Question

A region R in the xy-plane is given. Find equations for a transformation T that maps a rectangular region S in the uv-plane onto R, where the sides of S are parallel to the u – and v -axes.

lies between the circles \({{x}^{2}}{ + }{{y}^{2}}{ = 1}\) and \({{x}^{2}}{ + }{{y}^{2}}{ = 2}\) in the first quadrant.

Short answer

The required equations are

\(\begin{array}{l}x = u\cos v\\y = u\sin v\end{array}\)

Step-by-step solution

Concept & Given data

In order forandto be parallel to their respective axis, they must be constants. Two things that are constant in the graph above are the distance to the origin (or radius) at the green regions, and the angles at the red regions. To find the distance to the curve or radius, I used the distance formula, or the fact the radius of a circle centered at the origin is.

For the angle, we know that so must be equal to .

\(\begin{aligned}{}u = \sqrt {{x^2} + {y^2}} \\v = \arctan (y/x)\end{aligned}\)

The four equations of the graph were those found in the previous step: \(u = 1,u = \sqrt 2 ,v = 0\),

and ..

Solve for x and y and simplify final calculation

Solve for x,

\(\begin{aligned}{}v &= \arctan \frac{y}{x}\\\tan v &= \frac{y}{x}\\y &= x\tan v\end{aligned}\)

\(\begin{aligned}{{}}u &= \sqrt {{x^2} + {y^2}} \\u &= \sqrt {{x^2} + {x^2}{{\tan ^2}v} }\\u &= x\sqrt {1 + {{\tan }^2v} }\end{aligned}\)

Use trigonometric identity,

\(\begin{aligned}{{}{}}{1 + {\tan }^2}v &= {{\sec }^2}v\\u &= x\sec v\\u &= \frac{x}{{\cos v}}\\x &= u\cos v\end{aligned}\)

Solve for y,

\(tan v =\frac{y}{u\cos v}\)

\(\begin{aligned}y&=u\cos v\tan v \\y&=u\cos v\frac{\sin v}{\cos v} \\y&=u\sin v \\\end{aligned}\)

Hence, the required equations are

\(\begin{align}& x=u\cos v \\ & y=u\sin v \\ \end{align}\)