THE JACOBIAN & CHANGE OF VARIABLES MATH 200 GOALS Be able to - PowerPoint PPT Presentation

MATH 200 WEEK 10 - WEDNESDAY THE JACOBIAN & CHANGE OF VARIABLES

MATH 200 GOALS ▸ Be able to convert integrals in rectangular coordinates to integrals in alternate coordinate systems

MATH 200 DEFINITION ▸ Transformation: A transformation, T, from the uv-plane to the xy-plane is a function that maps (u,v) points to (x,y) points. ▸ x = x(u,v); y = y(u,v)

MATH 200 EXAMPLE ▸ Consider the region in the r θ -plane bounded between r=1, r=2, θ =0, and θ = π /2 ▸ The transformation T:x=rcos θ , y=rsin θ maps the region in the r θ -plane into this region in the xy-plane ▸ Transformations need to be one-to-one and must have continuous partial derivatives

MATH 200 JACOBIAN ▸ The area of a cross section in the xy-plane may not be exactly the same as the area of a cross section in the uv- plane. We want to determine the relationship; that is, we want to determine the scaling factor that is needed so that the areas are equal.

MATH 200 IMAGE OF S UNDER T ▸ Suppose that we start with a tiny rectangle as a cross section in the uv-plane with dimensions u and v. The image will be roughly a parallelogram (as long as our partition is small enough). ▸ T: x=x(u,v), y=y(u,v) ▸ x and y are functions of u and v

MATH 200 ▸ Let’s label the corners of S as follows ▸ A(u 0 ,v 0 ) ▸ B(u 0 + Δ u,v 0 ): B is a a little to the right of A ▸ C(u 0 ,v 0 + Δ v): C is a little higher that A ▸ D(u 0 + Δ u,v 0 + Δ v): D is a little higher and to the right of A

MATH 200 ▸ What happens to these points under the transformation T? ▸ Well, we have transformation functions x(u,v) and y(u,v), so we can plug the coordinates of the points A, B, C, and D to get the corresponding points in the xy-plane ▸ For the image of a point P under T , we’ll write T(P) = P’, so we get: ▸ A’(x(u 0 ,v 0 ),y(u 0 ,v 0 )) ▸ B’(x(u 0 + Δ u,v 0 ), y(u 0 + Δ u,v 0 )) ▸ C’(x(u 0 ,v 0 + Δ v), y(u 0 ,v 0 + Δ v)) ▸ D’(x(u 0 + Δ u,v 0 + Δ v), y(u 0 + Δ u,v 0 + Δ v))

MATH 200 ▸ We want to know how the area of R compares to the area of S. ▸ R is a parallelogram, so its area is the cross-product of the a × � vectors a and b: || � b || � � � � � � A � B � a = = � x ( u 0 + ∆ u, v 0 ) � x ( u 0 , v 0 ) , y ( u 0 + ∆ u, v 0 ) � y ( u 0 , v 0 ) , 0 � � � � � � A � C � b = = � x ( u 0 , v 0 + ∆ v ) � x ( u 0 , v 0 ) , y ( u 0 , v 0 + ∆ v ) � y ( u 0 , v 0 ) , 0 � WE’VE ADDED THE Z- COMPONENT BECAUSE CROSS PRODUCTS ARE ONLY DEFINED FOR VECTORS IN 3-SPACE

MATH 200 ▸ To simplify these vectors down to something more manageable, we look to the limit definition of the partial derivative: f ( x + ∆ x ) − f ( x ) d f FROM CALC 1 dx = lim ∆ x ∆ x → 0 f ( x + ∆ x, y ) − f ( x, y ) ∂ f ∂ x = lim ∆ x ∆ x → 0 ▸ Notice that the first component of the vector a looks like the numerator of the limit definition of partial derivative of x with respect to u x ( u 0 + ∆ u, v 0 ) − x ( u 0 , v 0 ) ≈ ∂ x ∂ u ∆ u IT’S APPROXIMATE BECAUSE WE’RE MISSING THE LIMIT

MATH 200 ▸ Applying this idea to both vectors, we get � � a = � x ( u 0 + ∆ u, v 0 ) � x ( u 0 , v 0 ) , y ( u 0 + ∆ u, v 0 ) � y ( u 0 , v 0 ) , 0 � � ∂ x � ∂ u ∆ u, ∂ y � ∂ u ∆ u, 0 � � x u ∆ u, y u ∆ u, 0 � � � b = � x ( u 0 , v 0 + ∆ v ) � x ( u 0 , v 0 ) , y ( u 0 , v 0 + ∆ v ) � y ( u 0 , v 0 ) , 0 � � ∂ x � ∂ v ∆ v, ∂ y � ∂ v ∆ v, 0 � � x v ∆ v, y v ∆ v, 0 �

MATH 200 ▸ Now we can take the cross product → − − → → − � � i j k � � a × − → − → � � b = 0 x u ∆ u y u ∆ u � � � � 0 x v ∆ v y v ∆ v � � = [( x u ∆ u )( y v ∆ v ) − ( x v ∆ v )( y u ∆ u )] − → k = ( x u y v − x v y u ) ∆ u ∆ v − → k ▸ For the area of R, we get a × − → � � � − → � = | x u y v − x v y u | ∆ u ∆ v b � � ∆ A R = | x u y v − x v y u | ∆ u ∆ v

MATH 200 CONCLUSIONS ▸ We’ve shown that the area of R is ∆ A R = | x u y v − x v y u | ∆ u ∆ v ▸ The area of S is ∆ A S = ∆ u ∆ v ▸ So the scaling factor we were looking for is |x u y v - x v y u | ▸ We can write this as the determinate of a special matrix called the Jacobian Matrix: � x u � ∂ ( x, y ) x v ∂ ( u, v ) = y u y v

MATH 200 EXAMPLE ▸ Compute the Jacobian (i.e., the determinate of the Jacobian Matrix) for the transformation from the r θ -plane to the xy-plane � x ( r, θ ) = r cos θ T : y ( r, θ ) = r sin θ

MATH 200 � x ( r, θ ) = r cos θ T : y ( r, θ ) = r sin θ x r = cos θ x θ = − r sin θ y r = sin θ y θ = r cos θ � x r � ∂ ( x, y ) x θ ∂ ( r, θ ) = y v θ y r � cos θ � − r sin θ = sin θ r cos θ = cos θ ( r cos θ ) − ( − r sin θ )(sin θ ) = r cos 2 θ + r 2 sin 2 θ = r (cos 2 θ + sin 2 θ ) = r

MATH 200 THEOREM ▸ Given a transformation T from the uv-plane to the xy-plane, if f is continuous on R and the Jacobian is nonzero, we have � � ∂ ( x, y ) �� �� � � f ( x, y ) dA xy = f ( x ( u, v ) , y ( u, v )) � dA uv � � ∂ ( u, v ) � R S ▸ For example, when converting from rectangular to polar, we have �� �� f ( x, y ) dA xy = f ( x ( r, θ ) , y ( r, θ )) r dA r θ R S

MATH 200 EXAMPLE ▸ Consider the following double integral: x − y �� x + y dA, R : region enclosed by y = x, y = x − 1 , y = − x + 1 , y = − x + 3 R IT WOULD BE NICE IF WE COULD TRANSFORM THIS REGION INTO AN R UPRIGHT RECTANGLE

MATH 200 ▸ Let’s take the transformation T to be � u = x − y T : v = x + y ▸ To convert our xy-integral to a uv-integral, we want the Jacobian � � ∂ ( x, y ) x u x v ∂ ( u, v ) = y u y v ▸ We could solve for u and v in our transformations OR we could use the convenient fact that ∂ ( x, y ) 1 ∂ ( u, v ) = ∂ ( u,v ) ∂ ( x,y )

MATH 200 ▸ First, we need the partial derivatives: u x = 1 , u y = − 1 , v x = 1 , v y = 1 ▸ Plug it all in to the Jacobian Matrix � u x � ∂ ( u, v ) u y ∂ ( x, y ) = v y θ v x � 1 � − 1 = 1 1 = 2 ▸ Take the reciprocal to get ∂ ( x, y ) ∂ ( u, v ) = 1 2

MATH 200 ▸ For the bounds, we have  y = x = ⇒ x − y = 0 = ⇒ u = 0    y = x − 1 = ⇒ x − y = 1 = ⇒ u = 1  y = − x + 1 = ⇒ x + y = 1 = ⇒ v = 1     y = − x + 3 = ⇒ x + y = 3 = ⇒ v = 3 ▸ Finally, our integral becomes � 3 � 1 � � �� x − y u 1 � � x + y dA = � dudv � � v 2 � 1 0 R

MATH 200 � 3 � 1 � � x − y u �� 1 � � x + y dA = � dudv � � v 2 � 1 0 R � 3 � 1 u = 1 v dudv 2 1 0 � 3 1 u 2 � = 1 1 � dv � v 2 2 � 1 0 � 3 = 1 1 v dv 4 1 3 � = 1 � 4 ln | v | � � 1 = 1 4 ln 3

MATH 200 EXAMPLE 2 ▸ Use the transformation x = v/u, y=v to evaluate the integral � x � 1 y 2 x 2 e y/x dydx 0 0 ▸ Let’s first convert the region from the xy-plane to the uv-plane ▸ The region extends from the line y=0 to the line y=x ▸ From x=0 to x=1

MATH 200 ▸ Apply the conversion formulas x=v/u and y=v ▸ y=0 becomes v=0 ▸ y=x becomes v=v/u ▸ u=1 ▸ x=0 becomes v/u=0 ▸ v=0 ▸ x=1 becomes v/u=1 ▸ v=u ▸ In summary, the region extends from the line v=0 to the line v=u SO THE REGION LOOKS THE SAME IN THE UV-PLANE ▸ Bounded by u=1

MATH 200 ▸ For the Jacobian, we need ▸ Lastly, we need to convert the partials of x=v/u and the integrand (the function y=v we’re integrating) ▸ x u = -v/u 2 y 2 ( v ) 2 x 2 e y/x = ( v/u ) 2 e v/ ( v/u ) ▸ x v = 1/u ▸ y u = 0 ▸ y v = 1 y 2 x 2 e y/x = u 2 e u � � ∂ ( x, y ) x u x v � � ∂ ( u, v ) = � � y u y v ▸ Now we have all the � � � � − v/u 2 1 /u pieces we need to set up � � = � � 0 1 the integral � � = − v/u 2

MATH 200 � x � u � 1 � 1 y 2 � − v u 2 e u � � x 2 e y/x dydx = � dvdu � � u 2 0 0 0 0 � u � 1 ve u dvdu = 0 0 � 1 u � 1 2 v 2 e u � du = � � 0 0 � 1 1 2 u 2 e u du = 0 ▸ At this point we need to do integration by parts twice to finish the problem

MATH 200 � 1 � 1 1 � 1 2 u 2 e u du = 1 − 1 2 ue u du � 2 u 2 e u � 2 � 0 0 0 � 1 � � 1 � = 1 2 e − 1 2 e u du � 2 ue u � − 2 � 0 0 � � 1 � = 1 2 e − 1 � 2 e − 2 e u � 2 � 0 = 1 2 e − e + 1 2(2 e − 2) = 1 2 e − 1

MATH 200 EXAMPLE 3 ▸ Let R be the region enclosed by xy = 1, xy = 2, xy 2 = 1, and xy 2 = 2. Evaluate the following integral. �� y 2 dA R R

MATH 200 ▸ Let u=xy and v=xy 2 ▸ This gives us a nice rectangular region in the uv-plane ▸ u=1 to u=2 and v=1 to v=2 ▸ To find the Jacobian we’ll need to solve for x and y in terms of u and v ▸ Solve each equation for x to get x = u/y and x = v/y 2 ▸ Setting those equal: u/y = v/y 2 ▸ Solve for y: y = v/u ▸ Plug into the u-equation: u=x(v/u) ▸ x = u 2 /v

MATH 200 ▸ Jacobian: � � ∂ ( x, y ) x u x v � � ∂ ( u, v ) = � � y u y v � � � � − u 2 /v 2 2 u/v � � = � � − v/u 2 1 /u � � = 2 v − 1 v = 1 v ▸ Setup: � 1 � 2 � 2 � v 2 �� y 2 dA = v dvdu u 2 1 1 R