Exercises — Gradient as direction of steepest ascent

KYA: hume sirf do partial derivatives chahiye, phir point plug in karo. Partials kyun: gradient by definition yeh list hai ki $f$ kitni tezi se change hoti hai agar ek variable ko nudge karo aur doosre ko freeze karo. $f_x=\frac{\partial }{\partial x}(3x^2-5y)=6x,f_y=\frac{\partial }{\partial y}(3x^2-5y)=-5.$ $(2,1)$ plug in karo: $\nabla f(2,1)=(6\cdot 2,-5)===(12,-5)==$. Notice karo ki $f_y$ mein koi $x$ ya $y$ nahi bacha — yeh theek hai, ek partial constant ho sakta hai.

KYA: identify karo ki kaun sa candidate length $1$ ka hai. Yeh matter kyun karta hai: steepest ascent ki direction gradient hai apni length se divided — ek unit arrow. Raw gradient $(2,4)$ ki length $\sqrt{4+16}=\sqrt{20}\ne 1$ hai, toh yeh ek direction aur ek rate dono ek saath hai, pure direction nahi. $\frac{1}{\sqrt{5}}(1,2)$ check karo: length $=\frac{1}{\sqrt{5}}\sqrt{1+4}=\frac{\sqrt{5}}{\sqrt{5}}=1$. ✓ Answer: $==\frac{1}{\sqrt{5}}(1,2)==$.

True. Ek level curve un points ka set hai jahan $f$ ka wahi height rehta hai (dekho Level Curves and Contour Maps). Agar tum iske saath-saath chalte waqt height change nahi hoti, toh change ki rate $D_{\mathbf{u}}f$ exactly $0$ hai.

(a) Partials. Doosre variable ko frozen constant ki tarah treat karo. $f_x=2xy,f_y=x^2\Rightarrow \nabla f(2,3)=(2\cdot 2\cdot 3,2^2)=(12,4).$ (b) Normalise karo. Length $\Vert \nabla f\Vert =\sqrt{12^2+4^2}=\sqrt{144+16}=\sqrt{160}=4\sqrt{10}$. ${\mathbf{u}}^{\ast }=\frac{(12,4)}{4\sqrt{10}}=\frac{(3,1)}{\sqrt{10}}.$ (c) Max rate $=\Vert \nabla f\Vert =4\sqrt{10}\approx 12.649$. Kyun: $\nabla f\cdot \mathbf{u}$ ki sabse badi possible value $\Vert \nabla f\Vert \cos 0=\Vert \nabla f\Vert$ hai.

Step 1 — direction vector = tip minus tail: $(4-1,6-2)=(3,4)$. Step 2 — normalise karo (warna rate galat length se scale ho jaayegi): $\Vert (3,4)\Vert =5$, toh $\mathbf{u}=(3,4)/5$. Step 3 — gradient $\nabla f(1,2)=(2,4)$ ke saath dot karo: $D_{\mathbf{u}}f=(2,4)\cdot \left(\frac{3}{5},\frac{4}{5}\right)=\frac{6+16}{5}=\frac{22}{5}===4.4==.$ Sanity check: $\Vert \nabla f\Vert =2\sqrt{5}\approx 4.472>4.4$, toh hum maximum se neeche hain — sahi hai, gradient direction ko koi nahi haraata.

$e^x$ kyun rehta hai aur $\cos y$ kyun $-\sin y$ ban jaata hai: partial $\partial /\partial x$ $y$ ko freeze karta hai, toh $\cos y$ ek constant multiplier hai; partial $\partial /\partial y$ $x$ ko freeze karta hai, toh $e^x$ ek constant multiplier hai. $f_x=e^x\cos y,f_y=-e^x\sin y.$ $(0,0)$ par: $e^0=1$, $\cos 0=1$, $\sin 0=0$, toh $\nabla f=(1\cdot 1,-1\cdot 0)=(1,0)$. Max rate $=\Vert (1,0)\Vert ===1==$.

Step 1 — gradient: $f_x=y,f_y=x$, toh $\nabla f(2,3)=(3,2)$. Step 2 — curve ki tangent. $xy=6$ ko implicitly differentiate karo ($y$ ko $x$ ka function maano): $y+x{y}^{\prime }=0\Rightarrow y^{\prime }=-\frac{y}{x}=-\frac{3}{2}$. $-\frac{3}{2}$ slope ka matlab tangent direction $\propto (2,-3)$ hai (run $2$, rise $-3$). Step 3 — perpendicularity ke liye dot product test (do vectors $\perp$ tab hote hain jab unka dot product $0$ ho, from Dot Product and Angle): $\nabla f\cdot (2,-3)=(3)(2)+(2)(-3)=6-6===0==.✓$ Toh gradient level curve ke perpendicular hai, exactly jaise theory predict karti hai.

Geometric form kyun: hume angle pata hai, $\mathbf{u}$ ke components nahi, toh directly $D_{\mathbf{u}}f=\Vert \nabla f\Vert \cos \theta$ use karo. $\Vert \nabla f\Vert =\sqrt{9+16}=5$, $\cos 60^{\circ }=\frac{1}{2}$. $D_{\mathbf{u}}f=5\times \frac{1}{2}===2.5==.$ Interpretation: tum maximum steepness ka aadha pa rahe ho kyunki tum seedhe uphill line se $60^{\circ }$ hatke chal rahe ho.

$f_x=2x,f_y=-2y$, toh $\nabla f(0,0)=(0,0)$ — zero vector. Iska matlab: $\Vert \nabla f\Vert =0$, toh maximum rate of increase $0$ hai. Har direction pehle order par flat hai — yeh ek critical point hai (yahan ek saddle). Koi unique steepest-ascent direction nahi hai, kyunki $\nabla f/\Vert \nabla f\Vert$ zero se divide karta hai aur undefined hai. Yeh degenerate case hai: steepest-ascent recipe tab hi ek direction deta hai jab $\nabla f\ne 0$ ho.

Gradient: $T_x=-2x,T_y=-4y$, toh $\nabla T(3,1)=(-6,-4)$. (a) Sabse tezi se garam hona gradient direction mein hai, normalised. $\Vert \nabla T\Vert =\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$. ${\mathbf{u}}^{\ast }=\frac{(-6,-4)}{2\sqrt{13}}=\frac{(-3,-2)}{\sqrt{13}}.$ Bug neeche-aur-baayein ki taraf jaata hai (origin ki taraf, jahan sabse zyada garam hai — samajh mein aata hai, $T$ $(0,0)$ par peak karta hai). (b) Rate $=\Vert \nabla T\Vert =2\sqrt{13}\approx ==7.211==$. (c) Due east $\mathbf{u}=(1,0)$ already unit vector hai: $D_{\mathbf{u}}T=(-6,-4)\cdot (1,0)===-6==$. East mein chalte hue yeh $6$ per unit distance se thanda hota hai.

Gradient: $f_x=-2x,f_y=-2y$, toh $\nabla f(1,1)=(-2,-2)$. (a) Steepest descent = $-\nabla f$, phir normalise karo: $-\nabla f=(2,2)$, length $2\sqrt{2}$, toh unit direction $=\frac{(2,2)}{2\sqrt{2}}=\frac{(1,1)}{\sqrt{2}}$. (b) Gradient descent apne update mein raw gradient (normalised nahi) use karta hai: ${\mathbf{x}}_{\text{new}}=(1,1)-0.1\cdot (-2,-2)=(1,1)+(0.2,0.2)===(1.2,1.2)==.$ Note: descent gradient ko subtract karta hai, aur kyunki $\nabla f$ uphill point karta hai, $-\eta \nabla f$ hume is dome ke peak ki taraf nudge karta hai — consistent hai $f$ ke origin ki taraf badhne se... ruko: yahan $f$ origin par peak karta hai, aur hum $(1.2,1.2)$ ki taraf door gaye. Iska matlab hum outer bowl climb kar rahe hain? Nahi — recheck karo: $-\eta \nabla f=+(0.2,0.2)$ hume bade $x,y$ par le jaata hai, jahan $f=9-x^2-y^2$ chota hai. Toh $f$ decrease hua: gradient descent ne sahi se $f$ reduce kiya. ✓

Andar chain rule kyun: $f$ inner function $g=x^2+y^2$ ka $\ln$ hai; har partial $\frac{1}{g}$ ko $g$ ke partial se multiply karta hai (dekho Multivariable Chain Rule). $f_x=\frac{2x}{x^2+y^2},f_y=\frac{2y}{x^2+y^2}.$ $(1,2)$ par: $x^2+y^2=5$, toh $\nabla f=\left(\frac{2}{5},\frac{4}{5}\right)$. Max rate $=\Vert \nabla f\Vert =\sqrt{(2/5{)}^2+(4/5{)}^2}=\sqrt{\frac{4+16}{25}}=\sqrt{\frac{20}{25}}=\frac{2}{\sqrt{5}}\approx ==0.894==$.

Steepest ascent exactly wahan undefined hai jahan $\nabla f=0$ ho (zero vector ko normalise nahi kar sakte, per L3·Q3). $f_x=2x+y=0,f_y=x+2y=0.$ Solve karo: pehle se, $y=-2x$; doosre mein substitute karo: $x+2(-2x)=x-4x=-3x=0\Rightarrow x=0$, hence $y=0$. Sirf ek point: $==(0,0)==$. Kyunki $f=x^2+xy+y^2=(x+\frac{y}{2}{)}^2+\frac{3}{4}{y}^2\ge 0$ aur equality sirf origin par hai, yeh ek minimum hai. Toh exactly ek point hai jahan koi steepest-ascent direction nahi hai, aur woh bowl ka bottom hai.

Yeh parent ka key claim kyun reprove karta hai — lekin dot-product inequality ki jagah single-variable optimisation se. $D(\varphi )=\nabla f\cdot (\cos \varphi ,\sin \varphi )=2\cos \varphi +0\cdot \sin \varphi =2\cos \varphi$. Differentiate karo: $D^{\prime }(\varphi )=-2\sin \varphi$. $=0$ set karo $\Rightarrow \sin \varphi =0\Rightarrow \varphi =0$ ya $\pi$. Second derivative: $D^{\prime \prime }(\varphi )=-2\cos \varphi$. $\varphi =0$ par: $D^{\prime \prime }(0)=-2<0$ → maximum. $\varphi =\pi$ par: $D^{\prime \prime }(\pi )=+2>0$ → minimum (yeh steepest descent hai). Maximum value $D(0)=2\cos 0===2===\Vert \nabla f\Vert =\Vert (2,0)\Vert$. ✓ Gradient ki magnitude hi peak rate hai.

(a) $D_{\mathbf{u}}f=(a,b)\cdot \frac{1}{\sqrt{2}}(1,-1)=\frac{a-b}{\sqrt{2}}.$ (b) "Level curve ke saath" ka matlab height change nahi ho rahi: $D_{\mathbf{u}}f=0\Rightarrow \frac{a-b}{\sqrt{2}}=0\Rightarrow ==a=b==$. Geometric read: jab $a=b$ ho toh gradient $(a,a)$ $(1,1)$ ke saath point karta hai, jo exactly $\mathbf{u}\propto (1,-1)$ ke perpendicular hai — toh walker constant height par hill ko skirt kar raha hai.