to approximate 𝑓 123 : 𝑓 123 = ≈𝐿 123 = ≈5.0267
The difference between 𝑓(123) and 𝐿(123) is 𝑓 123 −𝐿 123 ≈ < 10 −3

19. Differentials
The notation 𝑑𝑦 𝑑𝑥 was Leibniz means of representing the derivative of a function
It is written as a ratio because Leibniz believed that it was a ratio (so did Newton)
But this was their thinking prior to the development of the limit concept almost 200 years later
Recall that we define this notation as lim Δ𝑥→0 Δ𝑦 Δ𝑥 = 𝑑𝑦 𝑑𝑥
It is not a ratio, then, but a limit

20. Differentials
This matters because it is a mistake to think that, since it is written as a ratio, then all the laws relating to ratios also apply
In mathematics, we can often decide what we want something to mean (within reason) by defining what it means
Our definition cannot be such that other theorems are contradicted by results obtained from our definition
We will define the “numerator” and “denominator” of 𝑑𝑦 𝑑𝑥 separately in a way that allows us to use them separately

21. Differentials DEFINITION:
Let 𝑦=𝑓(𝑥) be a differentiable function so that 𝑑𝑦 𝑑𝑥 =𝑓′(𝑥). The differential 𝒅𝒙 is an independent variable (that is, it’s a number) and the differential 𝒅𝒚 is
𝑑𝑦= 𝑓 ′ 𝑥 𝑑𝑥
Note that 𝑑𝑦 is a dependent variable and also a function dependent on both 𝑥 and 𝑑𝑥.

22. Example 5: Finding the Differential 𝑑𝑦
Find the differential 𝑑𝑦 for the given values of 𝑥 and 𝑑𝑥.
𝑦= 𝑥 5 +37𝑥, 𝑥=1, 𝑑𝑥=0.01
𝑦= sin 3𝑥 , 𝑥=𝜋, 𝑑𝑥=−0.02
𝑥+𝑦=𝑥𝑦, 𝑥=2, 𝑑𝑥=0.05

23. Example 5: Finding the Differential 𝑑𝑦
Find the differential 𝑑𝑦 for the given values of 𝑥 and 𝑑𝑥.
𝑦= 𝑥 5 +37𝑥, 𝑥=1, 𝑑𝑥=0.01
In differential form, we can write
𝑑𝑦=5 𝑥 4 𝑑𝑥+37𝑑𝑥= 5 𝑥 𝑑𝑥
Now, 𝑑𝑦= ) 0.01 =0.42
We could also take the derivative as we have in the past
𝑑𝑦 𝑑𝑥 =5 𝑥 4 +37
Then rewrite this is 𝑑𝑦= 5 𝑥 𝑑𝑥

24. Example 5: Finding the Differential 𝑑𝑦
Find the differential 𝑑𝑦 for the given values of 𝑥 and 𝑑𝑥.
𝑦= sin 3𝑥 , 𝑥=𝜋, 𝑑𝑥=−0.02
In differential notation, 𝑑𝑦=3 cos 3𝑥 𝑑𝑥
Then
𝑑𝑦=3 cos 3𝜋 ⋅−0.02=−3⋅−0.02=0.06

25. Example 5: Finding the Differential 𝑑𝑦
Find the differential 𝑑𝑦 for the given values of 𝑥 and 𝑑𝑥.
𝑥+𝑦=𝑥𝑦, 𝑥=2, 𝑑𝑥=0.05
We can begin by solving the equation for 𝑦, but we can also use implicit differentiation
𝑑 𝑥+𝑦 =𝑑 𝑥𝑦
𝑑𝑥+𝑑𝑦=𝑥𝑑𝑦+𝑦𝑑𝑥
Now gather like terms together on each side and solve for 𝑑𝑦
𝑑𝑦−𝑥𝑑𝑦=𝑦𝑑𝑥−𝑑𝑥
𝑑𝑦 1−𝑥 = 𝑦−1 𝑑𝑥
𝑑𝑦= 𝑦−1 1−𝑥 𝑑𝑥

26. Example 5: Finding the Differential 𝑑𝑦
Find the differential 𝑑𝑦 for the given values of 𝑥 and 𝑑𝑥.
𝑥+𝑦=𝑥𝑦, 𝑥=2, 𝑑𝑥=0.05
𝑑𝑦= 𝑦−1 1−𝑥 𝑑𝑥
If 𝑥=2, then using the original equation we have 𝑦=2. Therefore,
𝑑𝑦= 2−1 1− =−0.05

27. Differential Notation
Using differential notation can make our work more concise
𝑑 𝑢+𝑣 =𝑑𝑢+𝑑𝑣, which corresponds to 𝑑 𝑑𝑥 𝑢+𝑣 = 𝑑𝑢 𝑑𝑥 + 𝑑𝑣 𝑑𝑥
𝑑 𝑢𝑣 =𝑢𝑑𝑣+𝑣𝑑𝑢, which corresponds to 𝑑 𝑑𝑥 𝑢𝑣 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥
𝑑 𝑢 𝑣 = 𝑣𝑑𝑢−𝑢𝑑𝑣 𝑣 2 , which corresponds to 𝑑 𝑑𝑥 𝑢 𝑣 = 𝑣 𝑑𝑢 𝑑𝑥 −𝑢 𝑑𝑣 𝑑𝑥 𝑣 2

28. Example 6: Finding Differentials of Functions
𝑑 tan 2𝑥
𝑑 𝑥 𝑥+1

29. Example 6: Finding Differentials of Functions
𝑑 tan 2𝑥
We need the Chain Rule here to deal with the inner function, 2𝑥:
𝑑 tan 2𝑥 = sec 2𝑥 ⋅𝑑 2𝑥
=2 sec 2𝑥 𝑑𝑥

30. Example 6: Finding Differentials of Functions
𝑑 𝑥 𝑥+1
Use the Quotient Rule:
𝑑 𝑥 𝑥+1 = 𝑥+1 ⋅𝑑𝑥−𝑥⋅𝑑 𝑥+1 𝑥+1 2
= 𝑥 𝑑𝑥+𝑑𝑥−𝑥 𝑑𝑥−𝑥 𝑑 1 𝑥 = 𝑑𝑥 𝑥 = 1 𝑥 𝑑𝑥

31. Estimating Change with Differentials
We can use local linearity to approximate the change in some differentiable function 𝑓(𝑥); that is, to approximate Δ𝑓
The equation of the tangent line at some point 𝑎 is 𝑓 𝑥 ≈𝐿 𝑥 = 𝑓 ′ 𝑎 𝑥−𝑎 +𝑓(𝑎)
If 𝑑𝑥 is some small number, then 𝑎+𝑑𝑥 is a number near 𝑎 and
𝑓 𝑎+𝑑𝑥 ≈𝐿 𝑎+𝑑𝑥 = 𝑓 ′ 𝑎 𝑎+𝑑𝑥−𝑎 +𝑓 𝑎
𝑓 𝑎+𝑑𝑥 ≈ 𝑓 ′ 𝑎 𝑑𝑥+𝑓 𝑎
Δ𝑓=𝑓 𝑎+𝑑𝑥 −𝑓 𝑎 ≈ 𝑓 ′ 𝑎 𝑑𝑥+𝑓 𝑎 −𝑓 𝑎
Δ𝑓≈ 𝑓 ′ 𝑎 𝑑𝑥

32. Estimating Change with Differentials

33. Differential Estimate of Change
DEFINITION
Let 𝑓(𝑥) be differentiable at 𝑥=𝑎. The approximate change in the value of 𝑓 when 𝑥 changes from 𝑎 to 𝑎+𝑑𝑥 is
Δ𝑓≈𝑑𝑓= 𝑓 ′ 𝑎 𝑑𝑥

34. Example 7: Estimating Change with Differentials
The radius 𝑟 of a circle increases from 𝑎=10 m to 10.1 m. Use 𝑑𝐴 to estimate the change in the circle’s area 𝐴. Compare this estimate with the true change Δ𝐴, and find the approximation error (i.e., the difference between the true value and the approximate value).

35. Example 7: Estimating Change with Differentials
The actual change in area is Δ𝐴=𝐴 −𝐴(10) The approximate change in area (given 𝐴=𝜋 𝑟 2 ) is 𝑑𝐴= 𝐴 ′ 𝑑𝑟=2𝜋𝑟 𝑑𝑟 Note that 𝑎=10 and 𝑑𝑥=0.1. So the approximate change in area is Δ𝐴≈𝑑𝐴=2𝜋 =2𝜋 m 2 The actual change is Δ𝐴=𝜋 −𝜋 10 2 =2.01𝜋 m 2 The error of approximation is Δ𝐴−𝑑𝐴=2.01𝜋−2𝜋=0.01𝜋 m 2

36. Absolute, Relative, and Percentage Change
We can describe the change in 𝑓 as 𝑥 changes from 𝑎 to 𝑎+𝑑𝑥 in three ways
Absolute change: actual Δ𝑓=𝑓 𝑎+𝑑𝑥 −𝑓(𝑎); approximate 𝑑𝐴= 𝑓 ′ 𝑎 𝑑𝑥
Relative change: actual Δ𝑓 𝑓 𝑎 ; approximate 𝑑𝑓 𝑓 𝑎
Percentage change: actual Δ𝑓 𝑓 𝑎 ⋅100%; approximate 𝑑𝑓 𝑓 𝑎 ⋅100%

37. Example 8: Changing Tires
Inflating a bicycle tire changes its radius from 12 inches to 13 inches. Use differentials to estimate the absolute change, the relative change, and the percentage change in the perimeter of the tire.

38. Example 8: Changing Tires
We have 𝑎=12 and 𝑎+𝑑𝑥=13, so 𝑑𝑥=1. Since 𝑃 𝑟 =2𝜋𝑟, then the approximate change (absolute) is Δ𝑃≈𝑑𝑃= 𝑃 ′ 12 𝑑𝑥=2𝜋⋅1=2𝜋 The relative change is 2𝜋 24𝜋 = 1 12 ≈0.083 The percentage change is 1 12 ⋅100%≈8.333%

39. Example 9: Estimating the Earth’s Surface Area
Suppose the earth were a perfect sphere and we determined its radius to be 3959±0.1 miles. What effect would the tolerance of ±0.1 miles have on our estimate of the earth’s surface area?

40. Example 9: Estimating the Earth’s Surface Area

41. Example 9: Estimating the Earth’s Surface Area
We can use algebra to determine the upper and lower values of the interval where the “true” surface area lies. 𝑆 3959 =4𝜋 ≈ sq. mi. 𝑆 =4𝜋 ≈ sq. mi. 𝑆 =4𝜋 ≈ sq. mi. Δ 𝑆 + =𝑆 −𝑆 3959 ≈ sq. mi. Δ 𝑆 − =𝑆 3959 −𝑆 ≈ sq. mi. From this we see that the “true” surface area is within about ±9950 square miles of the measured value.

42. Example 9: Estimating the Earth’s Surface Area
Now let’s compare this with the estimate using differentials. We are looking for the approximation of Δ𝑆≈𝑑𝑆= 𝑆 ′ 𝑟 𝑑𝑟, where 𝑟=3959 and 𝑑𝑟=0.1 ΔS≈𝑑𝑆=8𝜋𝑟⋅𝑑𝑟=8𝜋 3959 ±0.1 ≈±9950 Your textbook claims that this is an area about the same size as the state of Maryland.

43. Example 10: Determining Tolerance
About how accurately should we measure the radius 𝑟 of a sphere to calculate the surface area 𝑆=4𝜋 𝑟 2 within 1% of its true value?

44. Example 10: Determining Tolerance

45. Example 10: Determining Tolerance
We see from the previous slide that Δ𝑆 ≤0.01𝑆=0.04𝜋 𝑟 2 Since Δ𝑆≈𝑑𝑆=8𝜋𝑟 𝑑𝑟, then 8𝜋𝑟 𝑑𝑟 ≤0.04𝜋 𝑟 2 𝑑𝑟 ≤ 0.04𝜋 𝑟 2 8𝜋𝑟 ≈0.005𝑟 The radius should be measured to within 0.5% of its true value.

46. Example 11: Unclogging Arteries
In the late 1830s, the French physiologist Jean Poiseuille, discovered the formula we use today to predict how much the radius of a partially clogged artery has to be expanded to restore normal flow. His formula, 𝑉=𝑘 𝑟 4 , says that the volume 𝑉 of fluid flowing through a small pipe or tube in a unit of time at a fixed pressure is a constant time the fourth power of the tube’s radius 𝑟. How will a 10% increase in 𝑟 affect 𝑉?

47. Example 11: Unclogging Arteries
First recognize that the percentage change is d𝑟 𝑟 =0.1, and that we seek Δ𝑉 𝑉 ≈ dV V . Δ𝑉 𝑉 ≈ 𝑑𝑉 𝑉 = 4𝑘 𝑟 3 𝑑𝑟 𝑉 = 4𝑘 𝑟 3 𝑘 𝑟 4 ⋅𝑑𝑟=4⋅ 𝑑𝑟 𝑟 =4 0.1 =0.4 So a 10% increase in the radius results in a 40% increase in blood flow.