Applications of Differentiation – Local Linearization – Local Linear Approximation – Equation of the Tangent Line at a Point

So far, we have learned about how to find the derivative, or find the slope of the tangent line, of a function. In order to find the equation of a tangent line, we need to first look at the equation of a secant line.

Let us examine the formula of the point – slope form of a line:

y – y1= m (x – x1)

  • y1 = y coordinate of a given point
  • x1 = x coordinate of a given point
  • m = slope of the line (average rate of change between two points on the function).

The equation of the tangent line will be very similar to the point slope of the secant line:

y – y1= m’ (x – x1)

  • y1 = y coordinate of the point of tangency
  • x1 = x coordinate of the point of tangency
  • m’ = instantaneous rate of change of the function at x1

Essentially, the only difference is that instead of using m, which is for the average rate of change between 2 points, we will use m’, or dy/dx , the instantaneous rate of change at x1.

Since m’ is the derivative of the function at x1 , we can rewrite m’ as F’( x1).

y1 can be written as F( x1).

Therefore, the new formula is:

y – F( x1) = F’( x1) (x – x1)

When solving for y, which we will rewrite as L(x),

L(x) = F’( x1) (x – x1)+ F( x1)

Formula: Local Linearization of a Function

If F(x) is differentiable at (x1,y1), then the approximating function

L(x) = F’( x1) (x – x1)+ F( x1)

is the linearization or tangent-line approximation of F(x) at any value x.

  • x1= the x coordinate of the point of tangency
  • x = the x value used for approximation

Applications of Linear Approximations

We can use the Linearization equation L(x) to approximate the value of F(x) for any given x value.

Essentially F(x)= L(x) near x= a

We can also select x values close to x = a and obtain a reasonable approximation of F(x) from L(x). However, the farther the x values are away from x = a, the worse the approximation is.

Guidelines for finding the Linear Approximation of F(x):

  1. Find x1, which will be given to you in the problem.
  2. Plug in x1into F(x) to obtain F( x1).
  3. Find the derivative of the function. Afterwards, plug in x1 into F’(x) to obtain F’(x1).
  4. Plug in x1, F( x1), and F’(x1) into the formula.
  5. Select an x value that you would like to approximate, which you will substitute for x inside the formula.

Example 1:

If F (x) = x2 , Find the following:

  • The equation of the tangent line at x = 3
  • The linear approximation L(3.01)
  • Compare F(3.01) to L(3.01)
  • Compare other values of L(x) and F(x) by substituting different x values.

Solution:

First, let’s find the equation of the tangent line.

We need x1. In the problem, x1= 3.

Next, we need F( x1):

F( x1) = (3)2= 9

Our point of tangency is (3,9).

Next, we need to find F’(x1). Using the power rule,

F’(x) = 2x2-1=2x

F’(x1) = 2(3) = 6

Now we can plug in x1, F( x1), and F’(x1) into the formula:

L(x) = F’( x1) (x – x1)+ F( x1)

L(x) = 6 (x – 3)+ 9

Doing further simplifications,

L(x) = 6(x – 3)+ 9

L(x) = 6x – 18 + 9

L(x) = 6x – 9 ←— Equation of the tangent line at (3,9)

Second, find the linear approximation of L(3.01).

L(3.01) = 6(3.01) – 9

L(3.01) =9.06

Third, find F(3.01).

F(3.01) = (3.01)2=9.0601

Notice how L(3.01) and F(3.01) are almost the same value.

Let’s examine the graph of L(x) and F(x) to confirm this:

Fourth, test other x values for L(x) and F(x) and compare the results.

Let’s try L(4) and F(4).

L(4) = 6(4) – 9

L(4) = 15

F(4) = (4)2=16

Notice how the value of L(4) is somewhat close to the value of F(4) but is a worse approximation.

Let’s try L(10) and F(10):

L(10) = 6(10) – 9

L(10) = 51

F(10) = (10)2=100

Notice how L(10) is not even close to the value of F(10). L(10) is a terrible approximation for F(10).

Notice that the further we are from the point of tangency, the worse the approximation.

Therefore, linear approximations are most accurate when you are approximating x values that are closest to the point of tangency.

If we choose x values to plug into L(x) and F(x), we will get the most accurate approximation the closer we are to x = 3. However, if we use x values that are further away from x = 3, the worse our approximation is.

Example 2:

If F (x) = cos x, at x = 0, Find the following:

  • The equation of the tangent line at x = 0
  • The linear approximation L(0.05)
  • Compare F(0.05) to L(0.05)
  • Compare other values of L(x) and F(x) by substituting different x values.

Solution:

First, let’s find the equation of the tangent line.

We need x1. In the problem, x1= 0.

Next, we need F( x1):

F( x1) = cos(0) = 1

Our point of tangency is (0,1).

Next, we need to find F’(x1):

F’(x) = -sin(x)

F’(x1) = -sin(0) = 0

Now we can plug in x1, F( x1), and F’(x1) into the formula:

L(x) = F’( x1) (x – x1)+ F( x1)

L(x) = 0 (x – 0)+ 1

Doing further simplification,

L(x) = 0x – 0 + 1

L(x) =1

In this case, L(x) is a constant. Therefore, any x value that we substitute inside L(x) will always equal 1.

Second, find the linear approximation of L(0.05).

L(0.05) = 1

As mentioned above, since L(x) is a constant, any x value that we substitute inside L(x) will always equal 1.

Third, find F(0.05).

F(0.05) = cos(0.05)=0.999

Notice how L(3.01) and F(3.01) are almost the same value.

Taking a closer look,

Fourth, test other x values for L(x) and F(x) and compare the results.

Let’s try L(π/2) and F(π/2).

L(π/2) = 1

F(π/2 ) = cos(π/2)=0

Notice how the value of L(π/2) is not anywhere near the value of F(π/2). L(π/2) is a very bad approximation for F(π/2).

Let’s try L(3π/2) and F(3π/2):

L(3π/2) = 1

F(3π/2) = cos(3π/2) = 0

Notice how L(3π/2) is not even close to the value of F(3π/2). L(3π/2) is a terrible approximation for F(3π/2).

Notice that the further we are from the point of tangency, the worse the approximation.

Therefore, linear approximations are most accurate when you are approximating x values that are closest to the point of tangency.

If we choose x values to plug into L(x) and F(x), we will get the most accurate approximation the closer we are to x = 0. However, if we use x values that are further away from x = 0, the worse our approximation is.

Summary of Section

Use the tangent line at (x1, y1) as an approximation for the curve y = F(x) when x is near x = a. The equation of the tangent line is

y = F’( x1) (x – x1)+ F( x1)

The linearization of F(x) at x = a is

L(x) = F’( x1) (x – x1)+ F( x1)

which is used to compute the linear approximation or tangent line approximation of any x value near x = a.

References:
http://tutorial.math.lamar.edu/Classes/CalcI/LinearApproximations.aspx
https://www.math24.net/linear-approximation/
https://educationcareeronline.com/linear-approximation-formula-example/
https://www.desmos.com/calculator/fmd64xbq0m
http://scidiv.bellevuecollege.edu/dh/Calculus_all/CC_2_8_LinAppandDiff
http://www.opentextbookstore.com/appcalc/Chapter2-10.pdf

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