Linearization & Differentials

Linearization & Differentials

Part 1 of 7 — The Tangent Line Approximation

Topic Overview

Local Linearization Formula

$\boxed{L(x) = f(a) + f'(a)(x - a)}$

Worked Example

Approximate $\sqrt{4.1}$ using linearization.

$f(x) = \sqrt{x}$, $a = 4$ (nearest perfect square).

$f(4) = 2$, $f'(x) = \frac{1}{2\sqrt{x}}$, $f'(4) = \frac{1}{4}$

$L(x) = 2 + \frac{1}{4}(x-4)$ $L(4.1) = 2 + \frac{1}{4}(0.1) = \boxed{2.025}$

Actual: $\sqrt{4.1} \approx 2.02485$. Error $\approx 0.00015$!

Key Fact: Choose $a$ to be a nearby value where $f(a)$ and $f'(a)$ are easy to compute. The closer $x$ is to $a$, the better the approximation.

Practice — Linearization 🎯

Build the linearization. 🔍

Approximate. ✍️

Key Takeaways — Part 1

• Linearization: $L(x) = f(a) + f'(a)(x-a)$
• Choose $a$ near $x$ where $f(a)$ and $f'(a)$ are easy
• The approximation improves as $x \to a$
• This is the tangent line at $x = a$ used as an approximation

Linearization & Differentials

Part 2 of 7 — Approximating Values

Common Linearizations at $a = 0$

Key Fact: These linearizations appear frequently on the AP exam, especially $e^x \approx 1+x$ and $\sin x \approx x$ for small $x$.

Over/Under Estimates from Concavity

$\boxed{\text{Concave up} \Rightarrow \text{Tangent line underestimates}}$ $\boxed{\text{Concave down} \Rightarrow \text{Tangent line overestimates}}$

Worked Example

Approximate $e^{0.1}$ using linearization. Is it an over- or underestimate?

$f(x) = e^x$, $a = 0$. $L(x) = 1 + x$. $L(0.1) = 1.1$.

$f''(x) = e^x > 0$ everywhere — concave up — so tangent line is an underestimate.

Actual: $e^{0.1} \approx 1.10517$. Indeed $1.1 < 1.10517$.

Practice — Approximations 🎯

Classify each approximation. 🔍

Calculate. ✍️

Key Takeaways — Part 2

• Memorize common linearizations: $\sin x \approx x$, $e^x \approx 1+x$, $\ln(1+x) \approx x$
• Concave up ($f'' > 0$) $\Rightarrow$ underestimate
• Concave down ($f'' < 0$) $\Rightarrow$ overestimate
• AP frequently asks "is this an over- or underestimate?"

Linearization & Differentials

Part 3 of 7 — Differentials

Differentials vs. Derivatives

$\boxed{dy = f'(x)\,dx}$

Relationship

$\Delta y \approx dy \quad \text{when } dx = \Delta x \text{ is small}$

The differential $dy$ is the change along the tangent line. The actual change $\Delta y$ is the change along the curve.

Worked Example

$y = x^3$. Find $dy$ when $x = 2$ and $dx = 0.01$.

$dy = f'(x)\,dx = 3x^2\,dx = 3(4)(0.01) = 0.12$

Actual change: $\Delta y = (2.01)^3 - 8 = 8.120601 - 8 = 0.120601$

$dy = 0.12$ vs $\Delta y \approx 0.1206$ — very close!

Key Fact: $dy$ is a linear approximation to $\Delta y$. The smaller $dx$, the better the approximation.

Practice — Differentials 🎯

Compare $dy$ and $\Delta y$. 🔍

Compute the differential. ✍️

Key Takeaways — Part 3

• $dy = f'(x)\,dx$ is the differential (change along tangent)
• $\Delta y = f(x+dx) - f(x)$ is the actual change
• $dy \approx \Delta y$ for small $dx$
• The derivative $dy/dx$ is the ratio of differentials

Linearization & Differentials

Part 4 of 7 — Error Analysis

Error Terminology

Over/Under with Error Bounds

$\boxed{\text{Error} \approx \frac{1}{2}f''(a)(x-a)^2}$

This is the next term in the Taylor expansion. For small $(x-a)$, the error is approximately quadratic in the displacement.

Worked Example

Approximate $\sqrt{26}$ using $a = 25$. Estimate the error.

$L(26) = 5 + \frac{1}{10}(1) = 5.1$

Error $\approx \frac{1}{2}f''(25)(1)^2 = \frac{1}{2}\left(-\frac{1}{4 \cdot 125}\right) = -\frac{1}{1000} = -0.001$

So $\sqrt{26} \approx 5.1 - 0.001 = 5.099$. Actual: $5.0990\ldots$ — extremely close!

AP Tip: The sign of $f''(a)$ tells you whether $L$ overestimates ($f'' < 0$) or underestimates ($f'' > 0$). The magnitude tells you how large the error is.

Practice — Error Analysis 🎯

Error classification. 🔍

Calculate the error. ✍️

Key Takeaways — Part 4

• Error in linearization $\propto (x-a)^2$ (quadratic)
• Concave up $\Rightarrow$ underestimate; concave down $\Rightarrow$ overestimate
• Absolute error $= |\Delta y - dy|$; relative error $= |\Delta y - dy|/|\Delta y|$
• Stay close to $a$ for smaller errors

Linearization & Differentials

Part 5 of 7 — Applications

Propagation of Error

If a measurement $x$ has uncertainty $dx$, then the uncertainty in $f(x)$ is approximately:

$\boxed{df = |f'(x)|\,dx}$

Worked Example: Sphere Volume

A sphere has radius $r = 5$ cm with measurement error $dr = 0.02$ cm. Estimate the error in the volume.

$V = \frac{4}{3}\pi r^3$, $V'(r) = 4\pi r^2$

$dV = 4\pi(25)(0.02) = 2\pi \approx 6.28 \text{ cm}^3$

Relative error: $\frac{dV}{V} = \frac{4\pi r^2\,dr}{\frac{4}{3}\pi r^3} = \frac{3\,dr}{r} = \frac{3(0.02)}{5} = 0.012 = 1.2\%$

Key Fact: For $V = \frac{4}{3}\pi r^3$, the relative error in volume is 3 times the relative error in radius: $\frac{dV}{V} = 3\frac{dr}{r}$.

Linearization from a Table

AP problems often give a table of $f$ and $f'$ values and ask you to approximate $f$ at a nearby point:

Approximate $f(2.1)$: $L(2.1) = f(2) + f'(2)(0.1) = 5 + 3(0.1) = 5.3$

Approximate $f(3.9)$: $L(3.9) = f(4) + f'(4)(-0.1) = 11 + (-1)(-0.1) = 11.1$

Practice — Applications 🎯

AP table problem. 🔍

Propagation of error. ✍️

Key Takeaways — Part 5

• Error propagation: $df \approx |f'(x)|\,dx$
• Relative error: $df/f$. For $V = cr^n$: $dV/V = n\,dr/r$
• Table-based linearization: $L(x) = f(a) + f'(a)(x-a)$ using given values
• AP loves "approximate $f(2.1)$ given $f(2)$ and $f'(2)$" questions

Linearization & Differentials

Part 6 of 7 — AP-Style Workshop

AP FRQ Patterns

Full Worked AP Problem

The table below gives values of a twice-differentiable function $f$.

$x$	$0$	$1$	$2$	$3$
$f(x)$	$4$	$7$	$3$	$1$
$f'(x)$	$2$	$-1$	$-3$	$-2$

(a) Write the linearization of $f$ at $x = 2$.

(b) Use your answer to approximate $f(1.8)$.

(c) Given $f''(2) = -1$, is your estimate an over- or underestimate? Justify.

Solution (a): $L(x) = f(2) + f'(2)(x-2) = 3 + (-3)(x-2) = 3 - 3(x-2)$

Solution (b): $L(1.8) = 3 - 3(1.8-2) = 3 - 3(-0.2) = 3 + 0.6 = 3.6$

Solution (c): Since $f''(2) = -1 < 0$, $f$ is concave down near $x = 2$. The tangent line lies above the curve, so $3.6$ is an overestimate.

AP Tip: For "justify," you must state the concavity ($f'' < 0$), explain what it means (concave down), and conclude (tangent above curve = overestimate).

AP-Style Practice 🎯

Justify your reasoning. 🔍

AP Problem. ✍️

Key Takeaways — Part 6

• AP FRQs often combine table data with linearization
• Always justify over/underestimate with concavity
• Three-part justification: sign of $f''$, concavity, conclusion
• The tangent line approximation is exact when $f$ is linear

Linearization & Differentials

Part 7 of 7 — Comprehensive Assessment

Formula Reference

Common Linearizations at $a = 0$

Common AP Mistakes

Quiz Set 1 — Core Skills 🎯

Quiz Set 2 — Applications 🎯

Final review. 🔍

Final Challenge. ✍️

🎉 Topic Complete!

You've mastered Linearization & Differentials:

Key Fact: Linearization $L(x) = f(a) + f'(a)(x-a)$ is the foundation for approximation in calculus. On the AP exam, always pair your estimate with "over/underestimate" justified by concavity.