Verifying Vector Operation Identities

Question

I'm having a hard time verifying these identities, anyone have any suggestions for any of them?

For each Identity $F$ and $G$ denote vector fields, $\phi$ denotes a scalar field, and $R=xi+yj+zk$. $A$ is any constant vector, and f is any differentiable function of a single variable.

The identities I am having trouble verifying are:

1) $\nabla \cdot (R-A)=3$

2) $\nabla \times (R-A)=0$

3) $\nabla(|R-A|^n)$ = $n|R-A|^{n-2}(R-A)$

4) $\nabla(A \cdot R)=A$

Thanks in advance for any and all help!

Answer 1

For #3)

$R=<x,y,z>$, Let $A=<a,b,c>$

$R-A=<x-a,y-b,z-c>$

$|R-A|=\sqrt((x-a)^2 + (y-b)^2 + (z-c)^2)$

$|R-A|^n=[(x-a)^2 + (y-b)^2 + (z-c)^2]^{n/2}$

Then take the derivative w.r.t x,y,z

x: $n/2[(x-a)^2 + (y-b)^2 + (z-c)^2]^{n/2-1} * 2(x-a)$

y: $n/2[(x-a)^2 + (y-b)^2 + (z-c)^2]^{n/2-1} * 2(y-b)$

z: $n/2[(x-a)^2 + (y-b)^2 + (z-c)^2]^{n/2-1} * 2(z-c)$

The 2 in $n/2$ and in each of the $2(x-a)$, $2(y-b)$, $2(z-c)$, will cancel out. Thus you get:

$\nabla |R-A|^n= n[(x-a)^2 + (y-b)^2 + (z-c)^2]^{(n-2)/2} <x-a, y-b, z-c>$

where $[(x-a)^2 + (y-b)^2 + (z-c)^2]^{1/2}$ = $|R-A|^{n-2}$ and

$<x-a, y-b, z-c>$ = $R-A$

Therefore you get:

$n|R-A|^{n-2} * (R-A)$

Answer 2

For #1)

$R-A=<x-a,y-b,z-c>$

Then in order to take $\nabla (R-A)$ you must take the derivative of $(x-a)$ wrt to x, $(y-b)$ wrt y, and $(z-c)$ wrt to z and add them together to get:

$\nabla (R-A) = 1+1+1=3$

Answer 3

For #2)

$R=<x,y,z>, A=<a,b,c>$

$\nabla \times (R-A)$= $curl (R-A)$

$curl (R-A)$ =

$$curl (R-A) = \left|\begin{matrix}i&j&k\\\frac{d}{dx}&\frac{d}{dy}&\frac{d}{dz}\\\ x-a&y-b&z-c\end{matrix}\right|$$

=$(0-0)i -(0-0)j + (0-0)k$=$0$

For #4)

Again $R=<x,y,z>, A=<a,b,c>$

$A \cdot R = ax + by + cz$

$\nabla (A \cdot R) = <\frac{d}{dx}(ax), \frac{d}{dy}(by), \frac{d}{dz}(cz)>$

=$<a,b,c>$ = $A$