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

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

Point-Slope Form Connection

Recall the tangent line equation: $y - f(a) = f'(a)(x - a)$

Solving for $y$: $y = f(a) + f'(a)(x - a)$

This is exactly $L(x)$!

Why It Works

Geometric Interpretation

• The tangent line at $x = a$ has slope $f'(a)$
• It passes through the point $(a, f(a))$
• For $x$ close to $a$, the function $f(x)$ is close to the tangent line $L(x)$

Visual: Zoom in on any smooth curve → it looks like a straight line!

The Approximation

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

The closer $x$ is to $a$, the better the approximation!

How to Find Linear Approximation

Step-by-Step Process

Step 1: Identify the function $f(x)$ and the point $a$

Step 2: Calculate $f(a)$ (the $y$-value at $a$)

Step 3: Find $f'(x)$ and calculate $f'(a)$ (the slope at $a$)

Step 4: Write the linear approximation: $L(x) = f(a) + f'(a)(x - a)$

Step 5: Use $L(x)$ to estimate $f(x)$ for $x$ near $a$

Example 1: Estimating a Square Root

Use linear approximation to estimate $\sqrt{26}$.

Step 1: Choose a nearby point

$\sqrt{25} = 5$ is easy to compute!

Let $f(x) = \sqrt{x}$ and $a = 25$

We want to estimate $f(26)$ using $a = 25$

Step 2: Calculate $f(a)$

$f(25) = \sqrt{25} = 5$

Step 3: Find $f'(a)$

$f'(x) = \frac{1}{2\sqrt{x}}$

$f'(25) = \frac{1}{2\sqrt{25}} = \frac{1}{2(5)} = \frac{1}{10}$

Step 4: Write the linear approximation

$L(x) = f(25) + f'(25)(x - 25)$

$L(x) = 5 + \frac{1}{10}(x - 25)$

Step 5: Estimate $\sqrt{26}$

$\sqrt{26} \approx L(26) = 5 + \frac{1}{10}(26 - 25)$

$= 5 + \frac{1}{10}(1) = 5 + 0.1 = 5.1$

Answer: $\sqrt{26} \approx 5.1$

Check: $\sqrt{26} \approx 5.099...$ (calculator) → Very close! ✓

Example 2: Estimating a Trig Value

Use linear approximation to estimate $\sin(0.1)$.

Step 1: Choose a point

Use $a = 0$ (since $\sin 0$ is known)

Let $f(x) = \sin x$ and $a = 0$

Step 2: Calculate $f(0)$

$f(0) = \sin 0 = 0$

Step 3: Find $f'(0)$

$f'(x) = \cos x$

$f'(0) = \cos 0 = 1$

Step 4: Linear approximation

$L(x) = 0 + 1(x - 0) = x$

Step 5: Estimate

$\sin(0.1) \approx L(0.1) = 0.1$

Answer: $\sin(0.1) \approx 0.1$

Check: $\sin(0.1) \approx 0.0998...$ → Very close for small angles!

This is why $\sin x \approx x$ for small $x$ (in radians)!

Differential Notation

Alternative Form

We can write the linear approximation as:

$\Delta y \approx dy$

where:

• $\Delta y = f(x) - f(a)$ (actual change)
• $dy = f'(a) \cdot dx$ (approximate change)
• $dx = x - a$ (change in $x$)

The Formula

$f(x) \approx f(a) + f'(a) \cdot dx$

This is the same as $L(x) = f(a) + f'(a)(x-a)$!

Example 3: Using Differentials

The radius of a sphere is measured to be $r = 10$ cm with a possible error of $\pm 0.1$ cm. Estimate the maximum error in the calculated volume.

Step 1: Volume formula

$V = \frac{4}{3}\pi r^3$

Step 2: Find $dV$

$\frac{dV}{dr} = 4\pi r^2$

$dV = 4\pi r^2 \cdot dr$

Step 3: Plug in values

$r = 10$ cm, $dr = \pm 0.1$ cm

$dV = 4\pi (10)^2 (0.1) = 4\pi (100)(0.1) = 40\pi$

$dV \approx 125.7 \text{ cm}^3$

Answer: The maximum error in volume is approximately $\pm 126$ cm³.

Step 4: Relative error

$\frac{dV}{V} = \frac{40\pi}{\frac{4}{3}\pi(10)^3} = \frac{40\pi}{\frac{4000\pi}{3}} = \frac{40 \cdot 3}{4000} = \frac{120}{4000} = 0.03$

Relative error: 3%

When Linear Approximation Works Best

Good Approximations

Linear approximation is most accurate when:

1. $x$ is very close to $a$
2. The function is smooth (no corners or discontinuities)
3. The function doesn't curve too sharply (second derivative is small)

Poor Approximations

Linear approximation can be inaccurate when:

1. $x$ is far from $a$
2. The function has high curvature (large second derivative)
3. There are discontinuities or sharp corners

Error in Linear Approximation

The error is the difference between actual and approximate values:

$E(x) = f(x) - L(x)$

Second Derivative Test

If $f''$ exists, the error is related to the second derivative:

$E(x) \approx \frac{f''(a)}{2}(x-a)^2$

Key insight:

• Error grows like $(x-a)^2$ (quadratically)
• If $|f''|$ is large, the error is larger
• Error is always small when $x \approx a$

Applications of Linear Approximation

Application 1: Mental Math

Estimate $\sqrt{101}$:

$\sqrt{100} = 10$, and $f'(100) = \frac{1}{2\sqrt{100}} = \frac{1}{20}$

$\sqrt{101} \approx 10 + \frac{1}{20}(1) = 10.05$

Application 2: Physics

Small oscillations of a pendulum: $\sin \theta \approx \theta$ (for small angles)

This simplifies differential equations!

Application 3: Engineering

Estimating how measurement errors propagate through calculations.

Application 4: Economics

Marginal cost/revenue approximations using derivatives.

Linear vs. Quadratic Approximation

Linear (First Order)

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

Uses tangent line (first derivative)

Quadratic (Second Order)

$Q(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2$

Uses curvature (second derivative) → more accurate!

This leads to Taylor series (later topic)!

⚠️ Common Mistakes

Mistake 1: Using Wrong Point

Choose $a$ where $f(a)$ and $f'(a)$ are easy to compute!

Don't use $a = 26$ to estimate $\sqrt{26}$ → use $a = 25$!

Mistake 2: Units in Error Problems

When finding $dV$, $dr$, etc., keep track of units!

Mistake 3: Confusing $\Delta y$ and $dy$

• $\Delta y$ is the actual change (exact)
• $dy$ is the approximate change (using derivative)
• They're only approximately equal!

Mistake 4: Using Far-Away Points

Linear approximation only works well when $x$ is close to $a$.

Don't use $a = 0$ to estimate $\sin(1)$ → error is too large!

Mistake 5: Wrong Derivative

Make sure to calculate $f'(a)$ correctly!

Important Approximations to Know

Small Angle Approximations (in radians!)

For $x$ near 0:

• $\sin x \approx x$
• $\cos x \approx 1$
• $\tan x \approx x$

Exponential and Logarithm

For $x$ near 0:

• $e^x \approx 1 + x$
• $\ln(1+x) \approx x$

Power Functions

For $x$ near 1:

• $(1+x)^n \approx 1 + nx$ (binomial approximation)

The Big Picture

From Local to Global

• Local behavior: Linear approximation uses tangent line (local info)
• Global behavior: To approximate over larger intervals, need more terms (Taylor series)

Foundation for Advanced Topics

Linear approximation is the first step toward:

• Taylor series
• Newton's method
• Numerical analysis
• Differential equations

📝 Practice Strategy

1. Choose $a$ wisely - pick where $f(a)$ and $f'(a)$ are easy to compute
2. Write the formula first: $L(x) = f(a) + f'(a)(x-a)$
3. Calculate carefully - show each step
4. Check your answer - does it make sense? Is it close to $f(a)$?
5. For error problems, use differentials: $dy = f'(a) \cdot dx$
6. Remember: Only accurate when $x$ is close to $a$!
7. Memorize common approximations ($\sin x \approx x$, etc.)

$16^{1/4} = 2$ is easy to compute!

Let $f(x) = x^{1/4}$ and $a = 16$

Step 2: Calculate $f(16)$

$f(16) = 16^{1/4} = 2$

Step 3: Find $f'(x)$ and $f'(16)$

$f'(x) = \frac{1}{4}x^{-3/4} = \frac{1}{4x^{3/4}}$

$f'(16) = \frac{1}{4(16)^{3/4}} = \frac{1}{4(8)} = \frac{1}{32}$

Step 4: Linear approximation

$L(x) = f(16) + f'(16)(x - 16)$

$L(x) = 2 + \frac{1}{32}(x - 16)$

Step 5: Estimate $(15.9)^{1/4}$

$L(15.9) = 2 + \frac{1}{32}(15.9 - 16)$

$= 2 + \frac{1}{32}(-0.1)$

$= 2 - \frac{0.1}{32} = 2 - 0.003125$

$= 1.996875$

Approximation: $(15.9)^{1/4} \approx 1.997$

Step 6: Find actual value

Using a calculator: $(15.9)^{1/4} \approx 1.99687...$

Step 7: Calculate error

$\text{Error} = |\text{Actual} - \text{Approximate}|$

$= |1.99687 - 1.996875| \approx 0.000005$

Answer:

• Linear approximation: $1.997$
• Actual value: $1.99687$
• Error: approximately $0.000005$ (very small!)

$f'(x) = -\sin x$

$f'(0) = -\sin 0 = 0$

Step 3: Linear approximation

$L(x) = f(0) + f'(0)(x - 0)$

$L(x) = 1 + 0 \cdot x = 1$

Step 4: Estimate $\cos(0.2)$

$\cos(0.2) \approx L(0.2) = 1$

Step 5: Compare with actual value

Using calculator: $\cos(0.2) \approx 0.9801$

Error: $|0.9801 - 1| = 0.0199$

Analysis:

The linear approximation gives 1, but the actual value is about 0.98.

Why is the error larger than in previous examples?

Look at $f''(x) = -\cos x$, so $f''(0) = -1 \neq 0$

The function has curvature at $x = 0$, so the linear approximation isn't perfect.

The error is approximately: $E(x) \approx \frac{f''(0)}{2}x^2 = \frac{-1}{2}(0.2)^2 = -0.02$

This matches our observed error! ✓

Better approximation: Use quadratic approximation $Q(x) = 1 + 0 \cdot x + \frac{-1}{2}x^2 = 1 - \frac{x^2}{2}$

$\cos(0.2) \approx 1 - \frac{(0.2)^2}{2} = 1 - \frac{0.04}{2} = 1 - 0.02 = 0.98$

Much better! ✓

Answer:

• Linear approximation: $\cos(0.2) \approx 1$
• Actual: $\cos(0.2) \approx 0.9801$
• The linear approximation has moderate error because the function has curvature