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
$\displaystyle \bold{r}=x\bold{i}+y\bold{j}+z\bold{k}$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.
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,
$\displaystyle f_x=\frac{x}{\sqrt{x^2+y^2+z^2}}$
$\displaystyle f_y=\frac{y}{\sqrt{x^2+y^2+z^2}}$
$\displaystyle f_z=\frac{z}{\sqrt{x^2+y^2+z^2}}$
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))
And welcome. I could argue this both ways. A negative radius would mean that it would be pointing in the opposite direction as a vector with a negative magnitude would. However I am aware that some polar coordinate systems have been defined so that r > 0 in which case you would have 0 as the lower limit.Originally Posted by sarahisme
I understand what your problem is. If I was doing this question on a test. I would divide those to regions (the bowtie) into two regions. Then the entire problem goes away.Originally Posted by sarahisme
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I think in some instances the radius can be negative.