# Emergency help

1. Aug 12, 2004

### kelvintc

I encountered many problems while doing exercises in text books. And i have stated it down in a word document attached in this post. Hope someone can help and teach me how to solve those problems. Answers are given. I just don't know how to get those answer.

Thanks a lot.

p/s : Emergency = next week test.

(2), (4), (5), (6) solved.. thanks

#### Attached Files:

• ###### em_questions.doc
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Last edited: Aug 13, 2004
2. Aug 12, 2004

### arildno

Welcome to PF!

You must show some of your own work here; don't expect your homework to be done for you!
Having said that, let's take a specific question, nr. 4:
Now, what are your problems with this particular exercise?
Make a detailed comment on this.

3. Aug 12, 2004

### kelvintc

r dot grad T should be a scalar ? i wonder y answer's a vector. I used formulas and can't get those answers. Just can't understand why. So seeking help. Thanks.

4. Aug 12, 2004

### arildno

The answer should be a vector, as it is given in the answer (i haven't checked if the answer given is correct)

$$grad=\vec{a}_{x}\frac{\partial}{\partial{x}}+\vec{a}_{y}\frac{\partial}{\partial{y}}+\vec{a}_{z}\frac{\partial}{\partial{z}}$$

For example, the divergence of a vector $$\vec{v}$$ is given by:
$$div\vec{v}=grad\cdot\vec{v}$$

Are you familiar with this notation?

5. Aug 12, 2004

### kelvintc

actually i dunno what means (r dot grad) ? i know div and grad as well
p/s: how you draw those symbols?

6. Aug 12, 2004

### arildno

You can click on the LATEX code to see how you write it .

OK, so you know the "grad", which I'll henceforth write as $$\nabla$$

Let's first review how we get the scalar known as "divergence"
($$\nabla\cdot\vec{v}$$)
Let $$\vec{v}=u\vec{a}_{x}+v\vec{a}_{y}+w\vec{a}_{z}$$

We then have that:
$$\nabla\cdot\vec{v}=\vec{a}_{x}\cdot(\frac{\partial}{\partial{x}}\vec{v})+\vec{a}_{y}\cdot(\frac{\partial}{\partial{y}}\vec{v})+\vec{a}_{z}\cdot(\frac{\partial}{\partial{z}}\vec{v})$$

This simplifies to, in our case:
$$\nabla\cdot\vec{v}=\frac{\partial{u}}{\partial{x}}+\frac{\partial{v}}{\partial{y}}+\frac{\partial{w}}{\partial{z}}$$

(Please comment if this doesn't make sense to you!)

Now, we're ready to tackle $$\vec{v}\cdot\nabla$$
This is also a "dot" product (scalar product), and looks like:
$$\vec{v}\cdot\nabla=u\vec{a}_{x}\cdot\nabla+v\vec{a}_{y}\cdot\nabla+w\vec{a}_{z}\cdot\nabla$$

Or, simplified:
$$\vec{v}\cdot\nabla=u\frac{\partial}{\partial{x}}+v\frac{\partial}{\partial{y}}+w\frac{\partial}{\partial{z}}$$

This is a "scalar" operator which you then apply on T.

7. Aug 12, 2004

### arildno

8. Aug 12, 2004

### kelvintc

oic thanks

9. Aug 12, 2004

### kelvintc

in q 5 : i found the surface integral = -2, but line integral = 7/6. it didn't match Stoke's theorem.. i wonder y....

10. Aug 12, 2004

### kelvintc

i never learn that inverse dot product b4.. hehe... thanks

11. Aug 12, 2004

### kelvintc

and q 2 i really have no idea

12. Aug 12, 2004

### arildno

OK, first:
Have you checked that you get no.4 right?

Secondly, try and group together a few questions you think "belong" to each other, which you would like to focus on.

13. Aug 12, 2004

### kelvintc

no.1 to 6 are about vectors. All about integrals.. I　ｊｕｓｔ　ｗｏｎｄｅｒ y can't get answers using formulas. No.2 is really dunno how to start also.

14. Aug 12, 2004

### kelvintc

no.5 curl = -x^2 in z direction.. then integral can't get 7/6

15. Aug 12, 2004

### arildno

OK, we'll look into 2 (but you did check 4, or what?).

2.
Now, you've been given equations for two surfaces.
In general, if you have a surface given on the form S(x,y,z)=0 (or constant),
you know that the normal on that surface at a point (x,y,z) is parallell to the gradient of S (evaluated on the same point).
Post what you get here in some detail.

16. Aug 12, 2004

### kelvintc

17. Aug 12, 2004

### arildno

Post what you've done. In detail.

18. Aug 12, 2004

### kelvintc

1) since z=0, $$a_{r}$$ has no $$a_{x}$$ component , $$a_{\phi} = -\sin{\phi}{a}_{r} + \cos{\phi}{a}_{\phi}$$ ... and i can't t the answer for (a) and so on

Last edited: Aug 12, 2004
19. Aug 12, 2004

### arildno

I meant on Q4; the one I started with.

20. Aug 12, 2004

### kelvintc

(r dot grad T ) = $$(\vec{r}\cdot\nabla){T}={{x}a_{x}}\frac{\partial{2zy}}{\partial{x}}+{{y}a_{y}}\frac{\partial{xy^2}}{\partial{y}}+{{z}a_{z}}\frac{\partial{x^2yz}}{\partial{z}}$$
is it?

Last edited: Aug 12, 2004