Linear approximation and differentials: Worked examples

4.1.32 · D3 · Maths › Calculus I — Limits & Derivatives › Linear approximation and differentials

Yahan sab kuch parent Linear approximation and differentials aur uske engine Derivative as a limit par tika hai. Agar koi symbol unfamiliar lage, woh wahan define kiya gaya tha.

Scenario matrix

Kuch bhi solve karne se pehle, chalte hain list karte hain kya vary kar sakta hai. Teen knobs ek problem ka flavour badal dete hain:

Cell	Kya vary karta hai	Sign / case	Example jo isko hit karta hai
A	Step direction	target upar base se ( $x > a$ , step $+$ )	Ex 1 ( $4.1$ )
B	Step direction	target neeche base se ( $x < a$ , step $-$ )	Ex 2 ( $3.9$ )
C	Curvature	curve neeche bend karti hai (concave), line over-shoot karti hai	Ex 1 & 2 (√ is concave)
D	Curvature	curve upar bend karti hai (convex), line under-shoot karti hai	Ex 3 ( $e^{x}$ )
E	Degenerate slope	$f^{'} (a) = 0$ — line flat hai	Ex 4 ( $cos$ near $0$ )
F	Base at zero	$a = 0$ , step collapse hokar $x$ ban jaata hai	Ex 5 ( $ln (1 + x)$ )
G	Word problem	measurement → error, real units	Ex 6 (cone volume)
H	Exam twist	apna $f, a$ khud choose karo; ideas combine karo	Ex 7 ( $tan 4 6^{\circ}$ )
I	Limiting behaviour	error step size ke saath kaise badhta hai	Ex 8 (error scaling)
J	Negative slope	$f^{'} (a) < 0$ — ek 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

$4.1$ estimate karo (target base se upar) Forecast: abhi guess karo — kya $2.025$ true $4.1$ ka slight over-estimate hoga ya under-estimate? (Figure s01 ke left panel dekho — yellow line curve ke upar ride kar rahi hai.)

Step 1. Maano $f (x) = x$ , base $a = 4$ . Yeh step kyun? $4 = 2$ exact hai aur $4.1$ sabse paas ka easy anchor hai — Tangent line apne touch point ke sabse paas sabse faithful hoti hai.

Step 2. Slope: $f^{'} (x) = \frac{1}{2 x}$ , toh $f^{'} (4) = \frac{1}{4}$ . Yeh step kyun? Slope woh direction hai jis mein hum "chalte" hain. Positive slope ⇒ right chalene se $y$ badhta hai.

Step 3. Line: $L (x) = 2 + \frac{1}{4} (x - 4)$ . Step hai $x - a = 4.1 - 4 = + 0.1$ (positive ⇒ Cell A). Yeh step kyun? $L$ hi tangent hai; hum ise $0.1$ ki doori tak daayein ride karte hain.

Step 4. $L (4.1) = 2 + \frac{1}{4} (0.1) = 2.025$ . Yeh step kyun? Yahan estimate janam leta hai: hum actual target $x = 4.1$ ko line mein substitute karte hain, "tangent ride karo" ko ek single number mein badal dete hain jo hum report kar sakte hain.

Verify: True $4.1 = 2.02485 \dots$ , toh $2.025$ too big hai $\approx 0.00015$ se. Yeh Cell C se match karta hai: $x$ concave hai ( $f^{''} = - \frac{1}{4} x^{- 3/2} < 0$ ), toh line curve ke upar ride karti hai. ✓

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

Worked example

$3.9$ estimate karo (target base se neeche) Forecast: ab step negative hai. Is baar over ya under?

Step 1–2. Same $f (x) = x$ , $a = 4$ , $f (4) = 2$ , $f^{'} (4) = \frac{1}{4}$ . Kyun reuse karein? $3.9$ utna hi $4$ ke paas hai jitna $4.1$ tha — same anchor, opposite direction.

Step 3. Step $x - a = 3.9 - 4 = - 0.1$ (negative ⇒ Cell B). Left chalene se $y$ ghat ta hai. Yeh step kyun? Formula dono signs ko automatically handle karta hai; ek negative step subtract karta hai.

Step 4. $L (3.9) = 2 + \frac{1}{4} (- 0.1) = 2 - 0.025 = 1.975$ . Yeh step kyun? Below-base target substitute karna dikhata hai ki machinery ko koi special "left-side" version nahi chahiye — negative run apne aap subtraction kar deta hai, reported estimate produce karta hai.

Verify: True $3.9 = 1.97484 \dots$ Hamara $1.975$ phir se too big hai $\approx 0.00016$ se. Concavity dono sides pe over-shoot karta hai — line har jagah curve ke upar hoti hai. ✓ Dhyan do error Ex 1 se lagbhag identical hai (same $∣ x - a ∣$ ).

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

Worked example

$e^{0.1}$ estimate karo Forecast: $e^{x}$ upar curve karta hai. Predict karo: too big ya too small? (Figure s01 ka right panel line ko curve ke neeche dip karte dikhata hai.)

Step 1. $f (x) = e^{x}$ , base $a = 0$ . $a = 0$ kyun? $e^{0} = 1$ exactly, aur $0.1$ paas hai.

Step 2. $f^{'} (x) = e^{x}$ , toh $f^{'} (0) = 1$ . Kyun? $e^{x}$ apna khud ka derivative hai — ek gift.

Step 3. $L (x) = 1 + 1 \cdot (x - 0) = 1 + x$ . Yahi parent ka $e^{x} \approx 1 + x$ hai. Yeh step kyun? $a = 0$ $(x - a)$ ko $x$ mein collapse kar deta hai (Cell F flavour bhi).

Step 4. $L (0.1) = 1 + 0.1 = 1.1$ . Yeh step kyun? Target $x = 0.1$ ko memorized line $1 + x$ mein plug karna pura payoff hai — haath se exponentiate karne ki jagah ek addition se estimate mil jaata hai.

Verify: True $e^{0.1} = 1.10517 \dots$ , toh $1.1$ too small hai $\approx 0.005$ se. Cell D confirm: $f^{''} = e^{x} > 0$ (convex), line curve ke neeche hai, toh hum under-estimate karte hain. ✓ Taylor series se compare karo: agla term $+ \frac{x ^{2}}{2} = 0.005$ precisely yeh gap band karta hai.

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

Worked example

$cos (0.1)$ estimate karo — kya hota hai jab slope zero ho? Forecast: ek hill ke top par tangent horizontal hoti hai. $L$ kya predict karega? (Figure s02 is flat yellow line ko $cos x$ ke peak par draw karta hai.)

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

Step 2. $f^{'} (x) = - sin x$ , toh $f^{'} (0) = - sin 0 = 0$ . Slope zero hai (Cell E). Yeh kyun matter karta hai? Zero slope ka matlab hai tangent line flat hai: $L$ bilkul koi change predict nahi karta.

Step 3. $L (x) = 1 + 0 \cdot (x - 0) = 1$ . Yeh step kyun? "Start + Slope × Step" mein Slope $= 0$ ⇒ sirf Start.

Step 4. $L (0.1) = 1$ . Yeh step kyun? Target par evaluate karna degeneracy ko concrete banata hai: line wohi value $1$ return karta hai chahe hum kahaan bhi chalein, toh estimate mein zero first-order information hoti hai.

Verify: True $cos (0.1) = 0.99500 \dots$ Estimate $1$ off hai $0.005$ se — ususal se zyada relative to tiny step! Lesson: flat point par linear term vanish ho jaata hai, toh error curvature se dominate hoti hai ( $f^{''} = - cos$ , yahan $- 1$ ). Yahi precisely woh reason hai jis se Newton's method flat spots ke paas stall karta hai: near-zero slope se divide karna explode karta hai. ✓

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

Worked example

$ln (1.05)$ estimate karo Forecast: kaun sa standard $a = 0$ approximation apply hota hai?

Step 1. $f (x) = ln (1 + x)$ , base $a = 0$ . Kyun? $ln (1 + 0) = ln 1 = 0$ exactly.

Step 2. $f^{'} (x) = \frac{1}{1 + x}$ , toh $f^{'} (0) = 1$ . Kyun? Anchor par slope line set karta hai.

Step 3. $L (x) = 0 + 1 \cdot x = x$ (Cell F: step $= x - 0 = x$ ). Deta hai $ln (1 + x) \approx x$ . Yeh step kyun? Base-at-zero trick approximation ko ek-symbol formula mein badal deta hai.

Step 4. Yahan $x = 0.05$ hai (kyunki $1.05 = 1 + 0.05$ ): $L = 0.05$ . Yeh step kyun? Small offset $x = 0.05$ substitute karna formula ko action mein demonstrate karta hai — estimate literally offset khud hai, isliye $ln (1 + x) \approx x$ yaad rakhne layak hai.

Verify: True $ln (1.05) = 0.04879 \dots$ , toh $0.05$ too big hai $\approx 0.0012$ se. Curvature check karo: $f^{''} = - \frac{1}{( 1 + x ) ^{2}} < 0$ (concave) ⇒ over-estimate. Cell C logic se consistent. ✓

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

Worked example Ek cone ki fixed height

$h = 12$ cm hai aur radius $r = 5$ cm measured hai possible error $d r = 0.1$ cm ke saath. Volume $V = \frac{1}{3} π r^{2} h$ mein error estimate karo. Forecast: volume $r^{2}$ ki tarah badhta hai. Kya $2%$ radius error $2%$ se zyada ya kam volume error dega?

Step 1. $V$ ko $r$ ka function maano (height constant): $V (r) = \frac{1}{3} π h r^{2} = 4 π r^{2}$ ( $\frac{1}{3} \cdot 12 = 4$ use karke). Yeh step kyun? Sirf $r$ uncertain hai, toh hum sirf $r$ mein linearize karte hain — yahi Error propagation hai.

Step 2. Differential: $d V = V^{'} (r) d r = 8 π r d r$ . Differentials yahan kyun? Hum true error nahi jaante, sirf yeh ki $d r$ small hai, toh $Δ V \approx d V$ (parent ka $d y = f^{'} (x) d x$ ).

Step 3. $r = 5$ , $d r = 0.1$ plug karo: $d V = 8 π (5) (0.1) = 4 π$ . Yeh numbers kyun? Measured value par slope evaluate karo.

Step 4. $d V = 4 π \approx 12.566 cm^{3}$ . Yeh step kyun? $4 π$ ko units ke saath decimal mein badalna woh hai jo answer ko ek engineer ke liye usable banata hai — abstract $4 π$ tumhe tolerance nahi batata, "about $12.6 cm^{3}$ " batata hai.

Verify (units + relative): $d V$ ke units hain $cm^{2} \cdot cm = cm^{3}$ ✓ — ek volume, jaisa hona chahiye. Relative error: $\frac{d V}{V} = \frac{8 π r d r}{4 π r ^{2}} = \frac{2 d r}{r} = 2 \cdot \frac{0.1}{5} = 0.04 = 4%$ . Toh $2%$ radius error ( $0.1/5$ ) double hokar $4%$ ban jaata hai kyunki $V \propto r^{2}$ — exponent $2$ multiplier hai. ✓

Example 7 — Cell H: exam twist, apna $f, a$ choose karo, mixed units

Worked example Bina calculator ke

$tan (4 6^{\circ})$ estimate karo. Forecast: $4 6^{\circ}$ , $4 5^{\circ}$ ke paas hai jahan $tan = 1$ hai. Lekin formula ko radians chahiye — kyun?

Step 1. $f (x) = tan x$ , base $a = 4 5^{\circ} = \frac{π}{4}$ rad. Radians kyun? Derivative $\frac{d}{d x} tan x = sec^{2} x$ tabhi valid hai jab $x$ radians mein ho; degrees ek hidden factor of $\frac{π}{180}$ smuggle kar lete.

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

Step 3. Step: $4 6^{\circ} - 4 5^{\circ} = 1^{\circ} = \frac{π}{180} \approx 0.017453$ rad. Convert kyun? Run us hi unit mein hona chahiye jo slope expect karta hai (radians).

Step 4. $L = 1 + 2 \cdot 0.017453 = 1.034907$ . Yeh step kyun? Slope ko radian run se multiply karna aur start add karna wahan hai jahan unit-care pay off karta hai: yeh truth ke $0.001$ ke andar ek number produce karta hai, jabki wrong-unit run blast kar deta.

Verify: True $tan 4 6^{\circ} = 1.03553 \dots$ Hamara $1.03491$ too small hai $\approx 0.0006$ se. $tan$ just past $4 5^{\circ}$ convex hai ( $f^{''} = 2 sec^{2} x tan x > 0$ wahan) ⇒ under-estimate. ✓ Wrong-unit trap (step $= 1$ use karna $0.01745$ ki jagah) ne $\approx 3$ diya hota — wildly off, jo mistake flag kar deta.

Example 8 — Cell I: limiting behaviour, error kaise scale karta hai

Worked example

$f (x) = x^{2}$ ke liye $a = 1$ par, true change $Δ y$ aur differential $d y$ ko steps $d x = 0.1$ aur $d x = 0.01$ ke liye compare karo. Forecast: agar step $10 \times$ shrink karo, toh kya error $10 \times$ shrink hoga ya $100 \times$ ? (Figure s03 dono step sizes draw karta hai aur vertical gap = error mark karta hai.)

Step 1. $f^{'} (x) = 2 x$ , toh $f^{'} (1) = 2$ . Isliye $d y = 2 d x$ . Yeh step kyun? $d y$ tangent-line rise hai, true rise ka hamara approximation.

Step 2. True change: $Δ y = f (1 + d x) - f (1) = (1 + d x)^{2} - 1 = 2 d x + d x^{2}$ . Expand kyun? Leftover dekhne ke liye: $Δ y - d y = d x^{2}$ exactly — error squared step hai.

Step 3. $d x = 0.1$ ke liye: $d y = 0.2$ , $Δ y = 0.21$ , error $= 0.01$ . $d x = 0.01$ ke liye: $d y = 0.02$ , $Δ y = 0.0201$ , error $= 0.0001$ . Compare kyun? Scaling law padhne ke liye.

Step 4. Step $10 \times$ shrink karne se ( $0.1 \to 0.01$ ) error $100 \times$ shrink hua ( $0.01 \to 0.0001$ ). Yeh step kyun? Yeh final comparison hi punchline hai: dono error numbers side by side padhna algebraic fact " $error = d x^{2}$ " ko ek felt rule of thumb mein convert karta hai — step half karo, error quarter ho jaata hai.

Verify: Error $= d x^{2}$ , toh step half/tenth karna shrink ko square karta hai. Yahi parent ka "small × small shrinks faster" hai jo exact ban gaya, aur isliye degree-2 Taylor series term $\frac{f ^{''}}{2} d x^{2}$ precisely yeh leftover hai. ✓

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

Worked example

$f (x) = \frac{1}{x}$ use karke $a = 2$ par $\frac{1}{2.1}$ estimate karo. Forecast: $1/ x$ girta hai jaise $x$ badhta hai, toh slope negative hai. Agar hum right step karte hain (to $2.1$ ), kya estimate $0.5$ se upar jaayega ya neeche?

Step 1. $f (x) = \frac{1}{x}$ , base $a = 2$ . Kyun? $1/2 = 0.5$ exact hai aur $2.1$ paas hai. Yeh step kyun? Hume ek anchor chahiye jahan value aur slope dono painless hon.

Step 2. Slope: $f^{'} (x) = - \frac{1}{x ^{2}}$ , toh $f^{'} (2) = - \frac{1}{4} = - 0.25$ . Negative slope (Cell J). Yeh step kyun? Negative slope ka matlab hai tangent daayein utarti hai — right chalena $y$ ko lower karna chahiye. Yeh pehla example hai jiska walk direction downhill hai.

Step 3. Line: $L (x) = 0.5 - 0.25 (x - 2)$ . Step $x - a = 2.1 - 2 = + 0.1$ (positive run, lekin slope neeche khichta hai). Yeh step kyun? Sign bookkeeping hi pura point hai: positive run times negative slope deta hai negative rise.

Step 4. $L (2.1) = 0.5 - 0.25 (0.1) = 0.5 - 0.025 = 0.475$ . Yeh step kyun? Evaluate karna dikhata hai estimate $0.5$ se neeche giri exactly jaisa ek decreasing function demand karta hai — confirming karta hai ki negative slope ne walk ko downward steer kiya.

Verify: True $\frac{1}{2.1} = 0.47619 \dots$ , toh $0.475$ too small hai $\approx 0.0012$ se. Curvature check: $f^{''} = \frac{2}{x ^{3}} > 0$ (convex) ⇒ line curve ke neeche ⇒ under-estimate, Cell D logic se consistent hai chahe slope negative ho. ✓ Key takeaway: slope ka sign walk ki direction control karta hai; $f^{''}$ ka sign independently over/under control karta hai — dono unrelated hain.

Active recall

Recall Kaun se cells over-estimate karte hain aur kaun se under-estimate?

Concave (neeche bend, $f^{''} < 0$ ): line upar ⇒ over. Convex (upar bend, $f^{''} > 0$ ): line neeche ⇒ under. Concave over/under? ::: Over-estimate. Convex over/under? ::: Under-estimate.

Recall Kya negative slope yeh badal deta hai ki hum over- ya under-estimate karte hain?

Nahi — over/under $f^{''}$ (curvature) se decide hota hai, $f^{'}$ ke sign se nahi. Slope ka sign sirf yeh set karta hai ki walk kis direction mein jaati hai. Over vs under kya control karta hai? ::: $f^{''}$ ka sign, $f^{'}$ nahi.

Recall

$tan 4 6^{\circ}$ ko radians kyun chahiye the? Kyunki $\frac{d}{d x} tan x = sec^{2} x$ sirf radians mein hold karta hai; $1$ ka degree step use karna nonsense deta hai. $1^{\circ}$ kitne radians hota hai? ::: $π /180 \approx 0.01745$ .

Recall Agar step

$10 \times$ shrink ho, toh error...? $100 \times$ shrink hoti hai — error $d x^{2}$ ki tarah scale karti hai. Linear approx ka error scale karta hai like? ::: $(x - a)^{2}$ .

Connections

Linear approximation and differentials — parent tool jin par yeh examples practice karte hain.
Derivative as a limit — upar use kiya gaya har slope $f^{'} (a)$ supply karta hai.
Tangent line — woh line $L$ jis par hum har case mein ride karte hain.
Concavity and second derivative — woh $f^{''}$ sign jo over/under-shoot decide karta hai.
Taylor series — leftover $\frac{f ^{''}}{2} (x - a)^{2}$ degree-2 term hai.
Newton's method — Example 4 dikhata hai flat slopes ( $f^{'} \approx 0$ ) ise kyun break karte hain.
Error propagation — Example 6 ka cone $d y = f^{'} (x) d x$ ka direct application hai.