 # Find the Jacobian of the transformation. x=u+4v , y=u^2-2v he298c 2021-09-19 Answered
Find the Jacobian of the transformation
$$\displaystyle{x}={u}+{4}{v},{y}={u}^{{2}}-{2}{v}$$

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Basic information:
given:
$$\displaystyle{x}={u}+{4}{v}$$
$$\displaystyle{y}={u}^{{2}}-{2}{v}$$
$$\text{ Jacobian}=\begin{vmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}$$
$$\displaystyle{\frac{{\partial{x}}}{{\partial{u}}}}={\frac{{\partial}}{{\partial{u}}}}{\left({u}+{4}{v}\right)}={u}$$
$$\displaystyle{\frac{{\partial{y}}}{{\partial{u}}}}={\frac{{\partial}}{{\partial{u}}}}{\left({u}^{{2}}-{2}{v}\right)}={2}{u}$$
we thear v as command.
$$\displaystyle{\frac{{\partial{x}}}{{\partial{v}}}}={\frac{{\partial}}{{\partial{v}}}}{\left({u}+{4}{v}\right)}={4}$$
$$\displaystyle{\frac{{\partial{y}}}{{\partial{v}}}}={\frac{{\partial}}{{\partial{v}}}}{\left({u}^{{2}}-{2}{v}\right)}=-{2}$$
we thear u as command.
Calculation of Jacobian
Now, $$\displaystyle{\frac{{\partial{x}}}{{\partial{u}}}}={1},{\frac{{\partial{x}}}{{\partial{v}}}}={4}$$
$$\displaystyle{\frac{{\partial{y}}}{{\partial{u}}}}={2}{u},{\frac{{\partial{y}}}{{\partial{v}}}}=-{2}$$
$$\text{ Jacobian}=\begin{vmatrix}1 & 4 \\2u & -2 \end{vmatrix}$$
$$\displaystyle={\left(-{2}\right)}{\left({1}\right)}-{\left({4}\right)}{\left({2}{u}\right)}$$
$$\displaystyle=-{8}{u}-{2}$$
$$\displaystyle\ \text{ Jacobian}=-{8}{u}-{2}$$