Linearization – Taming Curves with Lines

Approximating the Complicated with the Simple

Imagine you’re trying to calculate $$\sqrt{16.1}$$​ without a calculator. The graph of $$y=\sqrt{x}$$ is a curve, which is complicated. But what if, for values of x very close to 16, we could pretend the graph was a straight line? That’s the entire premise of linearization.

Linearization is the process of approximating a function near a specific point by using the tangent line to the function at that point.

Why is this useful?

• It simplifies complex calculations.

• It’s the foundational idea for differentials and Euler’s method in differential equations.

• It’s how many scientific and engineering models work for small changes.

use the Tangent Line

Recall that the derivative f′(a) gives the slope of the tangent line to the curve y=f(x) at the point $x=a$.

The equation of a line with slope $m$ passing through a point $$(x_{1},y_{1})$$is:

$$y-y_{1}=m(x-x_{1})$$

For our tangent line:

• The point is (a,f(a)).

• The slope m is f′(a).

So, the equation of the tangent line is:

$$y-f(a)=f^\prime(a)(x-a)$$

Solving for y, we get:

$$y=f(a)+f^\prime(a)(x-a)$$

This line, L(x), is our linear approximation of f(x) near $x=a$.

The Formal Definition:

The linearization of a function f at x=a is the function:

$$L(x)=f(a)+f^\prime(a)(x-a)$$

We say that f(x)≈L(x) for x close to $a$.

Finding a Linearization

Let’s make this concrete with an example.

Example 1: Find the linearization of $$f(x)=\sqrt{x}$$ at a=16, and use it to approximate $$\sqrt{16.1}$$.

1. Find the point: f(a)=f(16)=16=4. Our point is (16,4).

2. Find the slope (the derivative):

   • $$f(x)=x^{1/2}$$ then

   • $$f^\prime(x)=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}$$​ and

   • $$f^\prime(a)=f^\prime(16)=\frac{1}{2\sqrt{16}}=\frac{1}{8}$$​

3. Write the equation of the tangent line (the linearization):

   $$L(x)=f(16)+f^\prime(16)(x-16)=4+\frac{1}{8}(x-16)$$

4. Use the linearization to approximate:
   We want $$\sqrt{16.1}=f(16.1)$$. Since 16.1 is close to 16 then we can use L(16.1).

$$L(16.1)=4+\frac{1}{8}(16.1-16)$$

$$L(16.1)=4+\frac{1}{8}(0.1)$$

$$L(16.1)=4+0.0125=4.0125$$

∴ Our linear approximation is 16.1≈4.012516.1​≈4.0125

How good is this approximation? A calculator gives 16.1≈4.01248. Our approximation is off by only about 0.00002! This is incredibly accurate.

Connection to Differentials

Linearization is often presented using differentials, which provide beautiful notation for this process.

• We define the differential dx as a small change in x.

• We define the differential dy as the corresponding change along the tangent line: dy=f′(a)dx.

• The actual change in the function is Δy=f(a+dx)−f(a).

The linear approximation can then be written as:

$$f(a+dx)\approx f(a)+dy$$

The differential dy is an approximation for the true change $\Delta y$. The tangent line gives us dy, while the curve gives us $\Delta y$.

Slightly More Complex Example

Example 2: Find the linearization of $f(x)=\cos (x)$ at a=π2a=2π​ and approximate $\cos (1.6)$. (Note: π2≈1.57082π​≈1.5708)

1. Find the point: f(a)=cos⁡(π2)=0f(a)=cos(2π​)=0. Point: (π2,0)(2π​,0).

2. Find the slope: f′(x)=−sin⁡(x)f′(x)=−sin(x). So, f′(a)=−sin⁡(π2)=−1f′(a)=−sin(2π​)=−1.

3. Write the linearization:

   L(x)=0+(−1)(x−π2)L(x)=0+(−1)(x−2π​)L(x)=−x+π2L(x)=−x+2π​

4. Use it to approximate $\cos (1.6)$:

   cos⁡(1.6)≈L(1.6)=−1.6+π2≈−1.6+1.5708=−0.0292cos(1.6)≈L(1.6)=−1.6+2π​≈−1.6+1.5708=−0.0292

   A calculator gives $\cos (1.6)\approx -0.0292$. In this case, the approximation is perfect to 4 decimal places because the linearization is very good near π22π​.

VI. The Caveats: When Does It Fail?

Linearization is powerful, but it has limits.

1. It’s a Local Approximation: The approximation is only reliable for values of $x$ close to the point of linearization $a$. The further you go from $a$, the worse the approximation gets.

   Try using our xx​ linearization at $a=16$ to approximate 2525​:
   L(25)=4+18(9)=5.125L(25)=4+81​(9)=5.125. The true value is 5. That’s a much larger error.

2. It Fails Where the Derivative Doesn’t Exist: If the function is not differentiable at $x=a$, then there is no unique tangent line, and linearization is impossible. (e.g., $f(x)=|x|$ at $x=0$).

VII. Why This Matters: The Deeper Meaning

Linearization is more than just a trick for approximating numbers. It is the heart of differential calculus.

• It’s the “Calculus I” version of a Taylor Series. The linearization $L(x)$ is the first-order Taylor polynomial for $f$ at $a$.

• It’s the foundation for numerical methods. Euler’s Method for solving differential equations is essentially repeated linearization.

• It’s how we build complex models. In physics and engineering, for small oscillations or perturbations, we often linearize non-linear systems to make them solvable. The behavior of a pendulum for small angles is a classic example where $\sin (\theta )\approx \theta$, which is a linearization at $\theta =0$.

Summary

The Linearization Mantra: “Near a point $a$, a differentiable function and its tangent line are virtually indistinguishable.”

The Formula:

L(x)=f(a)+f′(a)(x−a)

The Process:

1. Compute f(a) and f′(a).

2. Plug them into the formula.

3. Use $L(x)$ to approximate $f(x)$ for x near a.

By mastering linearization, you learn to see the world in a powerful new way: understanding complex, curved behavior by first understanding its simple, linear essence.