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What is the operator between A A^ in A vec = A A^

  1. Oct 5, 2010 #1
    consider a vector A

    A = A(A dir)

    here A dir is A with hat on it. i dont how to do that with LATEX.

    is the operator between A and A dir is multiplication . if yes then why?

    could anybody plz derive the differentiation of A in terms of A and A dir?
     
  2. jcsd
  3. Oct 5, 2010 #2

    HallsofIvy

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    What does "A with a hat on it" mean? What is the difference between A and A?
     
  4. Oct 5, 2010 #3
    A is a vector .
    A is magnitude of vector A
    A dir or "A with a hat on it" is vector whose magnitude is 1 and direction is equal to direction of A

    i've solved and understood "the differentiation of A" , so no need to re-solve it, but i wont mind if u can provide some different approach.
     
  5. Oct 5, 2010 #4

    Fredrik

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    Let's see...you have solved the problem and don't want us to re-solve it. You want us to show you a different approach, but you haven't told us your approach. And the problem doesn't really make sense. Differentiate with respect to what? You're going to have to be more specific.

    Hit the quote button to see how I'm doing the LaTeX. (Note that there's a bug that makes the wrong images show up in previews most of the time. The workaround is to refresh and resend after each preview).

    [tex]\mathbf A=A\hat A[/tex]

    Is the problem to start with [tex]\vec r=r\hat r[/tex] and differentiate with respect to time? Is this a curve in [tex]\mathbb R^2[/tex] or [tex]\mathbb R^3[/tex]?
     
    Last edited: Oct 5, 2010
  6. Oct 5, 2010 #5
    i dont want u to show me a different approach (ofcourse then i would need to provide u my approach which is rather a intuitional approach). what i meant by that is u can resolve it with an approach and if that approach is different than mine then i will be glad.

    "differentiation of A" means differentiation of A with respect to x (let say) then
    either (A depends on x) or (A does not depend on x)

    if (A does not depend on x) then
    [tex]\frac{d\textbf{A}}{dx} = 0 [/tex]
    it will always be zero whatever the operator is.
    which ofcourse i am not asking.

    what i was (or am) asking is
    [tex]\frac{d\textbf{A}}{dx} = ? [/tex]
    when [tex]\textbf{A} = \textbf{A}(x)[/tex],

    what will be
    [tex]\frac{d\textbf{A}}{dx} = ? [/tex]
    when the operator between [tex]A[/tex] and [tex]\hat A[/tex] is not multiplication
     
  7. Oct 5, 2010 #6

    Fredrik

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    If it's not multiplication, then you would have to specify what it is. But why wouldn't it be? Isn't your [tex]\mathbf{\hat A}[/tex] defined by

    [tex]\mathbf A=|A|\frac{\mathbf A}{|A|}=|A|\mathbf{\hat A}\ \text{?}[/tex]

    I don't recommend using the expression A=A(x) to say "A is a function and we use the symbol x for its argument", because it looks like you're setting a function equal a number. (A(x) is a number in the range of the function A).

    The derivative is defined as always

    [tex]\frac{dA(x)}{dx}=\lim_{h\rightarrow 0}\frac{\mathbf{A}(x+h)-\mathbf{A}(x)}{h}[/tex]

    and the product rule for derivatives still holds: (fg)'(x)=f'(x)g(x)+f(x)g'(x)
     
    Last edited: Oct 5, 2010
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