Vector fields — definition, visualization: Worked examples

4.4.23 · D3 · Maths › Multivariable Calculus › Vector fields — definition, visualization

Yeh page Vector fields — definition, visualization ke liye "no gaps" drill hai. Hum har tarah ki situation se guzrenge jo ek vector field throw kar sakta hai: har quadrant, origin (jahan cheezein blow up ho sakti hain), constant fields, swirls, gradient fields, ek inverse-square law, aur ek word problem. Har worked example ko us cell se tag kiya gaya hai jo scenario matrix mein fill karta hai.

Pehli line se pehle, plain words mein chaar chhote reminders taaki koi symbol bina reason ke na aaye:

$a^{2} + b^{2}$ kyun aur kuch nahi? Kyunki arrow ke do legs $a$ (rightward) aur $b$ (upward) right angles par hain, toh straight-line length hypotenuse hai — Pythagorean theorem woh ek hi tool hai jo do perpendicular legs ko ek length mein badalta hai.

The scenario matrix

Har case class jo yeh topic present kar sakta hai, aur kaunsa example isse cover karta hai:

Cell	Scenario class	Covered by
A	Constant / degenerate field (arrow kabhi nahi badalti)	Ex 1
B	Radial field sabhī chaar quadrants mein (signs of $x, y$ )	Ex 2
B $^{'}$	On-axis edge cases ( $x = 0$ ya $y = 0$ , ek component vanish ho jaata hai)	Ex 2b
C	Field at the origin (zero / degenerate input)	Ex 3
D	Rotational field + perpendicularity test	Ex 4
E	Gradient field from a scalar (partial derivatives use karta hai)	Ex 5
F	Inverse-square field ki limiting behaviour ( $r \to 0$ , $r \to \infty$ )	Ex 6
G	Direction vs magnitude split (unit vector + length alag alag)	Ex 7
H	Real-world word problem (river current)	Ex 8
I	Exam-style twist (kya ek mystery field ek gradient field hai?)	Ex 9

Goal: Ex 9 ke baad tumne har sign, har quadrant, axes, origin, dono limits, aur "special" field ke do flavours dekh liye hain.

Ex 1 — Constant field (Cell A)

Worked example Statement

$F (x, y) = (3, 0)$ . $(0, 0)$ , $(5, - 2)$ , aur $(- 1, 7)$ par arrow dhundho. Poore field ko describe karo.

Forecast: padhne se pehle guess karo — kya arrows point se point par change hote hain?

Components padho. $P (x, y) = 3$ , $Q (x, y) = 0$ . Yeh step kyun? Rule sirf $P$ aur $Q$ ke through $x, y$ par depend karta hai. Agar na $P$ na $Q$ mein $x$ ya $y$ hai, toh output change nahi ho sakta.
Har point par evaluate karo. $F (0, 0) = (3, 0)$ , $F (5, - 2) = (3, 0)$ , $F (- 1, 7) = (3, 0)$ . Yeh step kyun? Alag points substitute karna kuch nahi badalata kyunki $3$ aur $0$ mein koi $x$ ya $y$ nahi hai.
Describe karo. Har arrow seedha right point karta hai, length $3^{2} + 0^{2} = 3$ . Ek uniform "steady east wind." Yeh step kyun? Direction $= (3, 0)$ normalised $= (1, 0)$ ; length $3$ har jagah.

Verify: length $= 9 + 0 = 3$ ✅. Yeh degenerate case hai — ek single vector har jagah clone hua (parent note ka "steel-man" trap).

Ex 2 — Radial field, sabhī chaar quadrants (Cell B)

Figure — Vector fields — definition, visualization

Worked example Statement

$F (x, y) = (x, y)$ . Har quadrant mein arrow dhundho aur describe karo: $(2, 1)$ , $(- 2, 1)$ , $(- 2, - 1)$ , $(2, - 1)$ par.

Forecast: in mein se har jagah origin ke relative arrow kis taraf point karta hai?

Quadrant I — $(2, 1)$ : arrow $(2, 1)$ . Dono parts positive → up-and-right point karta hai, origin se door. Yeh step kyun? Output position vector ke barabar hai, toh yeh origin se point ki taraf point karta hai, yaani outward.
Quadrant II — $(- 2, 1)$ : arrow $(- 2, 1)$ . Left aur up → phir bhi outward (up-left). Yeh step kyun? $x < 0$ se $P < 0$ (leftward), $y > 0$ se $Q > 0$ (upward): arrow phir se centre se door jaata hai.
Quadrant III — $(- 2, - 1)$ : arrow $(- 2, - 1)$ . Down-left, outward.
Quadrant IV — $(2, - 1)$ : arrow $(2, - 1)$ . Down-right, outward. Yeh steps kyun? Har quadrant mein $(P, Q)$ ke signs $(x, y)$ ke signs copy karte hain, toh arrow hamesha radially outward point karta hai — "explosion / source" pattern. Figure dekho: sabhī chaar arrows centre dot se door fan karte hain.

Verify: $(2, 1)$ par length $4 + 1 = 5 \approx 2.236$ hai. General length $= x^{2} + y^{2} = r$ : arrows jitna bahar jaao utni lambi hoti jaati hain ✅.

Ex 2b — Axes par: edge cases (Cell B $^{'}$ )

Worked example Statement

Abhi bhi $F (x, y) = (x, y)$ ke saath, un points ko dekho jo axes par baithe hain (kisi open quadrant mein nahi): $x$ -axis par $(3, 0)$ , $(- 3, 0)$ , aur $y$ -axis par $(0, 4)$ , $(0, - 4)$ . Har component ka kya hota hai?

Forecast: $x$ -axis par, arrow ka kaunsa part zero ho jaata hai?

$x$ -axis par ( $y = 0$ ). $F (3, 0) = (3, 0)$ purely right point karta hai; $F (- 3, 0) = (- 3, 0)$ purely left point karta hai. Yeh step kyun? Jab $y = 0$ toh second component $Q = y = 0$ vanish ho jaata hai, toh arrow ka koi vertical part nahi hota — yeh axis ke along flat lie karta hai. Surviving part $P = x$ ka sign origin cross karne par flip hota hai (right for $x > 0$ , left for $x < 0$ ).
$y$ -axis par ( $x = 0$ ). $F (0, 4) = (0, 4)$ purely upar point karta hai; $F (0, - 4) = (0, - 4)$ purely neeche point karta hai. Yeh step kyun? Ab $P = x = 0$ vanish ho jaata hai, toh arrow ka koi horizontal part nahi hota — yeh seedha $y$ -axis ke along lie karta hai, origin par sign flip karta hai.
Lesson padho. Axes exactly woh jagah hain jahan ek component zero hai — quadrants ke beech ki boundary lines. Arrow abhi bhi radial (outward) hai, bas axis ke along point karta hai. Koi sign contradiction nahi: ek axis cross karna exactly woh jagah hai jahan ek component smoothly $0$ se guzar kar sign change karta hai. Yeh step kyun? Yeh quadrants ke beech ka gap fill karta hai: har point ya toh ek quadrant mein hai (Ex 2), axis par hai (yahan), ya origin par hai (Ex 3).

Verify: $F (3, 0) = (3, 0)$ , length $3$ ; $F (0, - 4) = (0, - 4)$ , length $4$ ; har case mein ek component exactly $0$ hai ✅.

Ex 3 — Origin par: degenerate input (Cell C)

Worked example Statement

Radial field $F (x, y) = (x, y)$ ke liye, exactly origin $(0, 0)$ par arrow kya hai? Kya iska direction defined hai?

Forecast: kya tum zero length ka arrow ek direction ke saath draw kar sakte ho?

Evaluate karo. $F (0, 0) = (0, 0)$ — zero vector. Yeh step kyun? Rule mein $x = 0, y = 0$ plug karo; dono components vanish ho jaate hain (yeh woh point hai jahan Ex 2b ke dono axes ke zero-cases milte hain).
Length. $∣ F (0, 0) ∣ = 0^{2} + 0^{2} = 0$ . Yeh step kyun? Zero-length arrow ek dot hai, spear nahi — iska koi direction nahi hota.
Direction yahan undefined hai. Unit direction $\frac{F}{∣ F ∣} = \frac{( 0 , 0 )}{0}$ zero se divide karta hai. Yeh step kyun? Hum zero vector ko normalise nahi kar sakte — $0$ se division forbidden hai. Toh origin ek critical / equilibrium point hai: kuch push nahi karta wahan.

Verify: length $0$ ✅. Physically: ek particle jo exactly source par rakha hai use koi push feel nahi hoti — yeh still baitta hai (unstable, lekin phir bhi). Yeh general lesson hai: direction claim karne se pehle hamesha check karo ki tumhara field special points par zero (ya undefined) toh nahi hai.

Ex 4 — Rotation field + perpendicularity (Cell D)

Worked example Statement

$F (x, y) = (- y, x)$ . $(1, 0)$ , $(0, 1)$ , $(- 1, 0)$ , $(0, - 1)$ par arrows dhundho aur prove karo ki field ek pure swirl hai.

Forecast: clockwise hai ya counter-clockwise?

Evaluate karo. $F (1, 0) = (0, 1)$ upar; $F (0, 1) = (- 1, 0)$ left; $F (- 1, 0) = (0, - 1)$ neeche; $F (0, - 1) = (1, 0)$ right. Yeh step kyun? Har output position vector ko $9 0^{\circ}$ counter-clockwise turn kiya hua hai. Up → left → down → right follow karo: yeh ek counter-clockwise circle trace karta hai (figure dekho).
Perpendicularity test — dot product. Upar ke definition callout se dot-product formula $u \cdot v = u_{1} v_{1} + u_{2} v_{2}$ use karte hue, $u = F = (- y, x)$ aur $v =$ radius $(x, y)$ lo: $F \cdot (x, y) = (- y) (x) + (x) (y) = - y x + x y = 0.$ Yeh tool kyun aur koi doosra nahi? Dot product exactly ek geometric question ka jawab deta hai: "kya yeh do arrows right angles par hain?" — yeh exactly tab zero hota hai jab woh perpendicular hoon. Kyunki $F \cdot (x, y) = 0$ har jagah hai, har arrow radius se $9 0^{\circ}$ par hai, yaani circle ki tangent. Koi outward part nahi matlab koi spreading nahi, pure rotation.
Length. $∣ F ∣ = (- y)^{2} + x^{2} = x^{2} + y^{2} = r$ . Yeh step kyun? Arrows distance $r$ ke saath lambi hoti hain, bilkul radial field ki tarah, lekin fan out karne ki jagah curl karti hain.

Verify: dot product $- y x + x y = 0$ ✅. $(1, 0)$ par length $= 0 + 1 = 1$ ✅. $(3, 4)$ par: $F = (- 4, 3)$ , $(3, 4)$ ke saath dot $(- 4) (3) + (3) (4) = - 12 + 12 = 0$ ✅, length $16 + 9 = 5 = r$ ✅.

Ex 5 — Scalar se Gradient field (Cell E)

Worked example Statement

Maano $f (x, y) = x^{2} + 3 y^{2}$ . Gradient field $F = \nabla f$ build karo aur $(1, 1)$ aur $(- 2, 0)$ par evaluate karo.

Forecast: arrows kis direction mein point karenge — $f$ par uphill ya downhill?

$\nabla$ kya hai? Gradient $\nabla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ do partial derivatives collect karta hai. Yeh tool kyun? Partial derivative $\partial f / \partial x$ iska jawab deta hai "agar main sirf $x$ nudge karun toh $f$ kitni tezi se badhta hai?" Dono nudge-rates ko ek arrow mein stack karna steepest increase ki single direction deta hai — Gradient and Directional Derivatives dekho.
Differentiate karo. $\frac{\partial}{\partial x} (x^{2} + 3 y^{2}) = 2 x$ ( $y$ ko frozen constant maano, toh $3 y^{2}$ ka zero $x$ -rate hai). $\frac{\partial}{\partial y} (x^{2} + 3 y^{2}) = 6 y$ . Yeh step kyun? "Partial" ka matlab hai doosre variable ko freeze karo; phir yeh ordinary one-variable differentiation hai.
Field assemble karo. $F (x, y) = (2 x, 6 y)$ .
Evaluate karo. $F (1, 1) = (2, 6)$ ; $F (- 2, 0) = (- 4, 0)$ . Yeh step kyun? Direct substitution. $(- 2, 0)$ par arrow left point karta hai kyunki $2 x = - 4 < 0$ ; field origin par minimum se door point karta hai — uphill, jaisa promise kiya tha.

Verify: $F (1, 1) = (2 \cdot 1, 6 \cdot 1) = (2, 6)$ ✅. Wahan length $= 4 + 36 = 40 \approx 6.325$ . Yeh $F$ ek conservative (gradient) field hai kyunki yeh ek scalar $f$ se aaya — Ex 9 ke liye isse bookmark karo. Conservative Fields and Potential Functions dekho.

Ex 6 — Inverse-square field: dono limits (Cell F)

Worked example Statement

2D inverse-square-style field $F (r) = \frac{r}{∣ r ∣ ^{3}}$ ke liye jahan $r = (x, y)$ aur $∣ r ∣ = r = x^{2} + y^{2}$ (upar define ki gayi origin se distance), $∣ F ∣$ dhundho aur describe karo kya hota hai jab $r \to 0$ aur jab $r \to \infty$ .

Forecast: centre ke paas push zyada strong hoti hai ya weak?

Magnitude. $∣ F ∣ = \frac{∣ r ∣}{∣ r ∣ ^{3}} = \frac{1}{∣ r ∣ ^{2}} = \frac{1}{r ^{2}}$ . Yeh step kyun? $r /∣ r ∣$ outward unit arrow hai (length exactly $1$ ); $∣ r ∣^{2}$ se divide karna length ko $1/ r^{2}$ tak shrink karta hai. Toh strength one-over-distance-squared se fall off hoti hai.
Limit $r \to \infty$ . $\frac{1}{r ^{2}} \to 0$ . Yeh tool kyun — ek limit? Ek limit iska jawab deta hai "yeh value kya approach karti hai jab $r$ kisi aisi jagah jaata hai jo main directly plug nahi kar sakta?" Door se, $1/ r^{2}$ $0$ approach karta hai: push kuch nahi tak fade ho jaati hai (figure mein shrinking outer arrows dekho).
Limit $r \to 0^{+}$ . $\frac{1}{r ^{2}} \to + \infty$ . Yeh step kyun? Centre ke close, ek tiny squared number se $1$ divide karna ek huge value deta hai — field origin par blow up karta hai. Toh $r = 0$ domain se excluded hai (wahan undefined), unlike Ex 3 ka radial field jo sirf zero tha.

Verify: $r = 2$ par, $∣ F ∣ = 1/4 = 0.25$ ✅; $r = 1$ par, $∣ F ∣ = 1$ ✅; $r = 1/2$ par, $∣ F ∣ = 1/ (1/4) = 4$ ✅ — centre ki taraf strong hoti jaati hai. Is field ki flux kyun special hai iske liye Flux and the Divergence Theorem dekho.

Ex 7 — Direction ko magnitude se alag karna (Cell G)

Worked example Statement

$F (x, y) = (x, y)$ ke liye point $(3, 4)$ par, arrow ko (unit direction) $\times$ (magnitude) ke roop mein likho.

Forecast: length kya hai, aur pure "which-way" arrow kya hai?

Magnitude. $∣ F (3, 4) ∣ = 3^{2} + 4^{2} = 25 = 5$ . Yeh step kyun? Legs $3$ aur $4$ par Pythagoras — classic $3 - 4 - 5$ triangle.
Unit direction. $\frac{F}{∣ F ∣} = \frac{( 3 , 4 )}{5} = (\frac{3}{5}, \frac{4}{5}) = (0.6, 0.8)$ . Yeh step kyun? Ek arrow ko uski apni length se divide karna ek length- $1$ arrow deta hai jo same taraf point karta hai — yeh pure direction hai, magnitude se stripped.
Reassemble karo. $(0.6, 0.8) \times 5 = (3, 4) = F$ ✅. Yeh step kyun? Split confirm karta hai: direction times length original arrow rebuild karta hai. Exactly aise hi field-plotters equal-length arrows draw karte hain (step 2 use karo) lekin step 1 se colour karte hain.

Verify: $0. 6^{2} + 0. 8^{2} = 0.36 + 0.64 = 1$ (genuinely ek unit vector) ✅; $5 \times 0.6 = 3$ , $5 \times 0.8 = 4$ ✅.

Ex 8 — Word problem: river current (Cell H)

Worked example Statement

Ek river ka surface current $v (x, y) = (0, 1 - x^{2})$ metres per second se model kiya gaya hai, jahan $x$ centre-line se distance hai ( $x = 0$ mid-river) aur $∣ x ∣ \leq 1$ banks hain. Centre par, $x = 0.5$ par, aur banks par current dhundho. Yeh fastest kahan hai?

Forecast: paani middle mein zyada tezi se move karta hai ya edge par?

Model padho. $P = 0$ (koi sideways drift nahi), $Q = 1 - x^{2}$ (downstream speed sirf centre se kitni door hai us par depend karti hai). Yeh step kyun? $P = 0$ ka matlab hai paani sirf seedha down-river flow karta hai; sideways part har jagah zero hai.
Centre $x = 0$ . $v = (0, 1 - 0) = (0, 1)$ → downstream $1 m/s$ .
Half-way $x = 0.5$ . $v = (0, 1 - 0.25) = (0, 0.75)$ → $0.75 m/s$ .
Banks $x = \pm 1$ . $v = (0, 1 - 1) = (0, 0)$ → zero current (paani edge par ruk jaata hai). Yeh steps kyun? $1 - x^{2}$ mein given $x$ values substitute karna. Pattern: speed centre par sabse zyada hai aur banks par zero ho jaati hai.
Fastest point. $1 - x^{2}$ tab sabse bada hota hai jab $x^{2}$ sabse chhota ho, yaani $x = 0$ . Yeh step kyun? Sabse chhota possible non-negative number ( $x^{2} = 0$ ) subtract karna sabse bada result deta hai.

Verify: $1 - 0^{2} = 1$ , $1 - 0. 5^{2} = 0.75$ , $1 - 1^{2} = 0$ ✅. Units: $Q$ m/s mein hai, ek velocity field se match karta hai — yeh genuinely ek Parametric Curves and Velocity style flow hai. Direction $(0, 1)$ ki length $1$ hai, toh centre par magnitude exactly $1$ m/s hai ✅.

Ex 9 — Exam twist: kya yeh field ek gradient field hai? (Cell I)

Worked example Statement

Ek field $F (x, y) = (2 x, 6 y)$ tumhare saamne drop kiya jaata hai (Ex 5 se, lekin pretend karo tum iska origin nahi jaante). Kya yeh ek gradient field hai, yaani kya koi scalar $f$ hai jisme $\nabla f = F$ ? Agar haan, toh $f$ explicitly dhundho.

Forecast: yes ya no guess karo, phir check karo.

Consistency test. $F = (P, Q)$ ke gradient hone ke liye, cross partials agree karne chahiye: $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ . Yeh tool kyun? Agar $P = \partial f / \partial x$ aur $Q = \partial f / \partial y$ , toh $\partial P / \partial y$ aur $\partial Q / \partial x$ dono same mixed second derivative $\partial^{2} f / \partial x \partial y$ ke barabar hain — toh unhe match karna chahiye. Yeh fast screening test hai (Conservative Fields and Potential Functions dekho).
Cross partials compute karo. $\frac{\partial}{\partial y} (2 x) = 0$ aur $\frac{\partial}{\partial x} (6 y) = 0$ . Yeh match karte hain ( $0 = 0$ ) → candidate passes, toh ab hum $f$ dhundhne jaate hain. Yeh step kyun? $2 x$ mein koi $y$ nahi, toh iska $y$ -rate $0$ hai; $6 y$ mein koi $x$ nahi, toh iska $x$ -rate $0$ hai.
$f$ build karna shuru karne ke liye $P$ ko $x$ ke respect mein integrate karo. Hum chahte hain $\partial f / \partial x = P = 2 x$ , toh woh derivative undo karo: $f (x, y) = \int 2 x d x = x^{2} + g (y) .$ Yeh step kyun? $2 x$ ko $x$ mein anti-differentiate karne se $x^{2}$ milta hai. Lekin "constant of integration" sirf $x$ ke respect mein constant hai — yeh abhi bhi $y$ par depend kar sakta hai. Toh hum ek unknown ==function of $y$ alone==, $g (y)$ , likhte hain, sirf ek number nahi. Yeh key subtlety hai: constant of integration poora function $g (y)$ hai, kyunki $\partial / \partial x$ kisi bhi pure- $y$ term ko annihilate karta hai.
Second component $Q$ use karke $g (y)$ fix karo. Hume $\partial f / \partial y = Q = 6 y$ bhi chahiye. Apne $f$ ko $y$ mein differentiate karo: $\frac{\partial}{\partial y} (x^{2} + g (y)) = 0 + g^{'} (y) = g^{'} (y) .$ Ise $6 y$ ke barabar set karo: $g^{'} (y) = 6 y$ . Yeh step kyun? $x^{2}$ term $y$ -derivative mein kuch contribute nahi karta, toh jo bhi bacha hai use saara $Q$ supply karna hai. Isse $g$ pin down hota hai.
$g (y)$ recover karne ke liye integrate karo, apne constant ke saath. $g (y) = \int 6 y d y = 3 y^{2} + C$ , jahan $C$ ek genuine numerical constant hai (ab koi variable nahi chup sakta — $g$ sirf $y$ par depend karta hai, aur hum $y$ -condition use kar chuke hain). Yeh step kyun? Yeh doosra integration true constant of integration $C$ introduce karta hai. Yeh ek plain number hai kyunki potential hamesha sirf ek additive constant tak determined hota hai — $f$ ko kisi bhi constant se shift karna $\nabla f$ ko unchanged rakhta hai.
Answer. $f (x, y) = x^{2} + 3 y^{2} + C$ . $C = 0$ choose karne par clean potential $f (x, y) = x^{2} + 3 y^{2}$ milta hai — exactly Ex 5 ka original scalar. Toh $F$ conservative hai. ✅

Verify: $\nabla (x^{2} + 3 y^{2} + C) = (2 x, 6 y) = F$ kisi bhi constant $C$ ke liye ✅. Cross-partial check $\partial_{y} (2 x) = 0 = \partial_{x} (6 y)$ ✅.

Recall Quick self-test (cloze)

Zero vector ki direction undefined hai kyunki tum zero se divide nahi kar sakte. Dot product $u \cdot v = u_{1} v_{1} + u_{2} v_{2}$ ka zero hona prove karta hai ki do arrows perpendicular hain. $x$ -axis par radial field $(x, y)$ ka second component == $y = 0$ == hota hai, toh arrow axis ke along flat lie karta hai. Jab $r \to 0$ , inverse-square magnitude $1/ r^{2} \to$ infinity (field blow up karta hai). Ek gradient candidate $(P, Q)$ screen pass karta hai jab $\partial P / \partial y =$ == $\partial Q / \partial x$ ==. Potential recover karne ke liye $P$ ko $x$ mein integrate karte waqt, "constant" actually ek ==function of $y$ == akela hota hai.

Which quadrant ek naive radial description ke liye fail karta hai?

Koi nahi — radial field

(x, y)

sabhī chaar quadrants mein outward point karta hai kyunki

(P, Q)

ke signs

(x, y)

ke signs copy karte hain.

Radial field ke liye axes par kya hota hai?

Ek component vanish ho jaata hai (

Q = 0

x

-axis par,

P = 0

y

-axis par), toh arrow us axis ke along seedha point karta hai, abhi bhi outward.

Origin radial field ke liye degenerate kyun hai?

F (0, 0) = (0, 0)

: zero length, toh direction undefined hai (normalise nahi kar sakte).

Centre par inverse-square ko radial se kaunsa limit alag karta hai?

Inverse-square

\to \infty

(origin par undefined); radial

\to 0

(defined, bas zero).

Potential recover karte waqt, integration ka constant

y

ka function kyun hota hai?

Kyunki

\partial / \partial x

sirf

y

par depend karne wale kisi bhi term ko kill karta hai, toh

x

mein anti-differentiate karna use dekh nahi sakta.

Connections

Vector fields — definition, visualization — parent topic jo yeh drill support karta hai.
Gradient and Directional Derivatives — Ex 5 ek gradient field build karta hai.
Divergence and Curl — radial vs rotational contrast (Ex 2 vs Ex 4).
Conservative Fields and Potential Functions — Ex 9 mein $f$ ki recovery.
Flux and the Divergence Theorem — Ex 6 ka inverse-square field.
Parametric Curves and Velocity — Ex 8 ka river velocity field.

Worked examples — Vector fields — definition, visualization

The scenario matrix

Ex 1 — Constant field (Cell A)

Ex 2 — Radial field, sabhī chaar quadrants (Cell B)

Ex 2b — Axes par: edge cases (Cell B $^{'}$ )

Ex 3 — Origin par: degenerate input (Cell C)

Ex 4 — Rotation field + perpendicularity (Cell D)

Ex 5 — Scalar se Gradient field (Cell E)

Ex 6 — Inverse-square field: dono limits (Cell F)

Ex 7 — Direction ko magnitude se alag karna (Cell G)

Ex 8 — Word problem: river current (Cell H)

Ex 9 — Exam twist: kya yeh field ek gradient field hai? (Cell I)

Connections

Concept Map