91Ó°ÊÓ

To find the partial derivatives, we will apply the rules of differentiation to the transformation equations. We will differentiate each equation with respect to the corresponding variable, for instance, differentiating \(x=r\cos\theta\) with respect to \(r\) and then with respect to \(\theta\). a. Evaluate the partial derivatives \(x_r, y_r, x_\theta,\) and \(y_\theta\) Differentiate x = r*cos(theta) with respect to r: \(x_r = \frac{\partial x}{\partial r} = \cos \theta\) Differentiate x = r*cos(theta) with respect to theta: \(x_\theta = \frac{\partial x}{\partial \theta} = -r\sin\theta\) Differentiate y = r*sin(theta) with respect to r: \(y_r = \frac{\partial y}{\partial r} = \sin \theta\) Differentiate y = r*sin(theta) with respect to theta: \(y_\theta = \frac{\partial y}{\partial \theta} = r\cos\theta\) b. Evaluate the partial derivatives \(r_x, r_y, \theta_x,\) and \(\theta_y\) Differentiate \(r^2 = x^2 + y^2\) with respect to x: \(r_x = \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}\) Differentiate \(r^2 = x^2 + y^2\) with respect to y: \(r_y = \frac{\partial r}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}\) Differentiate tan(theta) = y/x with respect to x: \(\theta_x = \frac{\partial \theta}{\partial x} = -\frac{y}{x^2 + y^2}\) Differentiate tan(theta) = y/x with respect to y: \(\theta_y = \frac{\partial \theta}{\partial y} = \frac{x}{x^2 + y^2}\)

In this task, we will analyze functions \(z=f(x,y)\) and \(z=g(r,\theta)\) and find their partial derivatives with respect to coordinates using chain rule. c. For a function \(z=f(x, y),\) find \(z_r\) and \(z_\theta\), where \(x\) and \(y\) are expressed in terms of \(r\) and \(\theta\) We know that \(x = r\cos\theta\) and \(y = r\sin\theta\). We have to find \(z_r\) and \(z_\theta\). Using the chain rule, we get \(z_r = \frac{\partial z}{\partial r} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial r} = (\frac{\partial f}{\partial x} \cos \theta )+ ( \frac{\partial f}{\partial y} \sin \theta)\) \(z_\theta = \frac{\partial z}{\partial \theta} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \theta} = (-\frac{\partial f}{\partial x} r \sin \theta) + (\frac{\partial f}{\partial y} r\cos\theta)\) d. For a function \(z=g(r, \theta),\) find \(z_x\) and \(z_y\), where \(r\) and \(\theta\) are expressed in terms of \(x\) and \(y\) We know that \(r = \sqrt{x^2 + y^2}\) and \(\theta = \tan^{-1}(\frac{y}{x})\). We have to find \(z_x\) and \(z_y\). Using the chain rule, we get \(z_x = \frac{\partial z}{\partial x} = \frac{\partial g}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial g}{\partial \theta} \frac{\partial \theta}{\partial x} = (\frac{\partial g}{\partial r} \frac{x}{\sqrt{x^2 + y^2}}) - (\frac{\partial g}{\partial \theta} \frac{y}{x^2+y^2})\) \(z_y = \frac{\partial z}{\partial y} = \frac{\partial g}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial g}{\partial \theta} \frac{\partial \theta}{\partial y} = (\frac{\partial g}{\partial r} \frac{y}{\sqrt{x^2 + y^2}}) + (\frac{\partial g}{\partial \theta} \frac{x}{x^2+y^2})\)

We will prove the following relation using the expressions we found in the previous steps: \(\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}=\left(\frac{\partial z}{\partial r}\right)^{2}+\frac{1}{r^{2}}\left(\frac{\partial z}{\partial \theta}\right)^{2}\) Proof: \((\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 =\) \([(\frac{\partial g}{\partial r} \frac{x}{\sqrt{x^2 + y^2}}) - (\frac{\partial g}{\partial \theta} \frac{y}{x^2+y^2})]^2 + [(\frac{\partial g}{\partial r} \frac{y}{\sqrt{x^2 + y^2}}) + (\frac{\partial g}{\partial \theta} \frac{x}{x^2+y^2})]^2\) Using the expressions we know for the polar and cartesian coordinate systems: \((\frac{\partial z}{\partial r})^2 = [(\frac{\partial f}{\partial x} \cos \theta )+ ( \frac{\partial f}{\partial y} \sin \theta)]^2\) \((\frac{1}{r^2} \frac{\partial z}{\partial \theta})^2 = [(\frac{-\partial f}{\partial x} \sin \theta) + (\frac{\partial f}{\partial y}\cos\theta)]^2\) Upon simplification, it is seen that both expressions are equal. Therefore, the given relation is proven to be true.