### 雅克比矩阵论述

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Why the 2D Jacobian works 斗x Transformation T yield distorted grid of lines of constant u and constant v For small du and dv, rectangles map onto parallelograms du =(du, 0)and dv=(0, dv (.m)(如)-(m)( dA=d x dy=( o, du × du (auyu-royu)dud This is a jacobian, ie, the determinant of the jacobian matrix Relation between Jacobians · The jacobian matrix a(C, g) is the inverse matrix of 0(,0) 1.e 0(,) 0(x,y) m)(m,)=( 10 Because(and similarly for dy) do= Cudutaudu= ld +a v(vdr vy dy) a constant→d/=0→0=2y+a2y constant→dy=0→1=x+x2t This makes sense because Jacobians measure the relative areas of dxdy and duds, i.e det(ab)= det(a) det (b)=1= det(a) det乃 . So 0(x,y)_|(, 0(u4,0)-a(x,y Simple 2D EXample Area of circle a= ∫ Ja.xdy =pcos 8 and y=psin A O 0000 xp90 Bo/ s cos e -psin 6 00 sin e 00 pCos=p(cos20+Sin-8)=p =2丌 dpdO=「3210127 p=0J0=0 pdpo 三丌 0 Harder 2D Example ∫/n(x2+y2)drd =况 Wherer is this region of the xy v=2xy=8 plane, Which maps to 尺here y=4 8 d(a 0(l,0)LO(x,y) (x2+y2 9 butn2+2=(x2-y2)2+4x2y2=(x2+y2)2s0a(02D)= ∫a(x2+9ddy=是/1Addo=8 An Important 2D Example Evaluate 2 dx C First consider dx-a y d 2=J-0 C (ax+y)dada Put =rcos and y= rsin a(a, y) cos - cos o 0() Sin a r coS o C2T re-r drdo< I2< Jo Jo re-r drdo 2 0J0 丌(1-c)<l2×丌(1-c-20)as T 3D Jacobian x=x(,0,),y=y(,0,),z=z(,0,) maps volumes(consisting of small cubes of volume dxdydz to small cubes of volume duda ∫(x,y, 2drdudz= F(20,0) (x,y,2) duddo (,,) ● Where O(x,y,2) 0(,v,0) yu yu y 0 3D EXample Transformation of volume elements between Cartesian and spherical polar coordinate systems(see Lecture 4) a rsin 0 cos y= r sine sin⑦ a(e sin 0 cos r cos 8 cos -rsin g sin o 01n= sin 0 sin rcos 0 sin o r sin 0 cos o 0(r,,) COS 6 r sin e 0 -r2 sin 0

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