Linear Approximation and Differentials

By the end of this lesson, you will be able to find the linearization of a function at a point, use it for approximate calculations, compute differentials, and interpret them as approximate changes.

The Big Idea

Near a point $a$, a differentiable function looks like its tangent line. The linear approximation (or linearization) is this tangent line, and it provides excellent approximations for $f(x)$ when $x$ is close to $a$.

Linearization

The linearization of $f$ at $x = a$:

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

This is the tangent line equation at $(a, f(a))$.

For $x$ near $a$: $f(x) \approx L(x) = f(a) + f'(a)(x-a)$

Example 1 — Approximating $\sqrt{4.02}$

Let $f(x) = \sqrt{x}$, $a = 4$.

$f(4) = 2$; $f'(x) = 1/(2\sqrt{x})$; $f'(4) = 1/4$

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

$\sqrt{4.02} \approx L(4.02) = 2 + \dfrac{1}{4}(0.02) = 2 + 0.005 = 2.005$

Actual: $\sqrt{4.02} \approx 2.00499...$ ✓ (error < 0.0001)

Example 2 — Approximating Trig

Approximate $\sin(31°)$.

$31° = \pi/6 + \pi/180 \approx \pi/6 + 0.01745$

$f(x) = \sin x$ at $a = \pi/6$: $f'(x) = \cos x$; $f'(\pi/6) = \sqrt{3}/2 \approx 0.866$

$\sin(31°) \approx \sin(30°) + 0.866 \times 0.01745 \approx 0.5 + 0.01512 = 0.515$

Actual: $\sin(31°) \approx 0.5150$ ✓

Differentials

For $y = f(x)$, the differential $dy$:

$$dy = f'(x),dx$$

where $dx$ is an arbitrary change in $x$.

$dy$ approximates $\Delta y$ (the actual change in $y$) when $dx$ is small.

$$\Delta y = f(x+dx) - f(x) \approx f'(x),dx = dy$$

Example 3 — Error Propagation

A sphere has radius $r = 10$ cm, measured with possible error $dr = \pm 0.1$ cm. Approximate the maximum error in the volume.

$V = \frac{4}{3}\pi r^3$; $dV = 4\pi r^2,dr$

$dV = 4\pi(100)(0.1) = 40\pi \approx 125.7$ cm³

Relative error: $\dfrac{dV}{V} = \dfrac{4\pi r^2,dr}{\frac{4}{3}\pi r^3} = \dfrac{3,dr}{r} = \dfrac{3(0.1)}{10} = 3%$

Newton's Method (Application of Linearization)

To find roots of $f(x) = 0$, iterate: