Exercises — Vector fields — definition, visualization

(a) $P=2,Q=2$ mein $x$ ya $y$ nahi hai → arrow kabhi nahi badalta → constant field. $(1,0)$ pe arrow $(2,2)$ hai: $45^{\circ }$ pe upar-daayein point karta hai, length $\sqrt{2^2+2^2}=2\sqrt{2}$. (b) Output position vector ke barabar hai → radial (outward) field. $(1,0)$ pe arrow $(1,0)$ hai: seedha daayein point karta hai, origin se door. (c) Output position vector ko $90^{\circ }$ ghuma ke mila hai → rotational field. $(1,0)$ pe arrow $(-0,1)=(0,1)$ hai: seedha upar point karta hai (unit circle ki tangent, counter-clockwise).

$\mathbf{F}(3,4)=(3,4)$, isliye $|\mathbf{F}|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$. Kyun: yahan arrow ki length position vector ki length hai, yaani origin se distance $r$, jo ki $5$ hai.

Har point ko $(-y,x)$ mein plug karo:

• $(1,0)\to (-0,1)=(0,1)$ — upar
• $(0,1)\to (-1,0)$ — baayen
• $(-1,0)\to (0,-1)$ — neeche
• $(0,-1)\to (1,0)$ — daayein

Upar → baayen → neeche → daayein padhna, jab tum circle ke around counter-clockwise jaate ho, exactly ek counter-clockwise swirl hai. Figure dekho.

Direction $=\frac{\mathbf{F}}{|\mathbf{F}|}$. L1.2 se, $|\mathbf{F}|=5$, isliye $\hat{\mathbf{F}}=\frac{(3,4)}{5}=\left(\frac{3}{5},\frac{4}{5}\right)=(0.6,0.8).$ Magnitude se kyun divide karein? Ek vector ko uski apni length se divide karna usse length $1$ pe scale kar deta hai jabki direction same rehti hai — check karo: $\sqrt{0.6^2+0.8^2}=\sqrt{0.36+0.64}=\sqrt{1}=1.$ ✅

Gradient $\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$ partial derivatives collect karta hai (dekho Gradient and Directional Derivatives).

• $\frac{\partial }{\partial x}(x^{2}y)=2xy$ ($y$ ko constant treat karo).
• $\frac{\partial }{\partial y}(x^{2}y)=x^2$ ($x$ ko constant treat karo).

Toh $\nabla f=(2xy,x^2)$. Yeh automatically ek conservative (gradient) field hai — dekho Conservative Fields and Potential Functions.

Position vector ke saath dot product lo: $\mathbf{F}\cdot (x,y)=(-y)(x)+(x)(y)=-xy+xy=0.$ Dot product kyun? Yeh woh tool hai jo jawab deta hai "kya yeh dono arrows perpendicular hain?" — iska value exactly $0$ hota hai jab woh $90^{\circ }$ pe milte hain. Yahan yeh sabhi $x,y$ ke liye $0$ hai. Geometric meaning: arrow hamesha origin se line (radius) ke right angles pe hota hai, yaani us point se guzarne wale circle ki tangent → pure rotation, koi outward/inward push nahi.

Zero arrow ke liye dono components ek saath zero chahiye: $x-1=0\text{ and }y+2=0\Rightarrow x=1,y=-2.$ Toh $\mathbf{F}=0$ sirf $(1,-2)$ pe hai. $\mathbf{F}=((x-1),(y+2))$ likhne pe — yeh exactly radial field $(u,v)$ hai jo shift hoke apna center $(1,-2)$ pe rakhta hai. Arrows $(1,-2)$ se baahir point karte hain, aur door jaate hue lamba hote hain. Zero point ek source hai (equilibrium jahan kuch push nahi karta).

Yahan $\mathbf{r}=(0,2)$ hai toh $|\mathbf{r}|=\sqrt{0^2+2^2}=2$. Field ki magnitude hai: $|\mathbf{F}|=\frac{|\mathbf{r}|}{|\mathbf{r}{|}^3}=\frac{1}{|\mathbf{r}{|}^2}=\frac{1}{2^2}=\frac{1}{4}=\mathrm{0.25.}$ $1/r^2$ kyun? Numerator $\mathbf{r}/|\mathbf{r}|$ ek unit arrow hai (length $1$); extra $1/|\mathbf{r}{|}^2$ strength hai, jo distance ke square ke saath kam hoti hai — yeh gravity/electrostatics ki pehchaan hai.

Dono archetypes ko unknown weights $a,b$ ke saath jodo: $a(x,y)+b(-y,x)=(ax-by,ay+bx).$ Components ko $(x-y,x+y)$ se match karo:

• $x$-part: $ax-by=x-y\Rightarrow a=1,b=1.$
• $y$-part: $ay+bx=y+x$ ✅ with $a=1,b=1.$

Toh $\mathbf{F}=(x,y)+(-y,x)$: ek outward push aur ek counter-clockwise swirl — arrows baahir ki taraf spiral karte hain. Yeh "radial + rotational" split Divergence and Curl (spreading vs. swirling) ka preview hai.

Hamen chahiye $\frac{\partial f}{\partial x}=2xy$. $x$ mein integrate karo ($y$ constant treat karo): $f=\int 2xydx=x^{2}y+g(y),$ jahan $g(y)$ ek unknown "constant of integration" hai jo $y$ pe depend kar sakta hai. Ab $\frac{\partial f}{\partial y}=x^2$ demand karo: $x^{2}y+g(y)$ differentiate karne pe $x^2+g^{\prime }(y)$ milta hai. $x^2$ ke barabar set karo: $x^2+g^{\prime }(y)=x^2\Rightarrow g^{\prime }(y)=0\Rightarrow g(y)=C.$ Toh $f(x,y)=x^{2}y+C$. Kyunki ek potential exist karta hai, $\mathbf{F}$ conservative hai — L2.3 se match karta hai jahan humne exactly yahi field $f=x^{2}y$ se banaya tha. Dekho Conservative Fields and Potential Functions.

Counter-clockwise swirl $(-y,x)$ se shuru karo; sign flip karo spin reverse karne ke liye: $(y,-x)$. Iski magnitude $\sqrt{y^2+x^2}=r$ already hai, jo exactly hum chahte hain. Toh lo: $\boxed{\mathbf{F}(x,y)=(y,-x)}.$ Tangent check (radius ke perpendicular): $\mathbf{F}\cdot (x,y)=(y)(x)+(-x)(y)=xy-xy=0$ ✅ → circles ke tangent. Clockwise check: $(1,0)$ pe, $\mathbf{F}=(0,-1)$ neeche point karta hai — circle ke around clockwise move karta hai ✅. Magnitude $=r$ check: $|\mathbf{F}|=\sqrt{y^2+(-x{)}^2}=\sqrt{x^2+y^2}=r$ ✅.

(i) $\mathbf{F}(2,3)=(2,-3)$. (ii) $|\mathbf{F}|=\sqrt{2^2+(-3{)}^2}=\sqrt{4+9}=\sqrt{13}\approx 3.606$. (iii) Zero ke liye $x=0$ aur $-y=0$ chahiye → sirf $(0,0)$. (Yeh ek saddle-type point hai: arrows $x$-axis ke along baahir flow karte hain, $y$-axis ke along andar.) (iv) $f$ ke liye dekho jisme $\partial f/\partial x=x$ ho: integrate karo → $f=\frac{1}{2}{x}^2+g(y)$. Phir $\partial f/\partial y=g^{\prime }(y)$ ko $-y$ ke barabar hona chahiye, toh $g(y)=-\frac{1}{2}{y}^2$. Isliye $f(x,y)=\frac{1}{2}{x}^2-\frac{1}{2}{y}^2+C,$ aur $\mathbf{F}$ conservative hai. Check karo: $\nabla f=(x,-y)$ ✅.

L3.3 se, $|\mathbf{F}|=1/r^2$.

• $r\to 0^{+}$: $1/r^2\to +\infty$ — field origin ke paas blow up karta hai. Origin ek singularity hai (point mass / charge wahan baitha hai; field $r=0$ pe undefined hai).
• $r\to \infty$: $1/r^2\to 0$ — field door jaake fade ho jaata hai, par actually kabhi zero nahi hota. Matlab: gravity/electric pull bahut paas mein enormous hoti hai aur distance ke square ke saath kam hoti hai — planet ki surface pe strong feel hoti hai aur deep space mein almost bilkul nahi.