Linear approximation and differentials: Worked examples

Everything here rests on the parent Linear approximation and differentials and its engine Derivative as a limit. If any symbol looks unfamiliar, it was defined there.

The scenario matrix

Before solving anything, let's list what can vary. Three knobs change the flavour of a problem:

Cell	What varies	Sign / case	Example that hits it
A	Step direction	target above base ( $x > a$ , step $+$ )	Ex 1 ( $4.1$ )
B	Step direction	target below base ( $x < a$ , step $-$ )	Ex 2 ( $3.9$ )
C	Curvature	curve bends down (concave), line over-shoots	Ex 1 & 2 (√ is concave)
D	Curvature	curve bends up (convex), line under-shoots	Ex 3 ( $e^{x}$ )
E	Degenerate slope	$f^{'} (a) = 0$ — the line is flat	Ex 4 ( $cos$ near $0$ )
F	Base at zero	$a = 0$ , step collapses to $x$	Ex 5 ( $ln (1 + x)$ )
G	Word problem	measurement → error, real units	Ex 6 (cone volume)
H	Exam twist	pick your own $f, a$ ; combine ideas	Ex 7 ( $tan 4 6^{\circ}$ )
I	Limiting behaviour	how error grows with step size	Ex 8 (error scaling)
J	Negative slope	$f^{'} (a) < 0$ — a decreasing function	Ex 9 ( $1/ x$ at $a = 2$ )

Figure — Linear approximation and differentials

Example 1 — Cell A + C: step up, concave curve

Worked example Estimate

$4.1$ (target above base) Forecast: guess now — will $2.025$ be a slight over-estimate or under-estimate of the true $4.1$ ? (Look at the left panel of figure s01 — the yellow line rides above the curve.)

Step 1. Let $f (x) = x$ , base $a = 4$ . Why this step? $4 = 2$ is exact and $4.1$ is the nearest easy anchor — the Tangent line is most faithful nearest its touch point.

Step 2. Slope: $f^{'} (x) = \frac{1}{2 x}$ , so $f^{'} (4) = \frac{1}{4}$ . Why this step? The slope is the direction we "walk". Positive slope ⇒ walking right raises $y$ .

Step 3. Line: $L (x) = 2 + \frac{1}{4} (x - 4)$ . Step is $x - a = 4.1 - 4 = + 0.1$ (positive ⇒ Cell A). Why this step? $L$ is the tangent; we ride it a distance $0.1$ to the right.

Step 4. $L (4.1) = 2 + \frac{1}{4} (0.1) = 2.025$ . Why this step? This is where the estimate is born: we substitute the actual target $x = 4.1$ into the line, turning "ride the tangent" into a single number we can report.

Verify: True $4.1 = 2.02485 \dots$ , so $2.025$ is too big by $\approx 0.00015$ . That matches Cell C: $x$ is concave ( $f^{''} = - \frac{1}{4} x^{- 3/2} < 0$ ), so the line rides above the curve. ✓

Example 2 — Cell B + C: step down, concave curve

Worked example Estimate

$3.9$ (target below base) Forecast: now the step is negative. Over or under this time?

Step 1–2. Same $f (x) = x$ , $a = 4$ , $f (4) = 2$ , $f^{'} (4) = \frac{1}{4}$ . Why reuse? $3.9$ is just as near $4$ as $4.1$ was — same anchor, opposite direction.

Step 3. Step $x - a = 3.9 - 4 = - 0.1$ (negative ⇒ Cell B). Walking left lowers $y$ . Why this step? The formula handles both signs automatically; a negative step subtracts.

Step 4. $L (3.9) = 2 + \frac{1}{4} (- 0.1) = 2 - 0.025 = 1.975$ . Why this step? Substituting the below-base target shows the machinery needs no special "left-side" version — the negative run does the subtraction for us, producing the reported estimate.

Verify: True $3.9 = 1.97484 \dots$ Our $1.975$ is again too big by $\approx 0.00016$ . Concavity over-shoots on both sides — the line is above the curve everywhere. ✓ Note the error is nearly identical to Ex 1 (same $∣ x - a ∣$ ).

Example 3 — Cell D: convex curve, line under-shoots

Worked example Estimate

$e^{0.1}$ Forecast: $e^{x}$ curves upward. Predict: too big or too small? (The right panel of figure s01 shows the line dipping below the curve.)

Step 1. $f (x) = e^{x}$ , base $a = 0$ . Why $a = 0$ ? $e^{0} = 1$ exactly, and $0.1$ is close.

Step 2. $f^{'} (x) = e^{x}$ , so $f^{'} (0) = 1$ . Why? $e^{x}$ is its own derivative — a gift.

Step 3. $L (x) = 1 + 1 \cdot (x - 0) = 1 + x$ . This is the parent's $e^{x} \approx 1 + x$ . Why this step? $a = 0$ collapses $(x - a)$ to $x$ (Cell F flavour too).

Step 4. $L (0.1) = 1 + 0.1 = 1.1$ . Why this step? Plugging the target $x = 0.1$ into the memorized line $1 + x$ is the whole payoff — one add gives the estimate instead of exponentiating by hand.

Verify: True $e^{0.1} = 1.10517 \dots$ , so $1.1$ is too small by $\approx 0.005$ . Cell D confirmed: $f^{''} = e^{x} > 0$ (convex), the line lies below the curve, so we under-estimate. ✓ Compare Taylor series: the next term $+ \frac{x ^{2}}{2} = 0.005$ exactly closes this gap.

Example 4 — Cell E: flat tangent, $f^{'} (a) = 0$

Worked example Estimate

$cos (0.1)$ — what happens when the slope is zero? Forecast: the tangent at the top of a hill is horizontal. What will $L$ predict? (Figure s02 draws this flat yellow line sitting on the peak of $cos x$ .)

Step 1. $f (x) = cos x$ , base $a = 0$ . Why? $cos 0 = 1$ exactly.

Step 2. $f^{'} (x) = - sin x$ , so $f^{'} (0) = - sin 0 = 0$ . Slope is zero (Cell E). Why this matters? A zero slope means the tangent line is flat: $L$ predicts no change at all.

Step 3. $L (x) = 1 + 0 \cdot (x - 0) = 1$ . Why this step? "Start + Slope × Step" with Slope $= 0$ ⇒ just Start.

Step 4. $L (0.1) = 1$ . Why this step? Evaluating at the target makes the degeneracy concrete: the line returns the same value $1$ no matter where we walk, so the estimate carries zero first-order information.

Verify: True $cos (0.1) = 0.99500 \dots$ The estimate $1$ is off by $0.005$ — bigger relative to the tiny step than usual! Lesson: at a flat point the linear term vanishes, so the error is dominated by curvature ( $f^{''} = - cos$ , here $- 1$ ). This is precisely why Newton's method stalls near flat spots: dividing by a near-zero slope explodes. ✓

Example 5 — Cell F: base at zero, memorizable formula

Example 6 — Cell G: real-world word problem with units

Worked example A cone has fixed height

$h = 12$ cm and radius measured $r = 5$ cm with possible error $d r = 0.1$ cm. Estimate the error in volume $V = \frac{1}{3} π r^{2} h$ . Forecast: the volume grows like $r^{2}$ . Will a $2%$ radius error give more or less than $2%$ volume error?

Step 1. Treat $V$ as a function of $r$ (height constant): $V (r) = \frac{1}{3} π h r^{2} = 4 π r^{2}$ (using $\frac{1}{3} \cdot 12 = 4$ ). Why this step? Only $r$ is uncertain, so we linearize in $r$ alone — this is Error propagation.

Step 2. Differential: $d V = V^{'} (r) d r = 8 π r d r$ . Why differentials here? We don't know the true error, only that $d r$ is small, so $Δ V \approx d V$ (parent's $d y = f^{'} (x) d x$ ).

Step 3. Plug $r = 5$ , $d r = 0.1$ : $d V = 8 π (5) (0.1) = 4 π$ . Why these numbers? Evaluate the slope at the measured value.

Step 4. $d V = 4 π \approx 12.566 cm^{3}$ . Why this step? Turning $4 π$ into a decimal with units is what makes the answer usable by an engineer — an abstract $4 π$ doesn't tell you the tolerance, "about $12.6 cm^{3}$ " does.

Verify (units + relative): $d V$ has units $cm^{2} \cdot cm = cm^{3}$ ✓ — a volume, as it must be. Relative error: $\frac{d V}{V} = \frac{8 π r d r}{4 π r ^{2}} = \frac{2 d r}{r} = 2 \cdot \frac{0.1}{5} = 0.04 = 4%$ . So a $2%$ radius error ( $0.1/5$ ) doubles to $4%$ because $V \propto r^{2}$ — the exponent $2$ is the multiplier. ✓

Example 7 — Cell H: exam twist, choose your own $f, a$ , mixed units

Worked example Estimate

$tan (4 6^{\circ})$ without a calculator. Forecast: $4 6^{\circ}$ is near $4 5^{\circ}$ where $tan = 1$ . But the formula needs radians — why?

Step 1. $f (x) = tan x$ , base $a = 4 5^{\circ} = \frac{π}{4}$ rad. Why radians? The derivative $\frac{d}{d x} tan x = sec^{2} x$ is only valid when $x$ is in radians; degrees would smuggle a hidden factor of $\frac{π}{180}$ .

Step 2. $f (\frac{π}{4}) = tan 4 5^{\circ} = 1$ . Slope $f^{'} (x) = sec^{2} x$ , and $sec \frac{π}{4} = 2$ , so $f^{'} (\frac{π}{4}) = (2)^{2} = 2$ . Why? Start value $1$ , direction $2$ .

Step 3. Step: $4 6^{\circ} - 4 5^{\circ} = 1^{\circ} = \frac{π}{180} \approx 0.017453$ rad. Why convert? The run must be in the same unit the slope expects (radians).

Step 4. $L = 1 + 2 \cdot 0.017453 = 1.034907$ . Why this step? Multiplying slope by the radian run and adding the start is where the unit-care pays off: it produces a number within $0.001$ of truth, whereas the wrong-unit run would have blown up.

Verify: True $tan 4 6^{\circ} = 1.03553 \dots$ Our $1.03491$ is too small by $\approx 0.0006$ . $tan$ is convex just past $4 5^{\circ}$ ( $f^{''} = 2 sec^{2} x tan x > 0$ there) ⇒ under-estimate. ✓ The wrong-unit trap (using step $= 1$ instead of $0.01745$ ) would have given $\approx 3$ — wildly off, which flags the mistake.

Example 8 — Cell I: limiting behaviour, how error scales

Worked example For

$f (x) = x^{2}$ at $a = 1$ , compare the true change $Δ y$ and the differential $d y$ for steps $d x = 0.1$ and $d x = 0.01$ . Forecast: if you shrink the step by $10 \times$ , does the error shrink by $10 \times$ or $100 \times$ ? (Figure s03 draws both step sizes and marks the vertical gap = error.)

Step 1. $f^{'} (x) = 2 x$ , so $f^{'} (1) = 2$ . Thus $d y = 2 d x$ . Why this step? $d y$ is the tangent-line rise, our approximation of the true rise.

Step 2. True change: $Δ y = f (1 + d x) - f (1) = (1 + d x)^{2} - 1 = 2 d x + d x^{2}$ . Why expand? To see the leftover: $Δ y - d y = d x^{2}$ exactly — the error is the squared step.

Step 3. For $d x = 0.1$ : $d y = 0.2$ , $Δ y = 0.21$ , error $= 0.01$ . For $d x = 0.01$ : $d y = 0.02$ , $Δ y = 0.0201$ , error $= 0.0001$ . Why compare? To read off the scaling law.

Step 4. Shrinking the step $10 \times$ ( $0.1 \to 0.01$ ) shrank the error $100 \times$ ( $0.01 \to 0.0001$ ). Why this step? This final comparison is the punchline: reading the two error numbers side by side converts the algebraic fact " $error = d x^{2}$ " into a felt rule of thumb — halve the step, quarter the error.

Verify: Error $= d x^{2}$ , so halving/tenthing the step squares the shrink. This is the parent's "small × small shrinks faster" made exact, and it's why the degree-2 Taylor series term $\frac{f ^{''}}{2} d x^{2}$ is precisely this leftover. ✓

Example 9 — Cell J: negative slope, a decreasing function

Worked example Estimate

$\frac{1}{2.1}$ using $f (x) = \frac{1}{x}$ at $a = 2$ . Forecast: $1/ x$ falls as $x$ grows, so the slope is negative. If we step right (to $2.1$ ), will the estimate go up or down from $0.5$ ?

Step 1. $f (x) = \frac{1}{x}$ , base $a = 2$ . Why? $1/2 = 0.5$ is exact and $2.1$ is close. Why this step? We need an anchor where both value and slope are painless.

Step 2. Slope: $f^{'} (x) = - \frac{1}{x ^{2}}$ , so $f^{'} (2) = - \frac{1}{4} = - 0.25$ . Negative slope (Cell J). Why this step? A negative slope means the tangent descends to the right — stepping right must lower $y$ . This is the first example whose direction of walk is downhill.

Step 3. Line: $L (x) = 0.5 - 0.25 (x - 2)$ . Step $x - a = 2.1 - 2 = + 0.1$ (positive run, but slope pulls down). Why this step? The sign bookkeeping is the whole point: a positive run times a negative slope gives a negative rise.

Step 4. $L (2.1) = 0.5 - 0.25 (0.1) = 0.5 - 0.025 = 0.475$ . Why this step? Evaluating shows the estimate dropped below $0.5$ exactly as a decreasing function demands — confirming the negative slope steered the walk downward.

Verify: True $\frac{1}{2.1} = 0.47619 \dots$ , so $0.475$ is too small by $\approx 0.0012$ . Curvature check: $f^{''} = \frac{2}{x ^{3}} > 0$ (convex) ⇒ line below curve ⇒ under-estimate, consistent with Cell D logic even though the slope is negative. ✓ Key takeaway: the sign of the slope controls the walk's direction; the sign of $f^{''}$ independently controls over/under — the two are unrelated.

Active recall

Recall Which cells over-estimate and which under-estimate?

Concave (bends down, $f^{''} < 0$ ): line above ⇒ over. Convex (bends up, $f^{''} > 0$ ): line below ⇒ under. Concave over/under? ::: Over-estimate. Convex over/under? ::: Under-estimate.

Recall Does a negative slope change whether we over- or under-estimate?

No — over/under is decided by $f^{''}$ (curvature), not by the sign of $f^{'}$ . The slope's sign only sets which way the walk goes. What controls over vs under? ::: The sign of $f^{''}$ , not $f^{'}$ .

Recall Why did

$tan 4 6^{\circ}$ need radians? Because $\frac{d}{d x} tan x = sec^{2} x$ only holds in radians; using a degree step of $1$ gives nonsense. What step (in radians) is $1^{\circ}$ ? ::: $π /180 \approx 0.01745$ .

Recall If the step shrinks by

$10 \times$ , the error shrinks by...? $100 \times$ — the error scales like $d x^{2}$ . Error of linear approx scales like? ::: $(x - a)^{2}$ .

Connections

Linear approximation and differentials — the parent tool these examples exercise.
Derivative as a limit — supplies every slope $f^{'} (a)$ used above.
Tangent line — the line $L$ we ride in each case.
Concavity and second derivative — the $f^{''}$ sign that decides over/under-shoot.
Taylor series — the leftover $\frac{f ^{''}}{2} (x - a)^{2}$ is the degree-2 term.
Newton's method — Example 4 shows why flat slopes ( $f^{'} \approx 0$ ) break it.
Error propagation — Example 6's cone is a direct application of $d y = f^{'} (x) d x$ .