let r = xi + yj + zk and r = |r|. if f = r/r^p, find div(f). (enter your answer in terms of r and p.) div(f)…

Explanation:

Step1: Expresar $\mathbf{F}$

Dado $\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ y $r = |\mathbf{r}|=\sqrt{x^{2}+y^{2}+z^{2}}$, entonces $\mathbf{F}=\frac{\mathbf{r}}{r^{p}}=\frac{x}{r^{p}}\mathbf{i}+\frac{y}{r^{p}}\mathbf{j}+\frac{z}{r^{p}}\mathbf{k}$.

Step2: Calcular la divergencia de $\mathbf{F}$

La divergencia de un campo vectorial $\mathbf{F}=M\mathbf{i}+N\mathbf{j}+P\mathbf{k}$ está dada por $\text{div}(\mathbf{F})=\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}$. Aquí, $M = \frac{x}{r^{p}}$, $N=\frac{y}{r^{p}}$, $P=\frac{z}{r^{p}}$. Calculamos $\frac{\partial M}{\partial x}$: Usando la regla del cociente y $r=\sqrt{x^{2}+y^{2}+z^{2}}$, entonces $\frac{\partial r}{\partial x}=\frac{x}{r}$. $\frac{\partial M}{\partial x}=\frac{1\cdot r^{p}-x\cdot p r^{p - 1}\frac{\partial r}{\partial x}}{r^{2p}}=\frac{r^{p}-x\cdot p r^{p - 1}\frac{x}{r}}{r^{2p}}=\frac{r^{2}-px^{2}}{r^{p + 2}}$. Análogamente, $\frac{\partial N}{\partial y}=\frac{r^{2}-py^{2}}{r^{p + 2}}$ y $\frac{\partial P}{\partial z}=\frac{r^{2}-pz^{2}}{r^{p + 2}}$. Entonces $\text{div}(\mathbf{F})=\frac{3r^{2}-p(x^{2}+y^{2}+z^{2})}{r^{p + 2}}$. Dado que $r^{2}=x^{2}+y^{2}+z^{2}$, entonces $\text{div}(\mathbf{F})=\frac{3r^{2}-pr^{2}}{r^{p + 2}}=\frac{(3 - p)r^{2}}{r^{p+2}}=\frac{3 - p}{r^{p}}$.

Step3: Encontrar $p$ tal que $\text{div}(\mathbf{F}) = 0$

Igualamos $\text{div}(\mathbf{F})$ a 0: $\frac{3 - p}{r^{p}}=0$. Dado que $r\neq0$ (ya que $r = |\mathbf{r}|$), entonces $3 - p=0$, de donde $p = 3$.

Answer:

$\text{div}(\mathbf{F})=\frac{3 - p}{r^{p}}$ $p = 3$