Change of variables — general Jacobian: Question bank

4.4.21 · D5 · Maths › Multivariable Calculus › Change of variables — general Jacobian

True or false — justify

Jacobian determinant hamesha positive hota hai.

False. Isme ek sign hota hai jo orientation record karta hai: positive value matlab

T

orientation preserve karta hai, negative matlab wo plane ko flip karta hai (jaise ek mirror). Integrals mein hum

∣ J ∣

isliye lete hain kyunki raw determinant negative ho sakta hai.

d x d y = r d r d θ

isliye hai kyunki ek polar cell literally

d r

aur

r d θ

sides ka ek rectangle hota hai.

True. Radial edge ki length

d r

hai, arc edge ki length

r d θ

hai (arc = radius × angle), toh chhota sa wedge approximately ek rectangle hai jiska area

r d r d θ

hai — wohi

r

jo Jacobian produce karta hai.

Agar

det J_{T} = 0

kisi point par hai, toh change-of-variables formula wahan bina kisi issue ke apply hota hai.

False. Zero Jacobian matlab linearized map area ko zero mein collapse kar deta hai (parallelogram ek segment mein납 flat ho jaata hai), isliye

T

wahan locally invertible nahi hai. Theorem ko region ke interior par nonzero Jacobian chahiye.

u

aur

v

ke roles swap karna (relabelling) integral ki value change kar deta hai.

False. Jacobian matrix ke do columns swap karne se

det J

ka sign flip hota hai, lekin integral

∣ J ∣

use karta hai, isliye absolute value — aur answer — unchanged rehta hai.

Ek linear change of variables ka Jacobian poore region mein hamesha constant hota hai.

True. Linear map

x = A u

ke liye partials

\partial x_{i} / \partial u_{j}

sirf constant matrix

A

ki entries hain, isliye

det J = det A

har jagah same hai aur integral se seedha bahar aa jaata hai.

Polar coordinates ke liye,

det J = r

negative ho sakta hai jab

r < 0

, isliye hume

∣ r ∣

likhna chahiye.

Principle mein True. Agar tum

r < 0

allow karo toh Jacobian

r

hai aur area ko

∣ r ∣

chahiye; standard convention

r \geq 0

rakhta hai, isliye

∣ r ∣ = r

aur hum bars drop kar dete hain — lekin sirf us convention ki wajah se.

Change-of-variables formula require karta hai ki

f

positive ho.

False.

f

koi bhi sign le sakta hai; area element

∣ J ∣ d u d v

hi non-negative rehna chahiye, integrand nahi. Jacobian ki absolute value area handle karti hai;

f

ka sign untouched rehta hai.

\frac{\partial ( x , y )}{\partial ( u , v )}

aur

\frac{\partial ( u , v )}{\partial ( x , y )}

equal hote hain.

False. Ye reciprocals hain: unka product

1

hai kyunki

J_{T}

aur

J_{T^{- 1}}

inverse matrices hain aur

det J_{T} \cdot det J_{T^{- 1}} = det (I) = 1

Spot the error

"Maine

u = x + y

v = x - y

set kiya,

\frac{\partial ( u , v )}{\partial ( x , y )} = - 2

compute kiya, aur integrand ko

2

se multiply kar diya."

Error hai galat-direction Jacobian use karna. Integral ko old area per new cell chahiye, yaani

\partial (x, y) / \partial (u, v)

, jiska magnitude

\frac{1}{2}

hai —

∣ - 2∣

ka reciprocal.

2

se multiply karna area ko

4

ke factor se overcount karta hai.

"Mera map orientation flip karta hai, isliye

det J = - 3

hai, aur maine integrand ko

- 3

se multiply kar diya."

Tum absolute value bhool gaye. Area negative nahi ho sakta; tumhe

∣ - 3 ∣ = 3

se multiply karna chahiye. Minus sign sirf yeh bata raha tha ki plane flip hua, jo is baat se irrelevant hai ki ek cell kitna area cover karta hai.

"Maine

f

aur Jacobian sahi transform kiya lekin integration ke original

x, y

limits rakh diye."

uv

-space mein region

S

kabhi find hi nahi kiya. Ek baar jab tum variables change karte ho toh tumhe limits ko new coordinates mein re-express karna hoga; purane limits rakhna bilkul galat region par integrate karta hai.

x = r cos θ, y = r sin θ

ke liye mujhe Jacobian

cos^{2} θ - sin^{2} θ

mila."

Determinant mein ek sign slip ho gaya. Sahi tarike se,

det (cos θ sin θ - r sin θ r cos θ) = r cos^{2} θ - (- r sin^{2} θ) = r (cos^{2} θ + sin^{2} θ) = r

. Off-diagonal product apna khud ka minus sign carry karta hai, jo cancel ho jaata hai.

"Map

T (u, v) = (u^{2}, v)

- 1 \leq u \leq 1

par theek hai kyunki ye smooth hai."

Smoothness kaafi nahi hai —

T

injective hona chahiye. Yahan

u

aur

- u

dono same

x = u^{2}

par map karte hain, isliye

T

strip ko apne aap par fold karta hai aur area double-count karta hai. Tumhe

u \geq 0

tak restrict karna hoga (ya region split karna hoga).

"Kyunki parallelogram ke edges

r_{u} d u

aur

r_{v} d v

hain, iska area

∣ r_{u} ∣∣ r_{v} ∣ d u d v

hai."

Woh formula assume karta hai ki edges perpendicular hain. Generally area

∣ r_{u} \times r_{v} ∣ d u d v = ∣ x_{u} y_{v} - y_{u} x_{v} ∣ d u d v

hai, jisme edges ke beech angle ka

sin

shamil hai — exactly Jacobian.

Why questions

Mapped cell ek parallelogram kyun hota hai, koi curved blob nahi?

Kyunki hum corner ke paas

T

ka first-order (linear) Taylor approximation use karte hain. Microscope ke neeche koi bhi smooth map linear lagta hai, aur ek linear map ek rectangle ko parallelogram mein bheejta hai; curvature ek higher-order effect hai jo cell ke shrink hone par vanish ho jaata hai.

Hum specifically determinant kyun use karte hain, partials ke kisi aur combination ki jagah?

Kyunki determinant defined hi hai ek linear map ke signed area/volume-scaling factor ke roop mein — dekho Determinant as Volume Scaling. Ye unique quantity hai jo batata hai ki Jacobian matrix ke under ek unit cell ka area kitna change hota hai.

2D derivation mein cross product kyun aata hai jabki hum plane mein hain?

Do edge vectors ko 3D mein

(a_{x}, a_{y}, 0)

aur

(b_{x}, b_{y}, 0)

ke roop mein embed karo, unka cross product purely

z

direction mein point karta hai magnitude

∣ a_{x} b_{y} - a_{y} b_{x} ∣

ke saath — jo hai parallelogram ka area. Ye signed area extract karne ka sabse clean tarika hai.

Hum

T

ko har cell par uske linearization se replace kyun kar sakte hain bina accuracy khoye?

Kyunki hum infinitely fine cells ki limit le rahe hain. Linear approximation ki error faster shrink hoti hai (order

d u^{2}

) cell area se (order

d u d v

), isliye integral limit mein sirf linear part survive karta hai.

Reciprocal shortcut

\partial (x, y) / \partial (u, v) = 1/ [\partial (u, v) / \partial (x, y)]

kyun hold karta hai?

Multivariable chain rule se,

J_{T} J_{T^{- 1}} = I

, isliye

det J_{T} \cdot det J_{T^{- 1}} = 1

. Ye exactly Inverse Function Theorem hai jo guarantee karta hai ki inverse exist karta hai jahan Jacobian nonzero hai.

Hum change of variables choose kyun karte hain pehli jagah — ye kaunsi problem solve kar raha hai?

Ya toh region

R

ya integrand

f

ko simpler banane ke liye. Jaise Polar Coordinates ek disk ko rectangle mein turn karte hain, aur

u = x + y

ek two-variable integrand ko ek variable mein collapse karta hai, complexity ko Jacobian factor ke saath trade karke.

Spherical Jacobian

ρ^{2} sin ϕ

kyun hai, sirf

ρ^{2}

nahi?

ρ^{2}

do "sideways" arc-lengths ko scale karta hai jo radius ke saath grow karte hain, jabki

sin ϕ

poles ke paas longitude circles ke shrinking ko account karta hai — dekho Spherical and Cylindrical Coordinates.

ϕ = 0

(pole) par

sin ϕ = 0

aur cell ek point mein degenerate ho jaata hai.

Edge cases

Origin par,

r = 0

, polar Jacobian ka kya hota hai?

det J = 0

ho jaata hai: har angle

θ

single point

(0, 0)

par map karta hai, isliye

T

wahan injective nahi hai aur area collapse ho jaata hai. Origin ek measure-zero point hai, isliye integrals unaffected hain — lekin

T

genuinely singular hai exactly wahan.

Agar ek region

R

ka area zero hai (ek curve ya ek point), toh formula kya deta hai?

Zero. Ek degenerate region koi area contribute nahi karta, isliye

∣ J ∣

ki parwah kiye bina dono sides vanish ho jaate hain. Isliye

T

ke singular points (isolated zero-Jacobian points) integral ko kharab nahi karte.

Kya Jacobian har jagah negative ho sakta hai aur formula phir bhi kaam kare?

Haan. Ek globally orientation-reversing map (jaise ek reflection) ka

det J < 0

poore region mein hota hai; tum sirf

∣ J ∣ > 0

use karo aur theorem bilkul theek kaam karta hai. Sign sirf orientation ke baare mein bookkeeping hai, correctness ke baare mein nahi.

Kya hoga agar

T

interior par injective hai lekin boundary points ko glue kar deta hai (jaise

θ = 0

aur

θ = 2 π

Ye theek hai. Theorem ko interior par injectivity chahiye; boundaries measure-zero hain aur integral ko change kiye bina overlap kar sakti hain. Isliye full disk

θ = 0

par seam ke bawajood kaam karta hai.

3D mein, geometrically ek zero Jacobian ka kya matlab hai?

Mapped cell ke teen edge vectors coplanar (ya collinear) ho jaate hain, isliye parallelepiped zero volume mein flatten ho jaata hai. Map ek 3D neighbourhood ko ek lower-dimensional set mein crush kar deta hai — locally non-invertible.

Map

u = x^{2} - y^{2}, v = 2 x y

ke liye, ye invertible kahan fail karta hai?

Origin

(x, y) = (0, 0)

par, jahan

\partial (u, v) / \partial (x, y) = 4 (x^{2} + y^{2}) = 0

. Wahan reciprocal

\partial (x, y) / \partial (u, v) = 1/ [4 (x^{2} + y^{2})]

blow up karta hai, Jacobian singularity ka signal deta hua.

Agar do alag

uv

-cells same

x y

-patch par map karte hain (non-injective

T

), kya galat hota hai?

Tum us patch ke area ko double-count karte ho, integral ko inflate karte hue. Formula silently assume karta hai ki har old cell exactly ek baar cover hoti hai; overlaps ko domain restrict karke ya subtract karke remove karna hoga.

Identity map

x = u, y = v

ke liye

∣ J ∣

kya hai, aur kya ye sense banta hai?

∣ J ∣ = 1

: cells bilkul stretch nahi hote, isliye

d x d y = d u d v

. Ye sanity-check baseline hai — koi coordinate change nahi, koi scaling nahi.

Connections

Change of variables — general Jacobian — woh parent jise ye bank drill karta hai
Determinant as Volume Scaling — determinant kyun scaling factor hai
Cross Product and Area — parallelogram-ka-area engine
Inverse Function Theorem — reciprocal shortcut guarantee karta hai
Polar Coordinates · Spherical and Cylindrical Coordinates — classic Jacobians