Problem: Let OABC be a unit square in the xy-plane with O(0,0),A(1,0),B(1,1) and C(0,1). Let u=x2βy2 and v=2xy be a transformation of the xy-plane into the uv-plane. The transform (or image) of the square is:
Answer Choices:
A.
B.
C.
D.
E.
Solution:
The image of A(1,0) in the xy-plane is Aβ²(1,0) in the uvplane since u=12β02 =1 and v=2(1)(0)=0. The image of B(1,1) is Bβ²(0,2), the image of C(0,1) is Cβ²(β1,0), and the image of O(0,0) is Oβ²(0,0).
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The image of the straight line AB whose xyβ equation is x=1 is the parabolic arcAβ²Bβ² whose uv- equation is v2=4(1βu), a parabola, since u=1βy2 and v=2y. The image of the straight line OA (equation y=0 ) is the straight line Oβ²Aβ² (equations v=0,u=x2 ), and the image of the straight line OC (equation x=0 ) is the straight line Oβ²Cβ² (equations v=0,u=βy2 ).