Gradient as direction of steepest ascent — Hinglish notes

4.4.10 · Maths › Multivariable Calculus

Hum pooch kya rahe hain?

Hum kya chahte hain: saari possible directions $u$ mein se, kaun si direction mein $f$ sabse tezi se grow karta hai, aur kitni tezi se?

KAISE: directional derivative ko scratch se derive karein

Yeh step kyun? Hum har direction ke liye limit nahi lena chahte. Multivariable chain rule use karo taaki $D_{u} f$ ko un cheezoon se express kiya ja sake jo hum pehle se jaante hain (partials).

Ek single-variable function define karo $x$ se $u$ direction mein jaane wali line ke saath: $g (h) = f (x + h u), u = (u_{1}, u_{2})$

Phir definition se $D_{u} f = g^{'} (0)$ .

Chain rule apply karo ( $x = x_{0} + h u_{1}$ , $y = y_{0} + h u_{2}$ ):

= f_x\,u_1 + f_y\,u_2$$ **Yeh step kyun?** $\frac{dx}{dh}=u_1$ aur $\frac{dy}{dh}=u_2$ kyunki $x,y$ linearly $h$ ke saath move karte hain. $h=0$ rakho: $$\boxed{D_{\mathbf u}f = f_x u_1 + f_y u_2 = \nabla f \cdot \mathbf u}$$ > [!formula] Key result > $$D_{\mathbf u}f = \nabla f \cdot \mathbf u \quad (\|\mathbf u\|=1)$$ > Directional derivative bas gradient aur direction ka **dot product** hai. --- ## KAISE: ab *sabse steep* direction dhundho Hum **maximise** karna chahte hain $D_{\mathbf u}f = \nabla f \cdot \mathbf u$ ko saare unit $\mathbf u$ par. Dot product ka geometric form use karo: $$\nabla f \cdot \mathbf u = \|\nabla f\|\,\|\mathbf u\|\cos\theta = \|\nabla f\|\cos\theta$$ **Yeh step kyun?** $\|\mathbf u\|=1$ hai, isliye sirf $\nabla f$ aur $\mathbf u$ ke beech ka angle $\theta$ matter karta hai. Kyunki $\cos\theta \in [-1,1]$: | $\theta$ | $\cos\theta$ | $D_{\mathbf u}f$ | matlab | |---|---|---|---| | $0$ ($\nabla f$ ke saath) | $+1$ | $+\|\nabla f\|$ | **steepest ascent** ↑ | | $\pi$ (opposite) | $-1$ | $-\|\nabla f\|$ | steepest descent ↓ | | $\pi/2$ (perpendicular) | $0$ | $0$ | level curve ke saath (flat) | > [!formula] Teen badi baatein > - Steepest ascent direction: $\displaystyle \mathbf u^* = \frac{\nabla f}{\|\nabla f\|}$ > - Maximum rate of increase: $\displaystyle \max_{\mathbf u} D_{\mathbf u}f = \|\nabla f\|$ > - $\nabla f$ level curves $f=\text{const}$ ke **perpendicular** hai (kyunki level curve ke saath move karne par $D_{\mathbf u}f=0 \Rightarrow \cos\theta=0$). --- ## Worked examples > [!example] Example 1 — Steepest direction dhundho > Maano $f(x,y)=x^2+y^2$ point $(1,2)$ par. > > **Step 1.** $f_x = 2x,\ f_y=2y$ → *Kyun?* partials lo. > **Step 2.** $\nabla f(1,2) = (2,4)$ → *Kyun?* point plug in karo. > **Step 3.** Steepest ascent direction $= \dfrac{(2,4)}{\sqrt{20}} = \dfrac{1}{\sqrt5}(1,2)$ → *Kyun?* unit length ke liye normalise karo. > **Step 4.** Max rate $= \|\nabla f\| = \sqrt{4+16}=\sqrt{20}=2\sqrt5$ → *Kyun?* maximum directional derivative gradient ki magnitude ke barabar hoti hai. > [!example] Example 2 — Diye gaye direction mein directional derivative > Wahi $f$, $(1,2)$ par, direction $(4,6)$ ki taraf. > > **Step 1.** Direction vector $(4-1,\,6-2)=(3,4)$. *Kyun?* tip minus tail. > **Step 2.** Unit vector $\mathbf u = (3,4)/5$. *Kyun?* $\|\mathbf u\|$ 1 hona chahiye, warna rate galat scale ho jaata hai. > **Step 3.** $D_{\mathbf u}f = \nabla f\cdot\mathbf u = (2,4)\cdot(3/5,4/5)= \frac{6+16}{5}=\frac{22}{5}=4.4.$ *Kyun?* boxed formula apply karo. > Note karo $4.4 < 2\sqrt5\approx4.47$ — consistent hai: gradient direction ko koi beat nahi kar sakta. > [!example] Example 3 — Level curve ke perpendicular > $f(x,y)=xy$, point $(2,3)$. Usse hone wali level curve: $xy=6$. > > **Step 1.** $\nabla f=(y,x)=(3,2)$. *Kyun?* $xy$ ke partials. > **Step 2.** Tangent to $xy=6$: implicit diff $y + x\,y'=0 \Rightarrow y' = -3/2$, isliye tangent direction $\propto(2,-3)$. *Kyun?* slope $dy/dx$. > **Step 3.** Check karo $\nabla f\cdot(2,-3)=(3)(2)+(2)(-3)=0$. ✓ *Kyun?* gradient ⟂ level curve, bilkul jaisa predict kiya tha. --- ## Common mistakes > [!mistake] "Main raw direction vector use karoonga, unit vector nahi." > **Kyun sahi lagta hai:** $\nabla f\cdot\mathbf v$ ek directional derivative jaisa dikhta hai. > **Kyun galat hai:** agar $\|\mathbf v\|\ne 1$, to tum $\|\mathbf v\|$ times zyada paa lete ho — woh rate *$\mathbf v$ ki length ke per unit* hai, unit distance per nahi. > **Fix:** hamesha **normalise karo**: $\mathbf u = \mathbf v/\|\mathbf v\|$ dot karne se pehle. > [!mistake] "Gradient level curve ke *saath saath* point karta hai, jaise ek contour line." > **Kyun sahi lagta hai:** tum map par trails ke saath chalte ho, isliye gradients unhe follow karte honge. > **Kyun galat hai:** level curve ke saath $f$ nahi badalta, isliye *fastest-increase* arrow ko curve se **door** jaana chahiye — uske perpendicular. > **Fix:** yaad rakho $D_{\mathbf u}f=0$ contour ke saath $\Rightarrow \cos\theta=0 \Rightarrow \perp$. > [!mistake] "Steepest ascent aur steepest descent unrelated directions hain." > **Fix:** descent bas $-\nabla f$ hai (opposite arrow). Gradient descent (ML) literally $-\nabla f$ mein step karta hai. --- > [!recall]- Feynman: ek 12-saal ke bachche ko samjhao > Socho tum ghane kohere mein ek ऊबड़-खाबड़ pahaad par khade ho. Tum door nahi dekh sakte, lekin apne pairon ke neeche har direction mein dhaal feel kar sakte ho. **Gradient** ek aisi arrow jaisi hai jo jadui tarah se seedha uphill point karti hai — seedha uupar jaane ka sabse steep rasta. Poora ghumo aur tum sabse steep raste se **neeche** point kar rahe ho. Sideways chalo arrow ke right angle par aur zameen flat rehti hai (tum pahaad ke around same height par chal rahe ho). **Jitni lambi** arrow, wahan pahaad utna hi steep hai. > [!mnemonic] Yaad rakho > **"GUS"** — **G**radient = **U**phill, **S**teepest. Aur **"dot karne se pehle normalise karo."** > Magnitude $\|\nabla f\|$ = *kitna steep hai*; direction = *kaun sa rasta uupar hai*. --- ## Recall checkpoint > [!recall] Bina dekhe jawab de sakte ho? > 1. $D_{\mathbf u}f$ ko dot product ke roop mein likho. > 2. $D_{\mathbf u}f$ ki max value $\|\nabla f\|$ kyun hoti hai? > 3. $\nabla f$ level curve ke saath kaun sa angle banata hai? --- ### #flashcards/maths What is the directional derivative $D_{\mathbf u}f$ in terms of the gradient? ::: $D_{\mathbf u}f = \nabla f \cdot \mathbf u$ ek **unit** vector $\mathbf u$ ke liye. In which direction does $\nabla f$ point? ::: **Steepest ascent** ki direction mein ($f$ ki sabse tezi se increase). What is the maximum rate of increase of $f$ at a point? ::: $\|\nabla f\|$, gradient ki magnitude. What direction gives steepest descent? ::: $-\nabla f$ (gradient ke opposite). Why must $\mathbf u$ be a unit vector in $D_{\mathbf u}f=\nabla f\cdot\mathbf u$? ::: Taaki rate "per unit distance" ho; warna yeh $\|\mathbf u\|$ se scale ho jaata hai. What angle is the gradient to a level curve $f=\text{const}$? ::: $90°$ — yeh perpendicular hai, kyunki $D_{\mathbf u}f=0$ curve ke saath. Derive $D_{\mathbf u}f$: what trick is used? ::: Chain rule on $g(h)=f(\mathbf x+h\mathbf u)$, phir $g'(0)=f_xu_1+f_yu_2$. Why is $\nabla f\cdot\mathbf u$ maximised along $\nabla f$? ::: $=\|\nabla f\|\cos\theta$, maximum tab hota hai jab $\cos\theta=1$, yaani $\theta=0$. --- ## Connections - [[Directional Derivative]] - [[Partial Derivatives]] - [[Multivariable Chain Rule]] - [[Dot Product and Angle]] - [[Level Curves and Contour Maps]] - [[Gradient Descent (Machine Learning)]] - [[Tangent Plane and Linearization]] ## 🖼️ Concept Map ```mermaid flowchart TD F[Scalar field f] -->|partial derivatives| GRAD[Gradient grad f] U[Unit direction u] -->|step along| DD[Directional derivative] DD -->|defined as| LIM[Limit of difference quotient] GRAD -->|chain rule gives| KEY[D_u f = grad f dot u] U --> KEY KEY -->|geometric dot product| ANG[grad f times cos theta] ANG -->|cos theta = 1| ASC[Steepest ascent along grad f] ANG -->|cos theta = -1| DESC[Steepest descent] ANG -->|cos theta = 0| LEVEL[Along level curve, flat] ASC -->|max rate equals| MAG[Magnitude of grad f] LEVEL -->|implies| PERP[grad f perpendicular to level curves] ``` ## 🔬 Deep Dive > [!intuition] Aur gehraai mein jao — visual, zero se > Is topic ki step-by-step 3Blue1Brown-style breakdowns. - [[4.4.10 D1 Foundations|D1 · Foundations — har symbol zero se]] - [[4.4.10 D2 Visual Walkthrough|D2 · Visual walkthrough — derivation pictures mein]] - [[4.4.10 D3 Worked Examples|D3 · Worked examples — har scenario]] - [[4.4.10 D4 Exercises|D4 · Exercises — graded, full solutions]] - [[4.4.10 D5 Question Bank|D5 · Question bank — concept traps]]