Someone on another much quieter forum has requested help on these problems. I'm too far out of uni to remember. I've told her about this site so she may turn up here later but to save time please could you provide solutions for these

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- Aug 22nd 2006, 02:30 AMGlaysherCalculus questions
Someone on another much quieter forum has requested help on these problems. I'm too far out of uni to remember. I've told her about this site so she may turn up here later but to save time please could you provide solutions for these

- Aug 22nd 2006, 02:38 AMGlaysher
And this one. For (b) replace lower limit 0 with $\displaystyle \frac{1}{cos \theta}$ in answer

- Aug 22nd 2006, 07:36 AMThePerfectHackerQuote:

Originally Posted by**Glaysher**

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Then,

$\displaystyle ||\bold{r}||=\sqrt{x^2+y^2+z^2}$ because that is the definition of the**norm**.

Since, $\displaystyle \bold{v}$ is some vector in Euclidean space we can write,

$\displaystyle \bold{v}=a\bold{i}+b\bold{j}+c\bold{k}$

Then,

$\displaystyle \bold{v}\times \bold{r}=\left| \begin{array}{ccc} \bold{i}&\bold{j}&\bold{k}\\ a&b&c\\x&y&z \right|

$

Thus,

$\displaystyle \bold{v}\times \bold{r}=(bz-cy)\bold{i}+(az-cx)\bold{j}+(ay-bx)\bold{k}$

$\displaystyle ||\bold{v}\times \bold{r}||=\sqrt{(bz-cy)^2+(az-cx)^2+(ay-bx)^2}$

Now, here is the problem.

You cannot find,

$\displaystyle \nabla|\bold{v}\times\bold{r}|$

Because the gradient only exists for*scalar-functions*, i.e. real functions. Another problem is what do you mean by | | do you mean the norm? It cannot mean absolute value because the cross product is a vector function is has not such concept as positive or negative. - Aug 22nd 2006, 07:58 AMThePerfectHacker
For the second problem, we note that is the

**curl**.

Use the identity,

$\displaystyle \mbox{curl }(f\bold{F})=f\mbox{curl }\bold{F}+(\nabla f\times \bold{F})$

In this case,

$\displaystyle \bold{F}=\bold{v}$ is contant vector function therefore its circulation is a zero-vector.

Thus, you are left with,

$\displaystyle \nabla f(r)\times \bold{v}$

That is,

$\displaystyle \left| \begin{array}{ccc} \bold{i}&\bold{j}&\bold{k}\\ f_x&f_y&f_z\\ a&b&c \end{array} \right|=(cf_y-bf_z)\bold{i}+(cf_x-af_z)\bold{j}+(bf_x-af_y)\bold{k}$

Where, (see other post)

$\displaystyle f=\sqrt{x^2+y^2+z^2}$

Thus,

$\displaystyle f_x=\frac{x}{\sqrt{x^2+y^2+z^2}}$

$\displaystyle f_y=\frac{y}{\sqrt{x^2+y^2+z^2}}$

$\displaystyle f_z=\frac{z}{\sqrt{x^2+y^2+z^2}}$ - Aug 22nd 2006, 07:58 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

a scalar function.

RonL - Aug 22nd 2006, 08:13 AMThePerfectHacker
The only complaint for the second set of problems is the second integral.

Glaysher said to replace the lower limit with,

$\displaystyle \frac{1}{\cos x}$.

In fact, he probably meant to say,

$\displaystyle -\frac{1}{\cos x}$

If soo, it is correct. - Aug 22nd 2006, 08:22 AMGlaysher
Yes it did occur to me later that I had made a mistake by missing out the minus. Thanks.

- Aug 22nd 2006, 08:29 AMGlaysherQuote:

Originally Posted by**ThePerfectHacker**

- Aug 22nd 2006, 04:38 PMsarahismeQuote:

Originally Posted by**ThePerfectHacker**

Hello,

I am the one Glaysher is asking the questions for :P

I was wondering with regard to part (b) of the second question especially, can you really make the radius go 'through' 0? that is, how can the radius have a negative value in polar coordinates? (i.e. -1/cos(theta)) - Aug 22nd 2006, 04:39 PMThePerfectHackerQuote:

Originally Posted by**sarahisme**

- Aug 23rd 2006, 01:17 AMsarahismeQuote:

Originally Posted by**ThePerfectHacker**

this is my working for parts (b) and (c), what do you guys think?

**(b)**

http://img241.imageshack.us/img241/7764/picture3qq0.png

**(c)**

http://img215.imageshack.us/img215/903/picture6ax5.png - Aug 23rd 2006, 02:51 AMGlaysherQuote:

Originally Posted by**sarahisme**

__>__0 in which case you would have 0 as the lower limit. - Aug 23rd 2006, 06:45 AMThePerfectHackerQuote:

Originally Posted by**sarahisme**

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I think in some instances the radius can be negative. - Aug 23rd 2006, 05:43 PMsarahismeQuote:

Originally Posted by**ThePerfectHacker**

http://img201.imageshack.us/img201/7...ure1qt6.th.png

also, what do you think of this as an answer for part (c)?

http://img201.imageshack.us/my.php?i...icture2np1.png - Aug 23rd 2006, 06:38 PMThePerfectHackerQuote:

Originally Posted by**sarahisme**

$\displaystyle |x|\geq |y|$

$\displaystyle |x|\leq 1$