Gradient perpendicular to level curves - surfaces: Exercises

4.4.11 · D4 · Maths › Multivariable Calculus › Gradient perpendicular to level curves - surfaces

Yeh page ek hi skill train karti hai: is fact ka use karna ki == $\nabla f$ level set $f = c$ ke perpendicular hota hai== — taaki tangents, normals, aur geometry compute kar sako. Har problem ka ek full worked solution ek [!recall]- callout ke andar fold karke rakha hai — problem padho, khud try karo, phir unfold karo. Prerequisites: the parent result, Multivariable Chain Rule, Directional Derivative, aur Tangent Plane to a Surface.

Shuru karne se pehle, do symbols jo hum baar baar use karenge:

Level 1 — Recognition

Tumhe ek function aur ek point diya gaya hai. Bas $\nabla f$ compute karo aur dekho iska kya matlab hai.

Exercise 1.1

$f (x, y) = x^{2} + 3 y^{2}$ ke liye, $(1, 2)$ par $\nabla f$ nikalo. Yeh us point se guzarne wali level curve ke relative kis direction mein point karta hai?

Recall Solution

$f_{x} = 2 x$ , $f_{y} = 6 y$ . Toh $\nabla f = (2 x, 6 y)$ , aur $(1, 2)$ par yeh $== (2, 12) ==$ hai. Level curve $f = 13$ ek ellipse hai. Gradient usse bahar ki taraf point karta hai, us direction mein jismein $f$ badhta hai (jab tum origin se door jaate ho toh chadhte ho). Yeh $(1, 2)$ par ellipse ki tangent ke perpendicular hai.

Exercise 1.2

$f (x, y, z) = x^{2} + y^{2} + z^{2}$ ke liye, $(1, 2, 3)$ par level surface $f = 14$ ke normal vector nikalo.

Recall Solution

$\nabla f = (2 x, 2 y, 2 z) === (2, 4, 6) ==$ at $(1, 2, 3)$ . Kyunki $\nabla f$ level surface (radius $14$ wala sphere) ke perpendicular hai, yeh vector hi ek normal hai. Note karo yeh $2 (1, 2, 3)$ ke barabar hai — sphere ke liye expected radially outward direction, bilkul sahi.

Exercise 1.3

$f (x, y) = x y$ ki level curve par $(4, 1)$ par kaun sa vector tangent hai: $(1, 4)$ ya $(- 1, 4)$ ?

Recall Solution

$\nabla f = (y, x) = (1, 4)$ at $(4, 1)$ . Tangent ko $\nabla f$ ke saath dot karke $0$ aana chahiye.

$(1, 4) \cdot (1, 4) = 1 + 16 = 17 \neq = 0$ — yeh normal hai, tangent nahi.
$(- 1, 4) \cdot (1, 4) = - 1 + 16 = 15 \neq = 0$ — yeh bhi tangent nahi! Diya gaya koi bhi option tangent nahi hai; asli tangent $(1, 4)$ ke perpendicular hai, jaise $== (- 4, 1) ==$ (kyunki $(1, 4) \cdot (- 4, 1) = - 4 + 4 = 0$ ). Yeh ek trap-spotting drill hai: kisi option ko sirf isliye mat maano kyunki woh "sideways lagta hai."

Level 2 — Application

Ab $\nabla f$ ko normal ki tarah use karo taaki tangent lines aur planes bana sako.

Exercise 2.1

Ellipse $x^{2} + 4 y^{2} = 8$ par $(2, 1)$ par tangent line nikalo.

Recall Solution

Maano $f = x^{2} + 4 y^{2}$ , toh $\nabla f = (2 x, 8 y) = (4, 8)$ at $(2, 1)$ . Yeh normal hai. Point–normal form: $\nabla f \cdot ((x, y) - (2, 1)) = 0$ : $4 (x - 2) + 8 (y - 1) = 0 \Rightarrow 4 x + 8 y = 16 \Rightarrow == x + 2 y = 4 == .$ Check: $(2, 1)$ se $2 + 2 = 4$ . ✅

Exercise 2.2

Sphere $x^{2} + y^{2} + z^{2} = 6$ par $(1, - 1, 2)$ par tangent plane nikalo.

Recall Solution

$\nabla f = (2 x, 2 y, 2 z) = (2, - 2, 4)$ at $(1, - 1, 2)$ . $2 (x - 1) - 2 (y + 1) + 4 (z - 2) = 0 \Rightarrow 2 x - 2 y + 4 z = 12 \Rightarrow== x - y + 2 z = 6 == .$ Check: $1 - (- 1) + 2 (2) = 1 + 1 + 4 = 6$ . ✅

Exercise 2.3

Paraboloid level surface $z = x^{2} + y^{2}$ , yaani $g (x, y, z) = x^{2} + y^{2} - z = 0$ , par $(1, 1, 2)$ par tangent plane nikalo.

Recall Solution

Surface ko level set $g = 0$ ki tarah likho jahan $g = x^{2} + y^{2} - z$ hai, taaki gradient-normal trick use kar sako. $\nabla g = (2 x, 2 y, - 1) = (2, 2, - 1)$ at $(1, 1, 2)$ . $2 (x - 1) + 2 (y - 1) - 1 (z - 2) = 0 \Rightarrow== 2 x + 2 y - z = 2 == .$ Check: $2 + 2 - 2 = 2$ . ✅ (Dhyan do teesre slot mein $- 1$ hai — yeh exactly wahi graph-normal $(f_{x}, f_{y}, - 1)$ hai jiske baare mein parent note mein warning di gayi hai.)

Level 3 — Analysis

Perpendicularity ko independently verify karo, aur implicit slope se reconcile karo.

Figure — Gradient perpendicular to level curves - surfaces

Exercise 3.1

Hyperbola $x y = 6$ par $(3, 2)$ par: $\nabla f$ nikalo, do independent tareekon se tangent direction nikalo (gradient method aur implicit-slope method), aur confirm karo ki dono agree karte hain.

Recall Solution

Gradient way. $f = x y$ , $\nabla f = (y, x) = (2, 3)$ at $(3, 2)$ . Tangent $(2, 3)$ ke perpendicular hai, jaise $t = (- 3, 2)$ (kyunki $(2, 3) \cdot (- 3, 2) = - 6 + 6 = 0$ ). Slope $= \frac{2}{- 3} = - \frac{2}{3}$ . Implicit-slope way. $\frac{d y}{d x} = - \frac{f _{x}}{f _{y}} = - \frac{y}{x} = - \frac{2}{3} = - \frac{2}{3}$ . Dono slope $== - \frac{2}{3} ==$ dete hain. ✅ Figure mein dekhne par: red $\nabla f$ arrow curve ko right angle par "spear" karta hai; blue tangent uske along chalta hai.

Exercise 3.2

$f (x, y) = x^{2} + y^{2}$ ka $(3, 4)$ par level circle ke unit tangent ki direction mein directional derivative kya hoga? Use compute karo aur number explain karo.

Recall Solution

$\nabla f = (2 x, 2 y) = (6, 8)$ . $(3, 4)$ par circle ka tangent radius ke perpendicular hai, direction $(- 4, 3)$ ; unit vector ke roop mein $u = \frac{1}{5} (- 4, 3)$ . $D_{u} f = \nabla f \cdot u = (6, 8) \cdot \frac{1}{5} (- 4, 3) = \frac{1}{5} (- 24 + 24) === 0 == .$ Yeh zero hai kyunki level curve ke along move karne par $f$ constant rehta hai — tangent direction mein Directional Derivative zero hona hi chahiye. Yeh parent result ek number ki form mein dikh raha hai.

Exercise 3.3

Circle $x^{2} + y^{2} = 25$ par kis point par gradient $(1, 1)$ direction mein point karta hai?

Recall Solution

$\nabla f = (2 x, 2 y)$ , $(x, y)$ ke parallel hai. Hum chahte hain $(x, y) ∥ (1, 1)$ , toh $x = y$ , aur $x^{2} + y^{2} = 25$ se: $2 x^{2} = 25$ , $x = \frac{5}{2}$ . Point $== (\frac{5}{2}, \frac{5}{2}) ==$ (outward-pointing wala). Wahan $\nabla f = \frac{10}{2} (1, 1)$ , jo indeed $(1, 1)$ ke along hai.

Level 4 — Synthesis

Gradients combine karo: intersections, parallel gradients (Lagrange flavour), aur steepest descent.

Exercise 4.1

Do level surfaces $x^{2} + y^{2} + z^{2} = 6$ aur $x^{2} + y^{2} - z = 2$ ek curve mein milte hain. $(1, 1, 2)$ par us intersection curve ke tangent mein ek direction vector nikalo.

Recall Solution

Intersection curve dono surfaces mein hai, toh uska tangent dono normals ke perpendicular hai. Normals compute karo:

$\nabla f = (2 x, 2 y, 2 z) = (2, 2, 4)$ .
$\nabla g = (2 x, 2 y, - 1) = (2, 2, - 1)$ . Dono ke perpendicular vector ka kaam cross product $\nabla f \times \nabla g$ karta hai (cross product woh tool hai jo do given vectors ke orthogonal vector deta hai): $\nabla f \times \nabla g = i 22 j 22 k 4 - 1 = (2 (- 1) - 4 (2), 4 (2) - 2 (- 1), 2 (2) - 2 (2)) = (- 10, 10, 0) .$ Tangent direction $== (- 10, 10, 0) ==$ , ya simplified $(- 1, 1, 0)$ . Check karo dono dots zero hain: $(2, 2, 4) \cdot (- 1, 1, 0) = 0$ aur $(2, 2, - 1) \cdot (- 1, 1, 0) = 0$ . ✅

Exercise 4.2

Ellipse $\frac{x ^{2}}{4} + y^{2} = 1$ par, first quadrant mein woh point nikalo jahan tangent line $y = - x$ ke parallel ho (equivalently jahan $\nabla f$ , $(1, 1)$ ke parallel ho, kyunki tangent $⊥ \nabla f$ aur $y = - x$ ka normal $(1, 1)$ hai).

Recall Solution

$f = \frac{x ^{2}}{4} + y^{2}$ , $\nabla f = (\frac{x}{2}, 2 y)$ . Hume chahiye $\nabla f ∥ (1, 1)$ : $\frac{x}{2} = 2 y \Rightarrow x = 4 y$ . Ellipse mein substitute karo: $\frac{( 4 y ) ^{2}}{4} + y^{2} = 1 \Rightarrow 4 y^{2} + y^{2} = 1 \Rightarrow y^{2} = \frac{1}{5}$ . First quadrant: $y = \frac{1}{5}$ , $x = \frac{4}{5}$ . Point $== (\frac{4}{5}, \frac{1}{5}) ==$ .

Exercise 4.3

$f (x, y) = x^{2} + y^{2}$ par $(3, 4)$ se shuru karte hue, gradient descent steepest decrease ki direction mein length $1$ ka ek step leta hai. Yeh kahan land karta hai?

Recall Solution

Steepest decrease $- \nabla f$ hai. Yahan $\nabla f = (6, 8)$ , magnitude $10$ , toh unit descent direction $- \frac{1}{10} (6, 8) = (- 0.6, - 0.8)$ hai. Naya point $= (3, 4) + 1 \cdot (- 0.6, - 0.8) = (2.4, 3.2)$ . Landing point $== (2.4, 3.2) ==$ . Yeh seedha origin ki taraf move karta hai (level circle ke perpendicular), kyunki gradient yahan radial hai.

Level 5 — Mastery

Full proofs aur edge cases: degenerate gradients aur general normals.

Exercise 5.1

Prove karo ki kisi bhi smooth $f$ ke liye, $f = c$ ka $x_{0}$ par tangent plane (jahan $\nabla f (x_{0}) \neq = 0$ ) $\nabla f (x_{0}) \cdot (x - x_{0}) = 0$ hai, ek curve argument use karke.

Recall Solution

Maano $r (t)$ surface par koi bhi smooth curve hai jisme $r (0) = x_{0}$ ho, toh saare $t$ ke liye $f (r (t)) = c$ . Multivariable Chain Rule se differentiate karo: $\frac{d}{d t} f (r (t)) = \nabla f (r (t)) \cdot r^{'} (t) = \frac{d}{d t} c = 0.$ $t = 0$ par: $\nabla f (x_{0}) \cdot r^{'} (0) = 0$ . Har tangent vector $r^{'} (0)$ aisi kisi curve se aata hai, toh $\nabla f (x_{0})$ sabhi ke orthogonal hai — yeh normal hai. Tangent plane $x$ ka woh set hai jismein $x - x_{0}$ us normal ke perpendicular ho: $\nabla f (x_{0}) \cdot (x - x_{0}) = 0$ . $■$

Exercise 5.2 (Degenerate case)

$f (x, y) = x^{2} - y^{2}$ ke liye, origin $(0, 0)$ par level set $f = 0$ examine karo. Wahan $\nabla f$ kya hai, aur "gradient is normal" statement ek single tangent line kyun nahi de paata?

Recall Solution

$f = 0$ matlab $x^{2} = y^{2}$ , yaani do crossing lines $y = x$ aur $y = - x$ . $\nabla f = (2 x, - 2 y) = (0, 0)$ at origin — gradient vanish kar jaata hai. Jab $\nabla f = 0$ toh theorem ki hypothesis fail ho jaati hai: koi well-defined normal direction nahi hai, aur waakai level set ka ek crossing point hai (do tangent lines, ek nahi). Yeh exactly singular point case hai: "ek unique tangent ke perpendicular" tab hi matter karta hai jab $\nabla f \neq = 0$ .

Exercise 5.3

Graph surface $z = x^{2} + y^{2}$ (ek paraboloid) ke liye, $(1, 1, 2)$ par graph ka outward normal nikalo, aur uski tulna domain mein level curve $x^{2} + y^{2} = 2$ ke normal se $(1, 1)$ par karo.

Recall Solution

Graph normal. Graph ko level set $g (x, y, z) = f (x, y) - z = x^{2} + y^{2} - z = 0$ ki tarah likho. Toh $\nabla g = (f_{x}, f_{y}, - 1) = (2, 2, - 1)$ . Woh $3$ -vector $== (2, 2, - 1) ==$ $3$ D mein graph ke normal hai. Level-curve normal. $2$ D domain mein, $\nabla f = (2 x, 2 y) = (2, 2)$ at $(1, 1)$ , level curve $x^{2} + y^{2} = 2$ ke normal hai. Yeh alag alag objects hain: $(2, 2)$ flat input plane mein rehta hai aur contour ke across point karta hai; $(2, 2, - 1)$ $3$ D mein rehta hai aur graph se tilt karke nikalta hai. Pehle do entries match karte hain kyunki graph normal ka horizontal shadow hi level-curve normal hai — yeh ek satisfying consistency hai, coincidence nahi.

Recall Self-test checklist

Perpendicularity kis identity se aati hai? ::: $\nabla f (r (t)) \cdot r^{'} (t) = 0$ (chain rule on $f = c$ ). Do surfaces ke intersection curve ka tangent? ::: $\nabla f \times \nabla g$ (dono normals ke perpendicular). Steepest descent direction? ::: $- \nabla f$ , level curve ke perpendicular. Poora result kab break karta hai? ::: Jab $\nabla f = 0$ (singular point, koi unique tangent nahi).

Connections

Multivariable Chain Rule — yahan har proof ke peeche differentiation step.
Directional Derivative — Exercise 3.2 ki zero value.
Tangent Plane to a Surface — Level 2 aur 5.
Lagrange Multipliers — Exercise 4.2 ka "parallel gradients" idea.
Steepest Descent / Gradient Descent — Exercise 4.3.