1. I was reading on the geometric interpretation of the grad operator. I've did until the point where this particular relation was given.(adsbygoogle = window.adsbygoogle || []).push({});

[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....

: )

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# A few Vector calc questions.

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