Question

Given R= xi + yj + zk, and r=l R l, determine:

Answer 1

(a)\[\nabla r ^{n}\]
(b)\[\nabla . (r ^{n} R)\]

Answer 2

if I could read the notation i might have a clue :/

gradient r^n I got no idea what that even implies

Answer 3

lol i tried reading up materials from ka stroud, james, etc etc and i couldnt understand the question also.. T_T

Answer 4

im thinking it might be russian :)

Answer 5

http://www.wolframalpha.com/input/?i=grad+r%5En

Answer 6

since r = sqrt(x^2+y^2+z^2) hmmm

Answer 7

\[r^n = (x^2+y^2+z^2)^{n/2}\]
but i got no clue what thats spose to be

Answer 8

unless, we take the derivative of that

Answer 9

i got no idea how to make the funny d so thisll be:

\[grad(r^n) = grad(x^2+y^2+z^2)^{n/2}\]\[\hspace{5em}=\frac{n}{2}(x^2+y^2+z^2)^{(n-1)/2}*2x\frac{df}{dx},\frac{n}{2}(x^2+y^2+z^2)^{(n-1)/2}*2y\frac{df}{dy}\]\[\hspace{10em},\frac{n}{2}(x^2+y^2+z^2)^{(n-1)/2}*2z\frac{df}{dz}\]

Answer 10

\[(x(x^2+y^2+z^2)^{(n-1)/2}\frac{df}{dx})\ i+...\]\[(y(x^2+y^2+z^2)^{(n-1)/2}\frac{df}{dy})\ j+...\]\[(z(x^2+y^2+z^2)^{(n-1)/2}\frac{df}{dz})\ k\]

maybe

Answer 11

forgot my ns :) xn yn and zn is what x y z are spose to be

Answer 12

r^n R looks to be a scalar; then gradiented

Answer 13

@.@

Answer 14

oo

Answer 15

i got no idea how to make this make sense tho lol

Answer 16

the upside down triangle is notation for gradient; which is just a derivative notation as well

Answer 17

and i prolly got some spurious looking dfs in there in hindsight

Answer 18

\[r ^{n}=\left( x ^{2}+y ^{2}+z ^{2} \right)^{n/2}, R=<x,y,z>\]
the scalar product between them is it \[<x\left( x ^{2}+y ^{2}+z ^{2} \right)^{n/2}, y\left( x ^{2}+y ^{2}+z ^{2} \right)^{n/2},z\left( x ^{2}+y ^{2}+z ^{2} \right)^{n/2}>\]?
if yes, then the del of it is it\[<nx ^{2}\left( x ^{2}+y ^{2}+z ^{2} \right)^{(n/2)-1}, ny ^{2}\left( x ^{2}+y ^{2}+z ^{2} \right)^{(n/2)-1},nz ^{2}\left( x ^{2}+y ^{2}+z ^{2} \right)^{(n/2)-1}>\]?

Answer 19

amistre, sry to bother you, can check whether my answer is correct, given if part a is correct?

Answer 20

\[<x\left( x ^{2}+y ^{2}+z ^{2} \right)^{n/2}, y\left( x ^{2}+y ^{2}+z ^{2} \right)^{n/2},z\left( x ^{2}+y ^{2}+z ^{2} \right)^{n/2}>\]
the scalar part looks good, assuming we are onthe right track.

one issue is the you have\[\nabla . \vec {rR}\]

which looks more like its asking for the dot product of the gradient and the rR .. unless thats a typo

Answer 21

From:

http://www.mathrec.org/vector.html

The divergence operator is normally written as the dot product between a nabla and a vector function. When fonts do not allow, the divergence is usually written as div f. Verbally, the divergence of f is often refered to as "del dot f". The divergence is a scalar quantity.

A serious discussion of the physical significance of the divergence is beyond the scope of this page, but if you think of the vector field as a fluid flow or a current, the divergence shows where fluid or charge is created or destroyed. The Cartesian form is given by:

div f = (∂/∂x, ∂/∂y, ∂/∂z) · (fx, fy, fz) = ∂fx/∂x + ∂fy/∂y + ∂fz/∂z

Answer 22

http://books.google.com/books?id=I-x1MLny6y8C&pg=PA32&lpg=PA32&dq=del+dot+vector&source=bl&ots=ylI8Rx4tJZ&sig=Y3axEbBf3AtAYo9KIuxWPI0auWQ&hl=en&sa=X&ei=Ye9lT6OvDqaqsQK_jLW2Dw&ved=0CG8Q6AEwBw#v=onepage&q=del%20dot%20vector&f=false

this might be insightful as well

Answer 23

as long as we have defined r^n R correctly;

then we simply add up the partial derivatives of it is what it looks like to me.

\[\vec A = \vec {r^nR}\]
\[\nabla.\vec A = A_x+A_y+A_z \]
where those are the partial derivatives of A with respect to x y and z