Linear Approximations

Near a point $x_{0}$ , the tangent line to a curve is the best straight-line approximation to the curve. The linearization L(x)=f(x_0)+f'(x_0)(x-x_0) gives accurate estimates of $f (x)$ when $x$ is close to $x_{0}$ .

Term	Formula
Linearization	L(x)=f(x_0)+f'(x_0)(x-x_0)
Linear approximation	$f (x) \approx L (x)$ for $x$ near $x_{0}$
Increment form	\Delta y\approx f'(x)\,\Delta x

Derivation

From the definition of the derivative, for $Δ x$ close to $0$:

\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\approx f'(x_0)

Multiplying through by $Δ x$ :

f(x_0+\Delta x)-f(x_0)\approx f'(x_0)\,\Delta x

Setting $x = x_{0} + Δ x$ (so $Δ x = x - x_{0}$ ):

f(x)\approx f(x_0)+f'(x_0)(x-x_0)

The function

L(x)=f(x_0)+f'(x_0)(x-x_0)

is called the linearization of $f$ at $x_{0}$ , and the approximation $f (x) \approx L (x)$ is called the linear approximation (or tangent-line approximation) of $f$ at $x_{0}$ .

Geometric interpretation: The graph of $L$ is the tangent line to $y = f (x)$ at $(x_{0}, f (x_{0}))$ . When $x$ is close to $x_{0}$ , this tangent line stays close to the curve.

Tangent line L(x) approximating f(x) near x0

In delta notation: \Delta y\approx f'(x)\,\Delta x.

Example. For $f (x) = x^{2}$ , the linear approximation at $x$ is

(x + Δ x)^{2} \approx x^{2} + 2 x Δ x

This says $(Δ x)^{2}$ is negligible when $Δ x$ is small (e.g., if $Δ x = 0.001$ , then $(Δ x)^{2} = 10^{- 6}$ ).

Geometric interpretation: square with incremented side

Worked Examples

Use the linearization of $f (x) = \sqrt{x}$ at $x_{0} = 9$ to approximate $\sqrt{10}$ and $\sqrt{8}$ .

Solution

f (x) = \sqrt{x}

, f'(x)=\dfrac{1}{2\sqrt{x}}.

f (9) = 3

, f'(9)=\dfrac{1}{6}.

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

x = 10

L (10) = 3 + \frac{1}{6} \approx 3.16667

. (True value:

\sqrt{10} \approx 3.16228

.) At

x = 8

L (8) = 3 - \frac{1}{6} \approx 2.83333

. (True value:

\sqrt{8} \approx 2.82843

Graph of sqrt(x) and its linearization at x=9

Approximate $\sin (31^{\circ})$ and $\sin (29^{\circ})$ .

Solution

Use

f (x) = \sin x

linearized at

x_{0} = π / 6

(since

\sin (π / 6) = 1 / 2

is known). f'(x)=\cos x, f'(\pi/6)=\cos(\pi/6)=\sqrt{3}/2.

L (x) = \frac{1}{2} + \frac{\sqrt{3}}{2} (x - \frac{π}{6})

Since

1^{\circ} = π / 180

rad:

\sin (31^{\circ}) \approx L (\frac{π}{6} + \frac{π}{180}) = \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot \frac{π}{180} \approx 0.5151

\sin (29^{\circ}) \approx L (\frac{π}{6} - \frac{π}{180}) = \frac{1}{2} - \frac{\sqrt{3}}{2} \cdot \frac{π}{180} \approx 0.4849

(True values: $0.5150$ and $0.4848$, respectively.)

Graph of sin x and its linearization at x=pi/6

What is the linear approximation of $f (x) = \sin x$ at $x = 0$ ? For what values of $x$ is it accurate to within $0.01$?

Solution

f (0) = 0

, f'(0)=1. The linearization is

L (x) = x

. This famous approximation

\sin x \approx x

holds for small

x

(in radians). For accuracy within $0.01$: we need

| \sin x - x | < 0.01

. From the graph of

y = | \sin x - x |

, this holds when

- 0.39 rad < x < 0.39 rad

(approximately

- {22.4}^{\circ} < x < {22.4}^{\circ}

Graph of sin x and the approximation y=x

When Is Linear Approximation Accurate?

The linear approximation is accurate when the curve has little "bending" near $x_{0}$ . A sharp bend at $(x_{0}, f (x_{0}))$ means the tangent line quickly diverges from the curve.

Comparison of good and poor linear approximations

The "bending" of a curve is measured by the second derivative (covered in Chapter 4).

Application: The Pendulum

A simple pendulum of length $L$ and mass $m$ satisfies:

L \frac{d^{2} θ}{d t^{2}} = - g \sin θ

This nonlinear equation has no closed-form solution in terms of elementary functions. Using the linear approximation $\sin θ \approx θ$ (valid for small angles), we replace it by:

\frac{d^{2} θ}{d t^{2}} + \frac{g}{L} θ = 0

whose solution is

θ (t) = θ_{0} \cos (\sqrt{\frac{g}{L}} t)

This approximation is valid for $θ_{0} ≲ {22.5}^{\circ}$ .

Frequently Asked Questions

When is the linear approximation a good one?

It is good when

x

is close to

x_{0}

and

f

does not bend sharply near

x_{0}

. More precisely, the error

| f (x) - L (x) |

is approximately \frac{1}{2}|f''(x_0)|(x-x_0)^2, so it is small when

| x - x_{0} |

is small and |f''(x_0)| is not too large.

Is the linearization the same as the tangent line?

Yes. The graph of L(x)=f(x_0)+f'(x_0)(x-x_0) is precisely the tangent line to

y = f (x)

(x_{0}, f (x_{0}))

Can I use linearization to estimate cube roots?

Yes. For example, to estimate

\sqrt[3]{28}

, linearize

f (x) = x^{1 / 3}

x_{0} = 27

: f'(x)=\frac{1}{3}x^{-2/3},

f (27) = 3

, f'(27)=1/27. So

\sqrt[3]{28} \approx 3 + \frac{1}{27} (28 - 27) = 3 + \frac{1}{27} \approx 3.037

Derivation

Worked Examples

When Is Linear Approximation Accurate?

Application: The Pendulum

Frequently Asked Questions

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