Let $x = e ^ { u } \cos v$ and $y = e ^ { u } \sin v$ Then $\frac { \partial ( x , y ) } { \partial ( u , v ) }$ is

A) $u e ^ { u }$
B) $\boldsymbol { u }$
C) $u$
D) $e ^ { 2 u }$
E) $- e ^ { 2 u }$

Question

Let $x = e ^ { u } \cos v$ and $y = e ^ { u } \sin v$ Then $\frac { \partial ( x , y ) } { \partial ( u , v ) }$ is

A) $u e ^ { u }$
B) $\boldsymbol { u }$
C) $u$
D) $e ^ { 2 u }$
E) $- e ^ { 2 u }$

Quizplus · Accepted Answer

The Jacobian determinant $\frac { \partial ( x , y ) } { \partial ( u , v ) }$ is calculated as follows: $$\begin{vmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix}= \begin{vmatrix}e^u \cos v & -e^u \sin v \e^u \sin v & e^u \cos v\end{vmatrix}= (e^u \cos v)(e^u \cos v) - (-e^u \sin v)(e^u \sin v) = e^{2u}(\cos^2 v + \sin^2 v) = e^{2u}.$$

Let x=eucos⁡vx = e ^ { u } \cos vx=eucosv And y=eusin⁡vy = e ^ { u } \sin vy=eusinv

Let $x = e ^ { u } \cos v$ And $y = e ^ { u } \sin v$