The given equation is correct:
##\qquad\qquad(\vec A \times \vec B)\times(\vec C \times \vec D) = [ \vec A \cdot (\vec B \times \vec D) ] \vec C - [ \vec A \cdot (\vec B \times \vec C) ] \vec D##
The result I got was
##\qquad\qquad(\vec A \times \vec B)\times(\vec C \times \vec D) = [ \vec A...
The equation to be proven had been improperly written because the vectors in the third and fourth terms had not been properly grouped. Replacing the lower-case-letter vectors with upper case letters, we should have $$\vec \nabla (\vec A\cdot \vec B)=(\vec A \cdot \vec \nabla)\vec B + (\vec B...
Show that $$(1){~~~~~~~~~~~~~~~~~~~~~~~~~}\vec \nabla\cdot(\vec a \times \vec b) = \vec b \cdot (\vec \nabla \times \vec a) - \vec a \cdot (\vec \nabla \times \vec b){~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}$$ Using suffix notation, we get for the left hand side of eq. (1)...
The OP posted this problem-to-prove question:
As I already said in post #11, the correct simplification should lead to$$\vec{\nabla}(\vec{r} \cdot \vec{u}) = \vec{u} + \vec{r}(\vec{\nabla} \cdot \vec{u}) + [~(\vec r \times \vec{\nabla}) \times \vec{u}~]$$But I was wondering if the identity to...
I find the use of brute-force method and suffix notations both tedious and cumbersome. Why don't you try using first useful vector identities to simplify your equation. The BAC minus CAB rule is one such useful identity.
But I noticed in your OP where you wrote:
that you didn't properly group...
I think you're right there. Just two steps away from the answer.
Using ln, just doing two computations won't give you the answer yet. You'll need to do at least three; first the value of ln##~y~##, then the value of (ln##~y##)/##n~##, then that of ##exp[{\frac {\ln y}{n}}]~## before you finally...
@kuruman ... You know fairly well what we were talking about. We were not arguing about the acceptability of settling a $75,000 debt by paying back only $15,000 , which you mentioned in post #26 and that seem to imply, that according to my own point of view, should be acceptable because they...
@jbriggs444 ...
Of course I wasn't talking about any two numbers having the same order of magnitude in general. I was talking to kuruman about these numbers given by the OP
in post #1:
and these numbers that I wrote in post #20:
I know that they are not equal but for the purposes of my...
@kuruman again ... Your post #9 said:
There you mentioned nothing about ##|β|~,## that ##β## can be ##\pm~.## Then, suddenly in post #19, only after 10 messages had already been posted, you surprisingly come up with the idea that ##β~,## in the formula you used in #9 , can be ##\pm~.## Why...
@kuruman ... Of course they are, they both have the same order of magnitude
There are minimum and maximum values of the frequency for any given color. I used the average value.
I'm sorry but I disagree with this assesment in post #17 regarding this thread:
The solutions given in post #6 and #9 are somewhat arbitrary and has some inconsistencies:
I pointed these things out in post #8:
It seems that no attention was paid to my misgivings and they only fell on deaf...
I only use Insert in MS Word to choose the shape that I want and I paste it on the document. Once you click on it the main menu Shape Format pops out and you can do almost anything the way you want it, to almost any measure that you like. It just needs familiarity and lots of practice. If you...
I don't understand this part of your last explanations:
Why is the fraction ##~f~## not ##~f\neq a^2\Delta \phi/(πa^2)?## You used one that is half as small that I think it should be. But if one uses the formula for the area of a triangle ½×Base×Height, which I already mentioned in post #23...
In the second part of your arguments, the one with (math+physics), I don't get this:
Isn't ##~\Delta \phi~## on the plane ⊥ to the ##~z~## axis subtending an ∠ with the ##~x~## and ##~y~## axes? And why do you have the factor 1/2 in the volume enclosed? For an infinitesimal volume, its simply...
I'm sorry but I already disagree with you right at the beginning of your arguments.
$$\dots~{\rm {assumption~1:}} ~r = 0~ {\rm with} ~[E - ρr/(2ε_0)] \neq0~\Rightarrow~E\neq0~\dots$$ $$\dots~{\rm {assumption~2:}}~rE = 0~\Rightarrow~\begin{cases}\begin{align} & ~(i)~r = 0~{\rm...
@kuruman ... I know fully well what open surfaces and closed surfaces are and how they differ from lines. I also happen to know what coaxial cylinders are. I wasn't talking about reducing the ##~r~## of the imaginary Gaussian surface ##~\rm S~## to zero in post #7. I was referring to the range...
So, if one knows what happens in the neighborhood of the point in question, then one can make conclusions about the limit at exactly that point? Then, if one wants to talk next about the limit of the magnitude of the electric field ##~E~## at a point, one also has to examine the continuity of...
@Delta2 ... Thanks for your comments.
But if ##~E=\rho r/2\epsilon_0~## for ##~r\neq 0~##, how did you get ##~\lim_{r\to 0}E=0~## for ##~r = 0~?~## Which expression for ##E~## did you use in evaluating the limit?
Could somebody familiar with the way mathematicians argue please take a look at this and make a comment? Thank you. I'm not a mathematician. What follows is about the electric field at the axis of a long uniformly charged right circular cylinder.
Following his arguments in post #5, he obtained...
Okay, I was wrong. I'm sorry. I accept my mistake. But even without the mathematics, it is clear that ##E = 0## at the center of the uniformly charged cylinder because ##\vec E## run radially away from the axis so that pairs of them moving away opposite one another cancel out. I failed to see...
Given: ... a very long uniformly charged cylinder of radius ##R = 7.2~\rm cm = 0.072~\rm m~##... the volume charge density is ##ρ = 1264~\rm{nC} ⋅ \rm m^{-3} = 1.264~×~10^{-6}~\rm C ⋅ \rm m^{-3}~##...
Unknown: ... the values of##~A~##and##~n~##for the electric field with magnitude##~E =...
Again, ##f(x) = x^2~\Rightarrow~df/dx = 2x## means that the rate of change of the function ##f(x)## with the independent variable ##x## is given by the function ##g(x) = 2x##.
When ##x = 2##, the rate of change of the function ##f(x)## is given by ##g(2) = 2(2) = 4##.
When ##x = 3##, the rate...
Yes, its final value is 4.0 since ##f(x) = x^2~\Rightarrow~df/dx = 2x## and evaluating ##df/dx## at ##x_0 = 2## gives ##f'(x_0) = 2x_0 = 2(2) = 4 = f'(2)##.