# A few Vector calc questions.

1. Mar 14, 2005

### misogynisticfeminist

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

$$d\varphi=0=C_1-C_1=\Delta C=(\nabla \varphi)\bullet d \vec r$$

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

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

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

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

why is it equivalent to,

$$3 f(r)+\frac {x^2}{r} \frac {df}{dr}+\frac {y^2}{r} \frac {df}{dr}+\frac {z^2}{r} \frac {df}{dr}$$ ????

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$$\nabla \times f \vec v \vert _x = \frac {\partial}{\partial y} (f V_z)-\frac {\partial}{\partial z} (f V_y)$$. Also, how does it reduce to the final answer,

$$f \nabla \times \vec V \vert _x +\nabla f \times \vec V \vert _x$$??

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

: )

Last edited: Mar 14, 2005
2. Mar 14, 2005

### Galileo

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

So for example:

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

$$\frac{\partial}{\partial x}f(r(x,y,z))=\frac{df}{dr}\frac{\partial r}{\partial x}$$

3. Mar 14, 2005

### HallsofIvy

Staff Emeritus
"This is when we permit to take us from the surface $$\varphi (x,y,z)=C_1$$
to another adjacent surface $$\varphi (x,y,z)=C_1$$ 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 $$(\nabla \varphi)\bullet d \vec r= 0$$: 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], $$(\nabla \varphi)\bullet d \vec r$$ is largest when cos(&theta;) is 1: i.e. &theta;= 0 and the two vectors are parallel.

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

That first term is what gives you "3f(r)".

4. Mar 14, 2005

### misogynisticfeminist

Hey, thanks alot, that helped. Actually what i had problems with was the use of the product rule while evaluating $$\partial/ \ {\partial} x (xf(r))$$ and i also failed to keep in mind that $$r=\sqrt {x^2+y^2+z^2$$ 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
5. Mar 14, 2005

### dextercioby

In general

$$(\nabla\times\vec{f})_{x}=...?$$

Daniel.

6. Mar 14, 2005

### misogynisticfeminist

For, $$\nabla \times \vec f$$, it is,

$$det \left( \begin{array}{ccc} \hat x & \hat y & \hat z\\ {\partial}/{\partial x} & {\partial}/{\partial y} & {\partial}/{\partial z} \\ F_x & F_y & F_z \\ \end{array} \right)$$

(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,

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

Using the product rule

$$( f \frac {\partial{V_z}}{\partial {y}} + {V_z}\frac {\partial {f}}{\partial y})-(f \frac {\partial{V_y}}{\partial {z}} + {V_y}\frac {\partial {f}}{\partial z})$$

Simplifying,

$$\frac {\partial}{\partial {y}} (f V_z +V_z f)-\frac {\partial}{\partial {z}} (f V_y+ V_y f)$$

$$2f( \frac {\partial}{\partial {y}} V_z - \frac {\partial}{\partial {z}} V_y)$$

$$2f \nabla \times \vec {v} \vert _x$$

Where do i go from here? Or are my steps wrong?

Last edited: Mar 14, 2005
7. Mar 14, 2005

### dextercioby

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

Okay.

Daniel.

8. Mar 14, 2005

### misogynisticfeminist

hmmm, i've tried it in the edited post above.

9. Mar 14, 2005

### dextercioby

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

Daniel.

10. Mar 14, 2005

### misogynisticfeminist

lol, leibniz's rule did everything. But I've arrived at the solution already, thanks alot.