Question

Suppose \(S\) is the unit square in the first quadrant of the \(u v\) -plane. Describe the image of the transformation \(T: x=2 u, y=2 v\)

Short answer

In conclusion, the image of the transformation T applied to the unit square S in the uv-plane results in a square of area 4 in the first quadrant of the xy-plane. This transformed square has vertices at A'(0,0), B'(2,0), C'(2,2), and D'(0,2), and its sides are parallel to the x and y axes with length 2.

Step-by-step solution

Find the vertices of the unit square S in the uv-plane

In the first quadrant of the \(uv\)-plane, the unit square \(S\) has the vertices \(A(0,0), B(1,0), C(1,1),\) and \(D(0,1)\).

Apply the transformation T to each vertex

To find the corresponding vertices in the \(xy\)-plane after the transformation \(T\), apply the given equations \(x = 2u\) and \(y = 2v\) to each vertex: For vertex \(A(0,0)\): \(x = 2(0) = 0\) \(y = 2(0) = 0\) So, \(A'(0,0)\). For vertex \(B(1,0)\): \(x = 2(1) = 2\) \(y = 2(0) = 0\) So, \(B'(2,0)\). For vertex \(C(1,1)\): \(x = 2(1) = 2\) \(y = 2(1) = 2\) So, \(C'(2,2)\). For vertex \(D(0,1)\): \(x = 2(0) = 0\) \(y = 2(1) = 2\) So, \(D'(0,2)\).

Describe the region T(S) in the xy-plane

Now that we have the transformed vertices, we can describe the region \(T(S)\) in the \(xy\)-plane. Since the transformation scales the unit square along both axes by a factor of 2, the image of the unit square is now a square with vertices at \(A'(0,0), B'(2,0), C'(2,2),\) and \(D'(0,2)\). The sides of the transformed square are parallel to the \(x\) and \(y\) axes and have length 2. The resulting region is a square of area 4 in the first quadrant of the \(xy\)-plane.