Q. 32 - Calculus

Question

In Exercises 27–32, functions x = x(u, v) and y = y(u, v) are given that determine transformations from an XY-coordinate system to a UV-coordinate system in R2. Use these functions to determine a region in the XY-plane that has the image specified for the given values of u and v, and find the Jacobian of the transformation.

$x = u sec v and y = u \tan v for 0 \le u \le 2 and 0 \le v \le \frac{\mathrm{\pi }}{4}$

Step-by-Step Solution

Answer

The Jacobian is equal to $J=u sec v$.

Step 1: Given information

The functions are,

$x = u sec v and y = u \tan v for 0 \le u \le 2 and 0 \le v \le \frac{\mathrm{\pi }}{4}$

Step 2: Find the Jacobian

The Jacobian is computed as,

$$\begin{gathered} \frac{\partial (x, y)}{\partial (u,v)}=det\left[\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial y}{\partial u}\\ \frac{\partial x}{\partial v}& \frac{\partial y}{\partial v}\end{array}\right] \\ \frac{\partial (x, y)}{\partial (u,v)}=det\left[\begin{array}{cc}sec v& \tan v\\ u sec v \tan v& u se{c}^2 v\end{array}\right] \\ \frac{\partial (x, y)}{\partial (u,v)}=use{c}^3 v-u sec v {\tan }^{2}v \\ \frac{\partial (x, y)}{\partial (u,v)}=use{c}^3 v-u sec v \left(se{c}^{2}v-1\right) \\ \frac{\partial (x, y)}{\partial (u,v)}=use{c}^3 v-u se{c}^3 v+u sec v \\ \frac{\partial (x, y)}{\partial (u,v)}=u sec v \end{gathered}$$