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Lesson 2: Linear Approximation and Differentials

Estimated time: 30-40 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Linearization

Near a point x = a, a differentiable function is well approximated by its tangent line. This tangent-line approximation is called the linearization.

Linearization

The linearization of f at x = a is: L(x) = f(a) + f'(a)(x − a)

For x close to a: f(x) ≈ L(x).

Example 1: Approximating a Square Root

Approximate √4.1 using linearization of f(x) = √x at a = 4.

Step 1: f(4) = 2, f'(x) = 1/(2√x), so f'(4) = 1/4.

Step 2: L(x) = 2 + (1/4)(x − 4).

Step 3: L(4.1) = 2 + (1/4)(0.1) = 2 + 0.025 = 2.025.

Actual value: √4.1 ≈ 2.02485. The error is about 0.00015.

Example 2: Approximating sin(0.1)

Use linearization of f(x) = sin x at a = 0.

Step 1: f(0) = 0, f'(0) = cos 0 = 1.

Step 2: L(x) = 0 + 1(x − 0) = x.

Step 3: sin(0.1) ≈ 0.1. (Actual: 0.09983...)

Differentials

Differentials

If y = f(x), the differential is: dy = f'(x) dx

Here dx is an independent variable (a small change in x), and dy approximates the resulting change in y.

The actual change is Δy = f(x + dx) − f(x). The differential dy approximates Δy when dx is small.

Example 3: Using Differentials

A cube's side is measured as 5 cm with an error of ±0.02 cm. Estimate the error in the computed volume.

Step 1: V = s³, so dV = 3s² ds.

Step 2: dV = 3(25)(0.02) = 1.5 cm³.

The volume error is approximately ±1.5 cm³ out of V = 125 cm³, a relative error of 1.2%.

Example 4: Relative and Percentage Error

If a sphere's radius is measured as 10 cm with error dr = 0.1 cm, estimate the percentage error in volume.

Step 1: V = (4/3)πr³, dV = 4πr² dr.

Step 2: Relative error: dV/V = (4πr² dr)/((4/3)πr³) = 3 dr/r = 3(0.1/10) = 0.03.

Result: 3% error in volume from a 1% error in radius.

Key Takeaways

Check Your Understanding

1. Use linearization of f(x) = x1/3 at a = 8 to approximate 8.11/3.

Answer: f(8) = 2, f'(x) = (1/3)x−2/3, f'(8) = 1/12. L(8.1) = 2 + (1/12)(0.1) = 2.00833.

2. If y = x4 and x changes from 2 to 2.01, estimate dy.

Answer: dy = 4x³ dx = 4(8)(0.01) = 0.32. (Actual Δy = 2.014 − 16 ≈ 0.3224.)

3. A circle's radius is 20 cm with error ±0.5 cm. Estimate the error in the area.

Answer: dA = 2πr dr = 2π(20)(0.5) = 20π ≈ 62.8 cm².

4. For f(x) = cos x, find the linearization at a = π/2.

Answer: f(π/2) = 0, f'(π/2) = −sin(π/2) = −1. L(x) = −(x − π/2) = π/2 − x.

Ready for More?

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Lesson 3 covers extreme values and critical points.

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