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Chapter 16: Problem 21

Compute the Jacobian \(J(u, v)\) for the following transformations. $$T: x=(u+v) / \sqrt{2}, y=(u-v) / \sqrt{2}$$

Short Answer

Expert verified
Question: Compute the Jacobian J(u, v) for the given transformations T: $$x=(u+v) / \sqrt{2}$$ $$y=(u-v) / \sqrt{2}$$ Answer: The Jacobian J(u, v) for the transformations T is: $$J(u, v) = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$$

Step by step solution

01

Differentiate \(x(u, v)\) with respect to \(u\) and \(v\):

Take the partial derivatives of \(x(u, v)\) with respect to \(u\) and \(v\). $$\frac{dx}{du} = \frac{d}{du}\left(\frac{u+v}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}$$ $$\frac{dx}{dv} = \frac{d}{dv}\left(\frac{u+v}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}$$
02

Differentiate \(y(u, v)\) with respect to \(u\) and \(v\):

Now, take the partial derivatives of \(y(u, v)\) with respect to \(u\) and \(v\). $$\frac{dy}{du} = \frac{d}{du}\left(\frac{u-v}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}$$ $$\frac{dy}{dv} = \frac{d}{dv}\left(\frac{u-v}{\sqrt{2}}\right) = -\frac{1}{\sqrt{2}}$$
03

Compute the Jacobian matrix:

Now that we have the partial derivatives, we can construct the Jacobian matrix as follows: $$J(u, v) = \begin{bmatrix} \frac{dx}{du} & \frac{dx}{dv} \\ \frac{dy}{du} & \frac{dy}{dv} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} $$ The requested Jacobian J(u, v) for the transformations T is: $$J(u, v) = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$$

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