exercise 5.3.1. if $s(x,y,z)=(x^{2}+y^{2}+z^{2})^{-3/2}$, solve for the gradient of $s$.

Explanation:

Step1: Recall gradient formula

The gradient of a scalar - valued function $S(x,y,z)$ is given by $\nabla S=\left(\frac{\partial S}{\partial x},\frac{\partial S}{\partial y},\frac{\partial S}{\partial z}\right)$.

Step2: Find $\frac{\partial S}{\partial x}$

Let $u = x^{2}+y^{2}+z^{2}$, so $S = u^{-\frac{3}{2}}$. Using the chain - rule $\frac{\partial S}{\partial x}=\frac{dS}{du}\cdot\frac{\partial u}{\partial x}$. We have $\frac{dS}{du}=-\frac{3}{2}u^{-\frac{5}{2}}$ and $\frac{\partial u}{\partial x} = 2x$. Then $\frac{\partial S}{\partial x}=-\frac{3}{2}(x^{2}+y^{2}+z^{2})^{-\frac{5}{2}}\cdot2x=-3x(x^{2}+y^{2}+z^{2})^{-\frac{5}{2}}$.

Step3: Find $\frac{\partial S}{\partial y}$

Similarly, using the chain - rule with $u = x^{2}+y^{2}+z^{2}$ and $S = u^{-\frac{3}{2}}$. $\frac{dS}{du}=-\frac{3}{2}u^{-\frac{5}{2}}$ and $\frac{\partial u}{\partial y}=2y$. Then $\frac{\partial S}{\partial y}=-\frac{3}{2}(x^{2}+y^{2}+z^{2})^{-\frac{5}{2}}\cdot2y=-3y(x^{2}+y^{2}+z^{2})^{-\frac{5}{2}}$.

Step4: Find $\frac{\partial S}{\partial z}$

Using the chain - rule with $u = x^{2}+y^{2}+z^{2}$ and $S = u^{-\frac{3}{2}}$. $\frac{dS}{du}=-\frac{3}{2}u^{-\frac{5}{2}}$ and $\frac{\partial u}{\partial z}=2z$. Then $\frac{\partial S}{\partial z}=-\frac{3}{2}(x^{2}+y^{2}+z^{2})^{-\frac{5}{2}}\cdot2z=-3z(x^{2}+y^{2}+z^{2})^{-\frac{5}{2}}$.

Answer:

$\nabla S=-3(x^{2}+y^{2}+z^{2})^{-\frac{5}{2}}(x,y,z)$