Exercises — Directional derivative — definition, formula

Hum kya check karte hain: formula demand karta hai ek unit vector — bilkul $1$ length ka direction arrow. $\Vert \mathbf{v}\Vert =\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\ne 1.$ Toh $\mathbf{v}$ as-is valid nahi hai. Normalise karo (apni length se divide karo): $\mathbf{u}=\frac{\mathbf{v}}{\Vert \mathbf{v}\Vert }=\left(\frac{3}{5},\frac{4}{5}\right).$ Check: $\sqrt{(3/5{)}^2+(4/5{)}^2}=\sqrt{9/25+16/25}=\sqrt{25/25}=1.$ ✓

Gradient kya hai: partial derivatives ki list — ek slope per axis (dekho Gradient vector). $f_x=\frac{\partial }{\partial x}(x^{3}y-y^2)=3x^{2}y,f_y=\frac{\partial }{\partial y}(x^{3}y-y^2)=x^3-2y.$ Toh $\nabla f=(3x^{2}y,x^3-2y)$. $(1,-1)$ par: $\nabla f(1,-1)=(3(1)(-1),1-2(-1))=(-3,3).$

Jhooth. $D_{\mathbf{u}}f=\nabla f\cdot \mathbf{u}$ ek scalar hai (do vectors ka dot product ek single number hota hai — dekho Dot product and cosine of angle). Woh vector jo sabse zyada upar point karta hai woh $\nabla f$ khud hai, aur steepest slope woh number hai $\Vert \nabla f\Vert$.

G — Gradient. $f_x=2x+3y$, $f_y=3x+2y$, toh $\nabla f=(2x+3y,3x+2y)$. $\nabla f(1,1)=(2+3,3+2)=(5,5).$ U — Unit direction. Direction vector $=(4,5)-(1,1)=(3,4)$, length $\sqrt{3^2+4^2}=5$, toh $\mathbf{u}=(3/5,4/5)$. N — Number (dot product). $D_{\mathbf{u}}f=(5,5)\cdot \left(\frac{3}{5},\frac{4}{5}\right)=\frac{15+20}{5}=\frac{35}{5}=7.$

Angle ek unit vector free mein kyun deta hai: $(\cos \alpha ,\sin \alpha )$ ki length hamesha $1$ hoti hai, kyunki ${\cos }^2\alpha +{\sin }^2\alpha =1$. Toh $\mathbf{u}=(\cos 60^{\circ },\sin 60^{\circ })=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$. Gradient. $f_x=e^x\cos y$, $f_y=-e^x\sin y$. $(0,0)$ par: $e^0=1$, $\cos 0=1$, $\sin 0=0$: $\nabla f(0,0)=(1,0).$ Dot product. $D_{\mathbf{u}}f=(1,0)\cdot \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)=\frac{1}{2}.$ Yahan $y$-rate $0$ hai, toh sirf walk ka horizontal lean matter karta hai.

Normalise: $\Vert (1,2,2)\Vert =\sqrt{1+4+4}=\sqrt{9}=3$, toh $\mathbf{u}=\left(\frac{1}{3},\frac{2}{3},\frac{2}{3}\right)$. Gradient: $\nabla f=(yz,xz,xy)$. $(1,2,3)$ par: $\nabla f=(2\cdot 3,1\cdot 3,1\cdot 2)=(6,3,2)$. Dot product: $D_{\mathbf{u}}f=(6,3,2)\cdot \left(\frac{1}{3},\frac{2}{3},\frac{2}{3}\right)=\frac{6}{3}+\frac{6}{3}+\frac{4}{3}=2+2+\frac{4}{3}=\frac{16}{3}\approx \mathrm{5.33.}$

Gradient: $\nabla f=(2x,2y)=(6,8)$, with $\Vert \nabla f\Vert =\sqrt{6^2+8^2}=\sqrt{100}=10$. (a) Steepest ascent $\nabla f$ ke along hai: $\mathbf{u}=\frac{1}{10}(6,8)=(0.6,0.8)$, rate $=\Vert \nabla f\Vert =10$. (b) Steepest descent opposite hai: rate $=-\Vert \nabla f\Vert =-10$.

(c) Zero rate tab hota hai jab $\mathbf{u}\perp \nabla f$ — level curve ke along chalte hue, jahan $f$ constant hai. Hume $(6,8)$ ke perpendicular direction chahiye. Perpendicular systematically banane ka tarika ("$90^{\circ }$ turn" rule): do vectors tab perpendicular hote hain jab unka dot product $0$ ho. Agar $\nabla f=(p,q)$ hai, toh $(-q,p)$ try karo: $(p,q)\cdot (-q,p)=-pq+qp=0.✓$ Toh do components swap karo aur unme se ek ko negate karo — yeh hamesha vector ko right angle se turn kar deta hai. Isliye rule $(p,q)\mapsto (-q,p)$ kaam karta hai; yeh magic nahi, dot product zero hone se forced hai. $\nabla f=(6,8)$ ke saath: perpendicular $(-8,6)$, length $\sqrt{64+36}=10$, toh ${\mathbf{u}}_1=(-0.8,0.6).$ Doosra choice (edge case). Right-angle turn clockwise ya counter-clockwise dono taraf ja sakta hai, toh do opposite unit directions hain zero slope ke saath. Doosra sirf $-{\mathbf{u}}_1$ hai (equivalently rule $(p,q)\mapsto (q,-p)$): ${\mathbf{u}}_2=(0.8,-0.6).$ Dono check out hote hain: $(6,8)\cdot (-0.8,0.6)=-4.8+4.8=0$ aur $(6,8)\cdot (0.8,-0.6)=4.8-4.8=0.$ ✓ Geometrically ye dono $\mathbf{a}$ par level circle ki tangent directions hain — same line, opposite arrows.

Neeche figure ek saath saare char vectors dikhata hai: yellow gradient (ascent), red opposite (descent), aur do green tangents (zero slope) level circle par baithe hue.

Figure padhna. Blue circle level curve $x^2+y^2=25$ hai. Yellow arrow $\nabla f=(6,8)$ circle ke seedha bahar point karta hai — steepest ascent, rate $10$. Red arrow uska exact reverse hai — steepest descent, rate $-10$. Do green arrows circle ke along hain (tangent), yellow se $90^{\circ }$; kisi bhi taraf chalna tumhe same contour par rakhta hai, toh slope $0$ hai. Notice karo green yellow ke perpendicular hai — yeh "swap-and-negate" ka geometric meaning hai.

Gradient: $\nabla f=(2x,-2y)=(4,-2)$. $\mathbf{u}$ ke along: $(4,-2)\cdot \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)=\frac{4-2}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\approx 1.414>0$ — surface is taraf rise karta hai. $\mathbf{w}$ ke along: $(4,-2)\cdot \left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)=\frac{4+2}{\sqrt{2}}=\frac{6}{\sqrt{2}}=3\sqrt{2}\approx 4.243>0$ — zyada tezi se rise karta hai. Interpretation: dono positive hain, lekin $\mathbf{w}$ $\nabla f=(4,-2)$ (jo down-and-right point karta hai) ki taraf zyada lean karta hai, toh yeh zyada chadhai karta hai. Sign batata hai upar/neeche; size batata hai kitna steep hai.

Cosine kyun aata hai: dot product satisfy karta hai $\nabla f\cdot \mathbf{u}=\Vert \nabla f\Vert \Vert \mathbf{u}\Vert \cos \theta$, aur $\Vert \mathbf{u}\Vert =1$ (dekho Dot product and cosine of angle). Toh $D_{\mathbf{u}}f=\Vert \nabla f\Vert \cos \theta$ — sirf angle se tune hone wala knob. Yahan $\Vert \nabla f\Vert =\sqrt{3^2+4^2}=5$, toh $\cos \theta =\frac{D_{\mathbf{u}}f}{\Vert \nabla f\Vert }=\frac{2.5}{5}=0.5\Rightarrow \theta =60^{\circ }.$

Gradient: $\nabla f=(2x+y,x)=(4,1)$, toh $\Vert \nabla f\Vert =\sqrt{16+1}=\sqrt{17}$. Steepest ka aadha: hume chahiye $D_{\mathbf{u}}f=\frac{1}{2}\Vert \nabla f\Vert$. Kyunki $D_{\mathbf{u}}f=\Vert \nabla f\Vert \cos \theta$: $\Vert \nabla f\Vert \cos \theta =\frac{1}{2}\Vert \nabla f\Vert \Rightarrow \cos \theta =\frac{1}{2}\Rightarrow \theta =60^{\circ }.$ Gradient se $60^{\circ }$ par ek heading banana (rotation kyun kaam karta hai). Plane mein koi bhi unit vector do perpendicular unit vectors ka mix likha ja sakta hai: $\hat{\mathbf{n}}$ gradient ke along point karta hai, aur ${\hat{\mathbf{n}}}^{\perp }$ usse $90^{\circ }$ cross karta hai. Agar hum lete hain $\mathbf{u}=\cos \theta \hat{\mathbf{n}}+\sin \theta {\hat{\mathbf{n}}}^{\perp },$ toh $\mathbf{u}$ automatically ek unit vector hai (kyunki $\hat{\mathbf{n}}\perp {\hat{\mathbf{n}}}^{\perp }$ se $\Vert \mathbf{u}{\Vert }^2={\cos }^2\theta +{\sin }^2\theta =1$ milta hai), aur yeh $\hat{\mathbf{n}}$ se $\theta$ angle banata hai kyunki $\mathbf{u}\cdot \hat{\mathbf{n}}=\cos \theta$. Yeh exactly $\theta$ se rotation hai — isi tarah rotation matrix ek perpendicular axis at a time banaya jata hai. Yahan $\hat{\mathbf{n}}=\frac{1}{\sqrt{17}}(4,1)$, aur Problem 3.1 ke swap-and-negate rule se, ${\hat{\mathbf{n}}}^{\perp }=\frac{1}{\sqrt{17}}(-1,4)$. $\theta =60^{\circ }$ ke saath: $\mathbf{u}=\frac{1}{2}\hat{\mathbf{n}}+\frac{\sqrt{3}}{2}{\hat{\mathbf{n}}}^{\perp }.$ Numerically $\hat{\mathbf{n}}\approx (0.970,0.243)$, ${\hat{\mathbf{n}}}^{\perp }\approx (-0.243,0.970)$, toh $\mathbf{u}\approx 0.5(0.970,0.243)+0.866(-0.243,0.970)\approx (0.274,0.962).$ Slope check karo: $(4,1)\cdot (0.274,0.962)=1.096+0.962=2.058\approx \frac{\sqrt{17}}{2}=\mathrm{2.06.}$ ✓ ($-\sin \theta$ choose karne se doosri valid heading milti hai — doosri taraf rotate karke.)

Key link: $(3,4)$ se guzarne wali level curve circle $x^2+y^2=25$ hai. Gradient $\nabla f=(6,8)$ us circle ke perpendicular hai. Isliye ek tangent direction $(6,8)$ ke perpendicular hai. Perp unit vector (swap-and-negate, phir normalise): $(-8,6)/10=(-0.8,0.6)$. $D_{\mathbf{u}}f=(6,8)\cdot (-0.8,0.6)=-4.8+4.8=0.$ Zero kyun mantaaq hai: level curve ke along $f$ constant hai, toh uska rate of change $0$ hai — chahe tum dono tangent directions mein se koi bhi chuno.

Idea: kisi axis ke along directional derivative us partial derivative ke barabar hoti hai (gradient ko $(1,0)$ se dot karne par $f_x$ milta hai). $D_{(1,0)}f=f_x=2x=10\Rightarrow x=5,D_{(0,1)}f=f_y=2y=6\Rightarrow y=3.$ Toh $\mathbf{a}=(5,3)$, aur $\nabla f=(10,6)$, se $\Vert \nabla f\Vert =\sqrt{10^2+6^2}=\sqrt{136}=2\sqrt{34}\approx \mathrm{11.66.}$

$\nabla f=(2x,2y)=(0,0)$ origin par. Phir kisi bhi unit $\mathbf{u}$ ke liye: $D_{\mathbf{u}}f=(0,0)\cdot (u_1,u_2)=0.$ Geometry: origin bowl ka bottom hai (ek minimum). Har direction momentarily flat hai — surface first order mein har direction mein level hai. Yeh degenerate case hai jahan "steepest ascent" ki koi preferred direction nahi hai kyunki instantaneously koi ascent hai hi nahi.

Direct limit. Unit $\mathbf{u}=(a,b)$ lo, toh $a^2+b^2=1$. Ray $(ha,hb)$ ke along: $\frac{f(ha,hb)-f(0,0)}{h}=\frac{1}{h}\cdot \frac{(ha{)}^2(hb)}{(ha{)}^2+(hb{)}^2}=\frac{1}{h}\cdot \frac{h^3{a}^{2}b}{h^2(a^2+b^2)}=\frac{a^{2}b}{a^2+b^2}=a^{2}b,$ $a^2+b^2=1$ use karke. Toh $D_{\mathbf{u}}f(0,0)=a^{2}b$ — har direction ke liye exist karta hai. Partials. $\mathbf{u}=(1,0)$ rakho: $f_x(0,0)=1^2\cdot 0=0$. $\mathbf{u}=(0,1)$ rakho: $f_y(0,0)=0^2\cdot 1=0$. Toh $\nabla f(0,0)=(0,0)$, aur formula predict karta $\nabla f\cdot \mathbf{u}=(0,0)\cdot (a,b)=0\text{for all directions.}$ Lekin true value $a^{2}b$ hai, jo e.g. $\mathbf{u}=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ ke liye nonzero hai: wahan $a^{2}b=\frac{1}{2}\cdot \frac{1}{\sqrt{2}}=\frac{1}{2\sqrt{2}}\approx 0.354\ne 0$. Moral: dot-product shortcut require karta hai ki $f$ us point par differentiable ho (dekho Tangent planes and differentiability). Yahan $f$ nahi hai, toh formula fail karta hai chahe saari directional derivatives exist karti hain.

Tool: dot product ka cosine form (dekho Dot product and cosine of angle): $D_{\mathbf{u}}f=\nabla f\cdot \mathbf{u}=\Vert \nabla f\Vert \Vert \mathbf{u}\Vert \cos \theta =\Vert \nabla f\Vert \cos \theta ,$ kyunki $\Vert \mathbf{u}\Vert =1$. Yahan $\theta$ $\mathbf{u}$ aur $\nabla f$ ke beech ka angle hai. Yeh settle kyun karta hai: $\Vert \nabla f\Vert$ ek fixed positive number hai, toh $D_{\mathbf{u}}f$ tab sabse bada hoga jab $\cos \theta$ sabse bada hoga. Aur $\cos \theta \le 1$ equality ke saath iff $\theta =0$, yani $\mathbf{u}$ $\nabla f$ ke same direction mein point karta hai. Woh direction $\mathbf{u}=\nabla f/\Vert \nabla f\Vert$ hai, aur phir $D_{\mathbf{u}}f=\Vert \nabla f\Vert \cdot 1=\Vert \nabla f\Vert .■$ Similarly minimum $\theta =\pi$ par hai, $-\Vert \nabla f\Vert$ deta hai (steepest descent).