Exercises — Linear approximation and differentials

Kya chahiye: $L(x)=f(a)+f^{\prime }(a)(x-a)$, toh do numbers chahiye: height $f(2)$ aur slope $f^{\prime }(2)$.

• Height: $f(2)=2^3=8$.
• Slope: $f^{\prime }(x)=3x^2$, toh $f^{\prime }(2)=3(4)=12$.

Assemble karo: $L(x)=8+12(x-2)$. Yeh $y=x^3$ ki tangent line hai point $(2,8)$ par.

$f(1)=\ln 1=0$. Slope $f^{\prime }(x)=\frac{1}{x}$, toh $f^{\prime }(1)=1$. $L(x)=0+1(x-1)=x-1.$ Ab $x=1+u$ rakho (toh "step" $x-1=u$ ho jaata hai): $\ln (1+u)\approx u$. Yeh standard near-zero rule hai.

$dy$ kyun: $dy=f^{\prime }(x)dx$ matlab "tangent ke saath rise = slope × run." $f^{\prime }(x)=-\sin x$, toh $dy=-\sin xdx$. $x=\frac{\pi }{3}$ par: $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}\approx 0.8660$. $dy=-0.8660\times 0.01=-\mathrm{0.008660.}$ Minus sign keh raha hai wahan cosine decrease ho rahi hai, toh thoda sa step right lene se $y$ girta hai.

$f,a$ chunno: $f(x)=x^{1/3}$, $a=8$ kyunki $\sqrt[3]{8}=2$ exactly aata hai aur $8.2$ paas mein hai. Height: $f(8)=2$. Slope: $f^{\prime }(x)=\frac{1}{3}{x}^{-2/3}$, toh $f^{\prime }(8)=\frac{1}{3}\cdot 8^{-2/3}=\frac{1}{3}\cdot \frac{1}{4}=\frac{1}{12}$. Line & evaluate: step $=8.2-8=0.2$. $\sqrt[3]{8.2}\approx 2+\frac{1}{12}(0.2)=2+0.016\overline{6}=\mathrm{2.01667.}$ (True value $2.01653\dots$ — approximately $0.00014$ ka fark.)

Chunno: $f(x)=x^5$, base $a=2$ (kyunki $2^5=32$ exact hai), $dx=1.97-2=-0.03$. Slope: $f^{\prime }(x)=5x^4$, $f^{\prime }(2)=5(16)=80$. Differential: $dy=80\times (-0.03)=-2.4$. $(1.97{)}^5\approx f(2)+dy=32-2.4=\mathrm{29.6.}$ (True $\approx 29.584$.) Dhyan do $dx<0$ cleanly handle hua — step bas left point karta hai.

Convert karo: $45^{\circ }=\frac{\pi }{4}$, aur step hai $1^{\circ }=\frac{\pi }{180}\approx 0.017453$ rad. Chunno: $f(x)=\tan x$, $a=\frac{\pi }{4}$ (wahan $\tan =1$ exactly). Slope: $f^{\prime }(x)={\sec }^{2}x=\frac{1}{{\cos }^{2}x}$. $\frac{\pi }{4}$ par, $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$, toh ${\cos }^2=\frac{1}{2}$ aur $f^{\prime }=2$. Line: $\tan (46^{\circ })\approx 1+2(0.017453)=1+0.034907=1.034907$. (True $\tan 46^{\circ }=1.03553\dots$ — $0.0006$ ka fark; tangent tezi se curve karta hai upar, isliye thoda bada miss.)

Shape ka second derivative test: $f^{\prime }(x)=\frac{1}{2}{x}^{-1/2}$, $f^{\prime \prime }(x)=-\frac{1}{4}{x}^{-3/2}<0$ $x>0$ ke liye. Negative second derivative $\Rightarrow$ curve concave down hai — yeh apni tangent lines ke neeche bend karti hai. Figure mein dekho: teal curve $a$ ke right mein orange tangent ke neeche baith jaati hai.

Toh tangent line $\sqrt{x}$ ke upar hai, aur $L(4.1)=2.025$ true $\sqrt{4.1}=2.02485\dots$ ka over-estimate hai. Confirmed: $2.025>2.02485$. Yeh Concavity and second derivative connection action mein hai.

Error law kahan se aata hai (Taylor's theorem with Lagrange remainder): agar $f$ ke do derivatives hain, toh kisi point $c$ ke liye jo strictly $a$ aur $x$ ke beech hai, $f(x)=\underset{L(x)}{\underbrace{f(a)+f^{\prime }(a)(x-a)}}+\underset{\text{remainder}=\text{error}}{\underbrace{\frac{1}{2}{f}^{\prime \prime }(c)(x-a{)}^2}}.$ Yeh standard Taylor series remainder hai: linear part ke baad jo bachta hai woh exactly $\frac{1}{2}{f}^{\prime \prime }(c)(x-a{)}^2$ form ka ek term hota hai. Toh $\text{error}=f(x)-L(x)=\frac{1}{2}{f}^{\prime \prime }(c)(x-a{)}^2$, jo $\propto (x-a{)}^2$ hai. Reasoning: kyunki error $\propto (x-a{)}^2$ hai, $(x-a)$ ko half karne se square ${\left(\frac{1}{2}\right)}^2=\frac{1}{4}$ se multiply hota hai. Toh error 4 ke factor se girta hai. $L$ banao (yaad karo $\frac{1}{4}$ kahan se aata hai): $f(x)=\sqrt{x}$, $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$, toh $f^{\prime }(4)=\frac{1}{2\cdot 2}=\frac{1}{4}$; $f(4)=2$ ke saath yeh deta hai $L(x)=2+\frac{1}{4}(x-4)$. Numeric check:

• Step $0.1$: $L(4.1)=2.025$, true $=2.0248457\dots$, error $=0.0001543$.
• Step $0.05$: $L(4.05)=2.0125$, true $=2.0124612\dots$, error $=0.0000388$. Ratio $=0.0001543/0.0000388\approx 3.98\approx 4$. Quadratic-error prediction se match karta hai — exactly isliye Taylor series ek $\frac{1}{2}{f}^{\prime \prime }(a)(x-a{)}^2$ term add karta hai is leading error ko khatam karne ke liye.

Differentials kyun: hum true error nahi jaante, sirf itna pata hai ki $ds$ small hai, toh $\Delta V\approx dV=V^{\prime }(s)ds$. $V^{\prime }(s)=3s^2$, toh $dV=3s^{2}ds=3(25)(0.02)=1.5{\text{cm}}^3$ (absolute). Relative error: $\frac{dV}{V}=\frac{3s^{2}ds}{s^3}=3\frac{ds}{s}=3\cdot \frac{0.02}{5}=3(0.004)=0.012=1.2\%.$ Neat pattern: power $s^n$ ke liye, relative error $n$ se multiply hota hai. (Sphere $r^3$ ne parent mein $3\frac{dr}{r}$ diya — same rule.) Dekho Error propagation.

Pehle, woh tool jis par hum lean karte hain — $(1+u{)}^n$ rule, yahan derived: $h(u)=(1+u{)}^n$ lo aur $u=0$ par linearize karo. Tab $h(0)=1$, aur $h^{\prime }(u)=n(1+u{)}^{n-1}$ toh $h^{\prime }(0)=n$. Hence $L(u)=1+nu$, yani $(1+u{)}^n\approx 1+nu(u\text{ small}).$ Piece 1: $\sqrt{4.1}\approx 2+\frac{1}{4}(0.1)=2.025$ ($f(x)=\sqrt{x}$, $f^{\prime }(4)=\frac{1}{4}$ use karte hue jaise 3.2 mein). Piece 2: boxed rule apply karo $u=0.02,n=3$ ke saath: $(1.02{)}^3=(1+0.02{)}^3\approx 1+3(0.02)=1.06$. Combine karo: product $\approx 2.025\times 1.06=2.1465$. (True: $\sqrt{4.1}=2.024846$, $1.02^3=1.061208$, product $=2.148741$.) Error $\approx 0.0022$. Yeh ek badi linearization se easier kyun hai: har factor ek clean, memorized near-zero form hai; do simple estimates multiply karna ek ugly product $\sqrt{x}(x{)}^3$ differentiate karne aur joint base point choose karne se bachata hai. Chhota cost yeh hai ki dono factors ke errors add up hote hain.

$f$ ko $x_0$ par linearize karo: $L(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)$. $f(x)=x^2-5$, $f(2)=4-5=-1$; $f^{\prime }(x)=2x$, $f^{\prime }(2)=4$. $L(x)=0$ solve karo: $-1+4(x-2)=0\Rightarrow x-2=\frac{1}{4}\Rightarrow x=2.25$. Toh $x_1=2.25$. Yeh exactly Newton's method update hai $x_1=x_0-\frac{f(x_0)}{f^{\prime }(x_0)}=2-\frac{-1}{4}=2.25$. (True $\sqrt{5}=2.23607$; ek step mein already $0.014$ tak accurate.) Key insight: Newton = axis tak tangent par baar baar sawari karna, yaani linear approximation ulta use karna — line evaluate karne ke bajaye solve karo.

$e^x$ ko $a=0$ par linearize karo: $f(0)=e^0=1$, $f^{\prime }(x)=e^x$ toh $f^{\prime }(0)=1$. Tab $L(x)=1+1\cdot (x-0)=1+x$. ✔ Estimates: $e^{0.05}\approx 1.05$; $\ln (1.05)=\ln (1+0.05)\approx 0.05$. Inverse check: $\ln (e^{0.05})$ ko $0.05$ hona chahiye. Estimates use karke: $\ln (1.05)\approx 0.05$. ✔ Yeh first order tak ek doosre ko undo karte hain — dono inverse functions ke degree-1 Taylor series hain, toh unka leading behaviour match karta hai.

$T$ ko $g$ ka function maano: $T(g)=2\pi \sqrt{L}{g}^{-1/2}$. Differentiate karo: $\frac{dT}{dg}=2\pi \sqrt{L}\cdot \left(-\frac{1}{2}\right)g^{-3/2}=-\pi \sqrt{L}{g}^{-3/2}$. Relative form (cleaner): kyunki $T\propto g^{-1/2}$ hai, $\frac{dT}{T}=-\frac{1}{2}\frac{dg}{g}$. Numerically: $\frac{dT}{T}=-\frac{1}{2}\cdot \frac{0.05}{9.81}=-0.002548=-0.2548\%.$ Absolute: pehle $T=2\pi \sqrt{1/9.81}=2\pi (0.31944)=2.00709\text{ s}$, toh $dT=T\cdot (-0.002548)=-0.005114\text{ s}$. Interpret karo: minus sign keh raha hai bada $g$ period ko shorten karta hai; $g$ mein $0.5\%$ error sirf $0.25\%$ error $T$ mein produce karta hai kyunki square root relative error ko half kar deta hai.

Second derivative: $f^{\prime \prime }(x)=-\frac{1}{4}{x}^{-3/2}$. Toh $|f^{\prime \prime }(x)|=\frac{1}{4}{x}^{-3/2}$, jo $x$ ki decreasing function hai (bada $x$ ⇒ chhota $x^{-3/2}$). Isliye interval $[4,4.1]$ par uska maximum left endpoint $x=4$ par hai: $|f^{\prime \prime }(4)|=\frac{1}{4}\cdot 4^{-3/2}$. Ab $4^{-3/2}=\frac{1}{4^{3/2}}=\frac{1}{8}$, toh $|f^{\prime \prime }(4)|=\frac{1}{4}\cdot \frac{1}{8}=\frac{1}{32}=0.03125$. Step $0.1$ par bound: $\frac{1}{2}(0.03125)(0.1{)}^2=\frac{1}{2}(0.03125)(0.01)=0.00015625$. Yeh $1.56\times 10^{-4}$ hai — $10^{-5}$ se bada, toh guarantee fail hoti hai (chahe actual error $1.5\times 10^{-4}$ bound ke kaafi paas raha). Required step: solve karo $\frac{1}{2}(0.03125)h^2\le 10^{-5}$: $h^2\le \frac{2\times 10^{-5}}{0.03125}=6.4\times 10^{-4}$, toh $h\le 0.0253$. Conclusion: $10^{-5}$ guarantee karne ke liye tumhe $x$ ko $4$ ke lagbhag $0.0253$ ke andar rakhna hoga (yaani $\sqrt{4.025}$ ya usse paas) linear model se — ya phir, step rakho aur quadratic Taylor series term add karo.

$L$ banao $a=100$ par: $f(100)=10$; $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$, $f^{\prime }(100)=\frac{1}{20}=0.05$. $\boxed{\sqrt{x}\approx 10+0.05(x-100)}$ Estimates:

• $\sqrt{102}\approx 10+0.05(2)=10.1$. (True $10.09950$; over-estimate.)
• $\sqrt{99}\approx 10+0.05(-1)=9.95$. (True $9.94987$; over-estimate.) Error ka sign: $f^{\prime \prime }(x)=-\frac{1}{4}{x}^{-3/2}<0$ ⇒ concave down ⇒ tangent curve ke upar dono sides par ⇒ dono estimates over-estimates hain. Upar ke tiny positive gaps se match karta hai.