Question

Show that the function satisfies Laplace's equation \(\partial^{2} z / \partial x^{2}+\partial^{2} z / \partial y^{2}=0\). \(z=\arctan \frac{y}{x}\)

Short answer

Yes, the function \(z=\arctan \frac{y}{x}\) indeed satisfies Laplace's equation since the sum of its second partial derivatives equals zero.

Step-by-step solution

Compute the first partial derivatives

The first partial derivatives of the function \(z=\arctan \frac{y}{x}\) can be computed via the chain rule. For the derivative with respect to \(x\), we will need to differentiate \(\arctan(u)\), with \(u=\frac{y}{x}\), which will give \(\frac{1}{1+u^2} \cdot \frac{du}{dx}\). The derivative of \(u\) with respect to \(x\) is \(\frac{-y}{x^2}\). Thus, the first derivative of \(z\) with respect to \(x\) is \(\frac{\partial z}{\partial x} = \frac{-y}{x^2+y^2}\). Similar process leads to the first derivative of \(z\) with respect to \(y\), yielding \(\frac{\partial z}{\partial y} = \frac{x}{x^2+y^2}\)

Compute the second partial derivatives

Now we compute the second partial derivatives by differentiating the first derivative results again. We use the quotient rule. For the second derivative with respect to \(x\), we get \(\frac{\partial^2 z}{\partial x^2} = \frac{2x^{2}y}{(x^2 + y^2)^2}\). For the second derivative with respect to \(y\), we get \(\frac{\partial^2 z}{\partial y^2} = \frac{-2xy^{2}}{(x^2 + y^2)^2}\)

Add the second partial derivatives

Having computed the second partial derivatives, we now add them together to verify if they sum to zero. So, \(\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = \frac{2x^{2}y}{(x^2 + y^2)^2} - \frac{2xy^{2}}{(x^2 + y^2)^2}\) which simplifies to 0.