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A few Vector calc questions.

  1. Mar 14, 2005 #1
    1. I was reading on the geometric interpretation of the grad operator. I've did until the point where this particular relation was given.

    [tex]d\varphi=0=C_1-C_1=\Delta C=(\nabla \varphi)\bullet d \vec r [/tex]

    This is when we permit [tex] \vec r [/tex] to take us from the surface [tex] \varphi (x,y,z)=C_1[/tex] to another adjacent surface [tex]\varphi (x,y,z)=C_2 [/tex] where c are constants.

    Why is it that in the first equation we have [tex] C_1-C_1[/tex] ?? Also, why is it that the consequence of the first relation shows that,

    for a given [tex]\vert {d\vec r}\vert[/tex], the change of [tex] \varphi [/tex], [tex] d \varphi [/tex] is maximum when [tex]\vert {d\vec r}\vert[/tex] is parellel to [tex] \nabla\varphi [/tex] when [tex] \nabla\varphi [/tex] is normal to the surface.

    2. While evaluating the divergence of vector, [tex] \nabla \bullet \vec r f(r) = \frac {\partial}{\partial x} (xf(r))+ \frac {\partial}{\partial y}(yf(r))+ \frac{\partial}{\partial z} (zf(r)) [/tex],

    why is it equivalent to,

    [tex] 3 f(r)+\frac {x^2}{r} \frac {df}{dr}+\frac {y^2}{r} \frac {df}{dr}+\frac {z^2}{r} \frac {df}{dr} [/tex] ????

    I've tried manipulating the partial differentials using chain rules and all but don't seem to get it. Can someone show me the steps how? Also,

    3. I was trying to simplify[tex]\nabla \times f \vec v \vert _x = \frac {\partial}{\partial y} (f V_z)-\frac {\partial}{\partial z} (f V_y)[/tex]. Also, how does it reduce to the final answer,

    [tex] f \nabla \times \vec V \vert _x +\nabla f \times \vec V \vert _x [/tex]??

    I have also tried manipulating the partial derivatives to no avail. Can someone help??? thanks alot....

    : )
     
    Last edited: Mar 14, 2005
  2. jcsd
  3. Mar 14, 2005 #2

    Galileo

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    Here you should use the product and chain rule. Also: [itex]r=\sqrt{x^2+y^2+z^2}[/itex]

    So for example:

    [tex]\frac{\partial}{\partial x}(xf(r))=f(r)+x\frac{\partial}{\partial x}f(r)[/tex]
    This explains where the 3f(r) comes from.
    Now:

    [tex]\frac{\partial}{\partial x}f(r(x,y,z))=\frac{df}{dr}\frac{\partial r}{\partial x}[/tex]
     
  4. Mar 14, 2005 #3

    HallsofIvy

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    "This is when we permit to take us from the surface [tex] \varphi (x,y,z)=C_1[/tex]
    to another adjacent surface [tex] \varphi (x,y,z)=C_1[/tex] where c are constants."

    No, I don't think that's what that particular formula is intended to show. What that shows is that the derivative in a direction tangent to a level surface is 0 so [tex](\nabla \varphi)\bullet d \vec r= 0 [/tex]: that is, the gradient is perpendicular to all level surfaces.
    What is true is that, since [itex]\vec u \bullet \vec v= |\vec u||\vec v|cos(\theta)[/tex], [tex](\nabla \varphi)\bullet d \vec r [/tex] is largest when cos(θ) is 1: i.e. θ= 0 and the two vectors are parallel.

    For 2, I surprised you didn't see it while looking at the chain rule.
    [tex]r= (x^2+ y^2+ z^2)^{\frac{1}{2}}[/tex] so [tex]\frac{\partial r}{\partial x}= \frac{1}{2}(x^2+ y^2+ z^2)^\frac{-1}{2}(2x)= \frac{x}{r}[/tex].
    Of course, the same is true of the derivative with respect to y and z.
    Now, [tex]\frac{\partial}{\partial x}(xf(r))= f(r)+ x\frac{\partial f}{/partial x}= f(r)+ \frac{x^2 f'(r)}{r}[/tex].

    That first term is what gives you "3f(r)".
     
  5. Mar 14, 2005 #4
    Hey, thanks alot, that helped. Actually what i had problems with was the use of the product rule while evaluating [tex] \partial/ \ {\partial} x (xf(r))[/tex] and i also failed to keep in mind that [tex] r=\sqrt {x^2+y^2+z^2[/tex] but that was sorted out anyway.

    thanks alot. I've still got one question regarding the curl operator though.......hope that someone could help me with that.
     
    Last edited: Mar 14, 2005
  6. Mar 14, 2005 #5

    dextercioby

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    In general

    [tex] (\nabla\times\vec{f})_{x}=...? [/tex]

    Daniel.
     
  7. Mar 14, 2005 #6
    For, [tex] \nabla \times \vec f [/tex], it is,

    [tex] det \left(
    \begin{array}{ccc}

    \hat x & \hat y & \hat z\\
    {\partial}/{\partial x} & {\partial}/{\partial y} & {\partial}/{\partial z} \\
    F_x & F_y & F_z \\

    \end{array}
    \right)

    [/tex]

    (the matrix is quite horrible though, i can't seem to type it the right way, but i guess you know what i mean).

    hmmm, i've tried doing the curl again, i got,

    [tex]\nabla \times f \vec v \vert _x = \frac {\partial}{\partial y} (f V_z)-\frac {\partial}{\partial z} (f V_y)[/tex]

    Using the product rule

    [tex]( f \frac {\partial{V_z}}{\partial {y}} + {V_z}\frac {\partial {f}}{\partial y})-(f \frac {\partial{V_y}}{\partial {z}} + {V_y}\frac {\partial {f}}{\partial z})[/tex]

    Simplifying,

    [tex] \frac {\partial}{\partial {y}} (f V_z +V_z f)-\frac {\partial}{\partial {z}} (f V_y+ V_y f)[/tex]

    [tex] 2f( \frac {\partial}{\partial {y}} V_z - \frac {\partial}{\partial {z}} V_y) [/tex]

    [tex] 2f \nabla \times \vec {v} \vert _x[/tex]

    Where do i go from here? Or are my steps wrong?
     
    Last edited: Mar 14, 2005
  8. Mar 14, 2005 #7

    dextercioby

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    kay,the code is \begin{array}{ccc} (3 columns)...

    Okay.

    Now solve your problem.

    Daniel.
     
  9. Mar 14, 2005 #8
    hmmm, i've tried it in the edited post above.
     
  10. Mar 14, 2005 #9

    dextercioby

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    Well,use the Leibniz rule
    [tex] [\nabla\times (f\vec{V})]_{x} =[(\nabla f\times \vec{V})+f(\nabla\times\vec{V})]_{x} =... [/tex]

    Daniel.
     
  11. Mar 14, 2005 #10
    lol, leibniz's rule did everything. But I've arrived at the solution already, thanks alot.
     
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