## The Jacobian of a two variable transformation is found from the determinant of the following matrix.\begin{vmatrix}\frac{\delta x}{\delta \mathcal{u}} & \frac{\delta x}{\delta \mathcal{v}}\\\frac{\delta y}{\delta \mathcal{u}} & \frac{\delta y}{\delta \mathcal{v}}\end{vmatrix}Using the definition of the determinant we find\begin{vmatrix}\frac{\delta x}{\delta \mathcal{u}} & \frac{\delta x}{\delta \mathcal{v}}\\\frac{\delta y}{\delta \mathcal{u}} & \frac{\delta y}{\delta \mathcal{v}}\end{vmatrix} = \frac{\delta x}{\delta u}\frac{\delta y}{\delta v} - \frac{\delta y}{\delta u}\frac{\delta x}{\delta v}Next Calculate all the partial derivatives.\begin{align} \frac{\delta x}{\delta u} = \frac{\delta}{\delta u}u\cos(v) = \cos(v) \\\frac{\delta y}{\delta v} = \frac{\delta}{\delta v}u\sin(v) = u\cos(v) \\\frac{\delta y}{\delta u} = \frac{\delta}{\delta u}u\sin(v) = \sin(v)\\\frac{\delta x}{\delta v} = \frac{\delta}{\delta v}u\cos(v) = -u\cos(v)\\ \end{align}Now substitute and calculate the answer.\begin{equation*}\frac{\delta x}{\delta u}\frac{\delta y}{\delta v} - \frac{\delta y}{\delta v}\frac{\delta x}{\delta v} = u\cos(v)\cos(v) -[-u\sin(v)\sin(v)]\end{equation*}Simplify using the trig identity \cos^2(x) + \sin^2(x) = 1 u[\cos^2(v) + \sin^2(v)] = u $\boxed{\mathcal{J}(u,v) = u}$Following this same process for part b) gives us the following\begin{align} \frac{\delta x}{\delta u} = \frac{\delta}{\delta u}u\sin(v) = \sin(v) \\\frac{\delta y}{\delta v} = \frac{\delta}{\delta v}u\cos(v) = -u\sin(v) \\\frac{\delta y}{\delta u} = \frac{\delta}{\delta u}u\cos(v) = \cos(v)\\\frac{\delta x}{\delta v} = \frac{\delta}{\delta v}u\sin(v) = u\cos(v)\\ \end{align} \begin{equation*}\frac{\delta x}{\delta u}\frac{\delta y}{\delta v} - \frac{\delta y}{\delta v}\frac{\delta x}{\delta v} = -[u\sin(v)\sin(v)] -[u\cos(v)\cos(v)]\end{equation*} -u[\sin^2(v)+\cos^2(v)] = u $\boxed{\mathcal{J}(u,v) = -u}$

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Baylor University

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