1 / 18

Consider a transformation T ( u,v ) = ( x ( u,v ) , y ( u,v )) from R 2 to R 2 .

Consider a transformation T ( u,v ) = ( x ( u,v ) , y ( u,v )) from R 2 to R 2 . Suppose T is a linear transformation T ( u,v ) = ( au + bv , cu + dv ) . Then the derivative matrix of T is D T ( u,v ) =.  x  x — —  u  v =  y  y — —  u  v. a b c d.

chiku
Download Presentation

Consider a transformation T ( u,v ) = ( x ( u,v ) , y ( u,v )) from R 2 to R 2 .

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Consider a transformation T(u,v) = (x(u,v) , y(u,v)) from R2 to R2. Suppose T is a linear transformation T(u,v) = (au + bv , cu + dv) . Then the derivative matrix of T is DT(u,v) = x x — — u v = y y — — u v abcd We now explore what the transformation does to the rectangle defined by points (vectors) (0 , 0), (u0 , 0), (u0 , v0), (0 , v0). v y (0 , v0) (u0 , v0) u x (0 , 0) (u0 , 0) (0 , 0)

  2. v y (au0 + bv0 , cu0 + dv0) (bv0 , dv0) (0 , v0) (u0 , v0) T(D) D (au0 , cu0) u x (0 , 0) (u0 , 0) (0 , 0) T(0,0) = (a(0) + b(0) , c(0) + d(0)) = (0,0) T(0 , v0) = (a(0) + b(v0) , c(0) + d(v0)) = (bv0 , dv0) T(u0 , v0) = (au0 + bv0 , cu0 + dv0) T(u0 , 0) = (a(u0) + b(0) , c(u0) + d(0)) = (au0 , cu0) The area of the rectangle D in the uv plane is u0v0 . The area of the parallelogram T(D) in the xy plane is

  3. T(0,0) = (a(0) + b(0) , c(0) + d(0)) = (0,0) T(0 , v0) = (a(0) + b(v0) , c(0) + d(v0)) = (bv0 , dv0) T(u0 , v0) = (au0 + bv0 , cu0 + dv0) T(u0 , 0) = (a(u0) + b(0) , c(u0) + d(0)) = (au0 , cu0) The area of the rectangle D in the uv plane is u0v0 . The area of the parallelogram T(D) in the xy plane is ||(au0 , cu0 , 0)(bv0 , dv0 , 0)|| = ||(0 , 0 , adu0v0–bcu0v0)|| = u0v0|ad–bc| . du dv = u0v0 and dx dy = u0v0|ad–bc|. Now we know that D T(D) Consequently, dx dy = |ad – bc| du dv = |det[DT]| du dv . T(D) D D

  4. This is a special case of a more general result. Suppose transformation T(u,v) = (x(u,v) , y(u,v)) from R2 to R2 is one-to-one. Then f(x,y) dx dy = f(x(u,v),y(u,v)) |det[DT]| du dv = T(D) D (x,y) f(x(u,v),y(u,v)) ——– du dv . (u,v) D This notation is used to represent x x — — u v y y — — u v det which is called the Jacobian determinant of T. One useful fact about one-to-one linear transformations is that a triangle is always mapped to a triangle, and a parallelogram is always mapped to a parallelogram.

  5. Example Consider the integral of xy over the parallelogram P formed by the lines y = 2x , y = 2x– 2 , y = x , and y = x + 1 . (a) v u–— 2 u–v —— , 2 Sketch the region P in the xy plane, and with T(u,v) = find the region D in the uv plane so that T(D) = P. (3,4) u (2,0) D (1,2) (2,2) T(D) = P x (0,–4) (2,–4) y v (x,y) = (0,0)  (u,v) = (0,0) (x,y) = (1,2)  (u,v) = (2,0) 2(y–x) u = v = (x,y) = (3,4)  (u,v) = (2,–4) x = (u–v)/2  y = u–v/2 2y– 4x (x,y) = (2,2)  (u,v) = (0,–4)

  6. x = (u – v)/2 y = u – v/2 (b) Evaluate by making the change of variables xydx dy P x x — — u v = y y — — u v 1/2 – 1/2 (x,y) ——– = (u,v) det det = 1/4 1 – 1/2 (u– v)(2u – v) (x,y) —————— ——– du dv = 4 (u,v) xydx dy = P D 0 2 (u – v)(2u – v) —————– du dv = 16 (2u2– 3uv + v2) ——————– du dv = 16 7 D –4 0

  7. Example Consider the following integral: (a) xy3 ——— dx dy (1 – x)4 (1/2 , 1/2) (0 , 1/2) D* D* (0 , 1/4) (3/4 , 1/4) where D is the region displayed in the graph. x y With the transformation T(u,v) = (u/v , v – u), there will be a trapezoidal region D in the uv plane so that T(D) = D*. Find this trapezoidal region D in the uv plane. (1/2 , 1) (3/4 , 1) u x = — u = v  y = v – uv = xy —— 1 – x (0 , 1/2) D y —— 1 – x (0 , 1/4) u v

  8. xy3 ——— dx dy (1 – x)4 (b) Evaluate by making the change of variables x = u/v y = v–u D* x x — — u v = y y — — u v 1/v – u/v2 (x,y) ——– = (u.v) v – u —— v2 det det = –1 1 xy3 ——— dx dy = (1 – x)4 Since v > u on D, we see that this is equal to its absolute value. uv du dv = D D*

  9. (1/2 , 1) (3/4 , 1) (0 , 1/2) D xy3 ——— dx dy = (1 – x)4 (0 , 1/4) uv du dv = u D v D* 1/2 v– 1/4 1 v – 1/4 239 —— 6144 uv du dv + uv du dv = 1/4 0 1/2 v – 1/2 1/2 1/2 + u 3/4 1 239 —— 6144 uv dv du + uv dv du = 0 1/4 + u 1/2 1/4 + u

  10. Recall: For each point (x,y) in R2, the polar coordinates (r,) are defined by x = r cos  and y = r sin  , where r =  x2 + y2 is the length of the vector (x,y) , and  = the angle that the vector (x,y) makes with the positive x axis . y (x,y)  (r,) r  x We have that 0  r and 0   < 2 .

  11. Example Describe the region x2 + y2 36 in terms of rectangular coordinates and in terms of polar coordinates.  6  x  6 –  6  x2  y   6  x2 or  6  y  6 –  6  y2  x   6  y2 r   < 6 0 2

  12. Example Describe the region inside the triangle with vertices (0,0), (2,0), and (2,–23) in terms of rectangular coordinates and in terms of polar coordinates. 5/3    r 2 (2,0) 2 / cos  ( or –/3    0 r 2 / cos  ) (2,–23) 0  x  2 –23  y  0 or – (3)x  y  0 – y / 3  x  2

  13.  / 2     r  3 / 2 Example Describe the region (x + 5)2 + y2 25 in terms of rectangular coordinates and in terms of polar coordinates. 0 – 10 cos

  14. Example Describe the region interior to both circles x2 + y2= 1 and x2 + (y + 1)2= 1 in terms of rectangular coordinates and in terms of polar coordinates. r = 1 r = 1  = r = 1  = 7 — 6 11 —— 6 r = – 2 sin 11 —— 6 7 — 6 7 — 6 11 —— 6     r      r      r   2   – 2 sin – 2 sin 0 0 0 1

  15. Example Suppose we want to integrate the function f(x,y) over the region D* = T(D) in the xy plane, where D is the same region described in the r plane and T(r,) = (r cos , r sin), that is, T is the polar coordinates transformation. Then, we know that (x,y) f( x(r,) , y(r,) ) ——— dr d (r,) f(x,y) dx dy = D* D y  (x,y) Find ——— . (r,) D D* r x cos – r sin (x,y) ——— = (r,) det = | r cos2 + r sin2 | = r sin r cos

  16. We see then how we can change from rectangular coordinates to polar coordinates: f(x,y) dx dy = f( r cos  , r sin ) r dr d . D* D x2 + y2 Example Consider the integral of the function f(x,y) = e over D* defined to be the region in the first quadrant of the xy plane between the two circles x2 + y2 = 4 and x2 + y2 = 25. (a) Sketch and describe the region of integration D* in the xy plane and the corresponding region D in the r plane. y  x2 + y2 = 25 (2 , /2) (5 , /2) D* x2 + y2 = 4 D r x (2 , 0) (5 , 0) 0     r   / 2 2 5

  17. x2 + y2 (b) Evaluate by making the change of variables edx dy D* x = r cos y = r sin /2 5 r2 r2 x2 + y2 er dr d = er dr d = edx dy = D* D 0 2 5 /2 /2 r2 e — d = 2 e25– e4 ——— d = 2 (e25– e4) ———— 4 0 0 r = 2

  18. –x2 Example Find edx by squaring and using polar coordinates. –     –x2 –y2 – x2 –y2 e dx e dy = e dx e dy = – – – –   2  2  –r2 –(x2 + y2) –r2 edx dy = e rdr d = e – —— d = 2 – – 0 0 0 r = 0 2  1 — d = 2 –x2  We see then that e dx =  0 –

More Related