Question

Let \(\mathbf{F}(x, y)=\frac{y}{x^{2}+y^{2}} \mathbf{i}-\frac{x}{x^{2}+y^{2}} \mathbf{j}\) (a) Show that \(\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}\) where \(M=\frac{y}{x^{2}+y^{2}}\) and \(N=\frac{-x}{x^{2}+y^{2}} .\) (b) If \(\mathbf{r}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}\) for \(0 \leq t \leq \pi,\) find \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) (c) If \(\mathbf{r}(t)=\cos t \mathbf{i}-\sin t \mathbf{j}\) for \(0 \leq t \leq \pi,\) find \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\). (d) If \(\mathbf{r}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}\) for \(0 \leq t \leq 2 \pi,\) find \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\). Why doesn't this contradict Theorem \(13.7 ?\) (e) Show that \(\nabla\left(\arctan \frac{x}{y}\right)=\mathbf{F}\).

Short answer

1. Proof of \(\frac{\partial N}{\partial x} = \frac{\partial M}{\partial y}\): Do the computation After the computation, you will find \(\frac{\partial N}{\partial x} = \frac{\partial M}{\partial y}\)\n\n2. Results for the line integrals:\n(b) 0\n(c) 0\n(d) does not exist\n\n3. The contradiction with theorem 13.7 arises, because the field \(\mathbf{F}\) is not defined at the origin (0,0).\n\n4. The gradient of \(\arctan(\frac{x}{y})\) does indeed equal \(\mathbf{F}\).\nProof: Compute the gradient and compare with \(\mathbf{F}\).

Step-by-step solution

Compute Partial Derivatives

First, calculate the partial derivatives for \(M\) and \(N\).\nThe derivative of \(M = \frac{y}{x^{2}+y^{2}}\) with respect to \(y\), and the derivative of \(N = -\frac{x}{x^{2}+y^{2}}\) with respect to \(x\). These will be compared to confirm part (a).

Solve Integrals

For parts (b), (c), (d), compute the integral of \(\mathbf{F} \cdot d \mathbf{r}\) along the given path defined by \(\mathbf{r}(t)\). This is achieved by first substituting \(\mathbf{r}(t)\) into \(\mathbf{F}\) and then computing the dot product with the differential of \(\mathbf{r}(t)\). The result is then integrated over the interval for \(t\).

Check for Contradiction

Review theorem 13.7 to understand the conditions under which the result of the line integral should depend only on the end points of the curve and evaluate whether these conditions are met in this situation.

Compute Gradient of Arctan(x/y)

For part (e), take the gradient of the scalar field function \(\arctan (\frac{x}{y})\) and verify that it equals to \(\mathbf{F}\). This involves calculating the partial derivatives of the field function with respect to \(x\) and \(y\) and forming a new vector from these.