Find the Jacobian of the transformation. $$x = 5 \alpha \sin \beta , y = 4 \alpha \cos \beta$$
A) $\frac { \partial ( x , y ) } { \partial ( \alpha , \beta ) } = 9 \alpha$
B) $\frac { \partial ( x , y ) } { \partial ( \alpha , \beta ) } = 36 \alpha$
C) $\frac { \partial ( x , y ) } { \partial ( \alpha , \beta ) } = - 20 \alpha$
D) $\frac { \partial ( x , y ) } { \partial ( \alpha , \beta ) } = - 20 \alpha \sin \beta \cos \beta$
E) $\frac { \partial ( x , y ) } { \partial ( \alpha , \beta ) } = - \alpha$

Question

Quizplus · Accepted Answer

The Jacobian of the transformation from $(\alpha, \beta)$ to $(x, y)$ is calculated as the determinant of the matrix formed by the partial derivatives of $x$ and $y$ with respect to $\alpha$ and $\beta$. For $x = 5 \alpha \sin \beta$ and $y = 4 \alpha \cos \beta$, the partial derivatives are:$$\frac{\partial x}{\partial \alpha} = 5 \sin \beta, \quad \frac{\partial x}{\partial \beta} = 5 \alpha \cos \beta, \quad \frac{\partial y}{\partial \alpha} = 4 \cos \beta, \quad \frac{\partial y}{\partial \beta} = -4 \alpha \sin \beta.$$The Jacobian is then:$$\begin{vmatrix}5 \sin \beta & 5 \alpha \cos \beta \4 \cos \beta & -4 \alpha \sin \beta\end{vmatrix}= (5 \sin \beta)(-4 \alpha \sin \beta) - (5 \alpha \cos \beta)(4 \cos \beta) = -20 \alpha (\sin^2 \beta + \cos^2 \beta) = -20 \alpha,$$since $\sin^2 \beta + \cos^2 \beta = 1$.

Find the Jacobian of the Transformation x=5αsin⁡β,y=4αcos⁡βx = 5 \alpha \sin \beta , y = 4 \alpha \cos \betax=5αsinβ,y=4αcosβ

Find the Jacobian of the Transformation $x = 5 \alpha \sin \beta , y = 4 \alpha \cos \beta$