Calculus questions

**Glaysher** · August 22nd 2006, 02:30 AM

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

**Glaysher** · August 22nd 2006, 02:38 AM

And this one. For (b) replace lower limit 0 with $\frac{1}{cos \theta}$ in answer

**ThePerfectHacker** · August 22nd 2006, 07:36 AM

Originally Posted by Glaysher

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

$\bold{r}=x\bold{i}+y\bold{j}+z\bold{k}$
---
Then,
$||\bold{r}||=\sqrt{x^2+y^2+z^2}$ because that is the definition of the norm.

Since, $\bold{v}$ is some vector in Euclidean space we can write,
$\bold{v}=a\bold{i}+b\bold{j}+c\bold{k}$
Then,
$\bold{v}\times \bold{r}=\left| \begin{array}{ccc} \bold{i}&\bold{j}&\bold{k}\\ a&b&c\\x&y&z \right| <br />$
Thus,
$\bold{v}\times \bold{r}=(bz-cy)\bold{i}+(az-cx)\bold{j}+(ay-bx)\bold{k}$
$||\bold{v}\times \bold{r}||=\sqrt{(bz-cy)^2+(az-cx)^2+(ay-bx)^2}$
Now, here is the problem.
You cannot find,
$\nabla|\bold{v}\times\bold{r}|$
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.

**ThePerfectHacker** · August 22nd 2006, 07:58 AM

For the second problem, we note that is the curl.
Use the identity,
$\mbox{curl }(f\bold{F})=f\mbox{curl }\bold{F}+(\nabla f\times \bold{F})$
In this case,
$\bold{F}=\bold{v}$ is contant vector function therefore its circulation is a zero-vector.

Thus, you are left with,
$\nabla f(r)\times \bold{v}$
That is,
$\left| \begin{array}{ccc} \bold{i}&\bold{j}&\bold{k}\\ f_x&f_y&f_z\\ a&b&c \end{array} \right|=(cf_y-bf_z)\bold{i}+(cf_x-af_z)\bold{j}+(bf_x-af_y)\bold{k}$
Where, (see other post)
$f=\sqrt{x^2+y^2+z^2}$
Thus,
$f_x=\frac{x}{\sqrt{x^2+y^2+z^2}}$

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

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

**CaptainBlack** · August 22nd 2006, 07:58 AM

Originally Posted by ThePerfectHacker

$\bold{r}=x\bold{i}+y\bold{j}+z\bold{k}$
---
Then,
$||\bold{r}||=\sqrt{x^2+y^2+z^2}$ because that is the definition of the norm.

Since, $\bold{v}$ is some vector in Euclidean space we can write,
$\bold{v}=a\bold{i}+b\bold{j}+c\bold{k}$
Then,
$\bold{v}\times \bold{r}=\left| \begin{array}{ccc} \bold{i}&\bold{j}&\bold{k}\\ a&b&c\\x&y&z \right| <br />$
Thus,
$\bold{v}\times \bold{r}=(bz-cy)\bold{i}+(az-cx)\bold{j}+(ay-bx)\bold{k}$
$||\bold{v}\times \bold{r}||=\sqrt{(bz-cy)^2+(az-cx)^2+(ay-bx)^2}$
Now, here is the problem.
You cannot find,
$\nabla|\bold{v}\times\bold{r}|$
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.

|.| is commonly used for the norm, and so this is the gradient of
a scalar function.

RonL

**ThePerfectHacker** · August 22nd 2006, 08:13 AM

The only complaint for the second set of problems is the second integral.
Glaysher said to replace the lower limit with,
$\frac{1}{\cos x}$ .
In fact, he probably meant to say,
$-\frac{1}{\cos x}$
If soo, it is correct.

**Glaysher** · August 22nd 2006, 08:22 AM

Yes it did occur to me later that I had made a mistake by missing out the minus. Thanks.

**Glaysher** · August 22nd 2006, 08:29 AM

Originally Posted by ThePerfectHacker

$\bold{r}=x\bold{i}+y\bold{j}+z\bold{k}$
---
Then,
$||\bold{r}||=\sqrt{x^2+y^2+z^2}$ because that is the definition of the norm.

Since, $\bold{v}$ is some vector in Euclidean space we can write,
$\bold{v}=a\bold{i}+b\bold{j}+c\bold{k}$
Then,
$\bold{v}\times \bold{r}=\left| \begin{array}{ccc} \bold{i}&\bold{j}&\bold{k}\\ a&b&c\\x&y&z \right| <br />$
Thus,
$\bold{v}\times \bold{r}=(bz-cy)\bold{i}+(az-cx)\bold{j}+(ay-bx)\bold{k}$
$||\bold{v}\times \bold{r}||=\sqrt{(bz-cy)^2+(az-cx)^2+(ay-bx)^2}$
Now, here is the problem.
You cannot find,
$\nabla|\bold{v}\times\bold{r}|$
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.

Yes I think it means the norm. That's how I took it. This question was the only one I remembered how to do and I needed a quick revision first. Thanks again.

**sarahisme** · August 22nd 2006, 04:38 PM

Originally Posted by ThePerfectHacker

The only complaint for the second set of problems is the second integral.
Glaysher said to replace the lower limit with,
$\frac{1}{\cos x}$ .
In fact, he probably meant to say,
$-\frac{1}{\cos x}$
If soo, it is correct.

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))

**ThePerfectHacker** · August 22nd 2006, 04:39 PM

Originally Posted by sarahisme

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))

That is not a radius, it is the angle.

**sarahisme** · August 23rd 2006, 01:17 AM

Originally Posted by ThePerfectHacker

That is not a radius, it is the angle.

but arn't we saying that we are integrating r from -1/cos(theta) to 1/cos(theta)? in which case, can we do that really?

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

(c)

**Glaysher** · August 23rd 2006, 02:51 AM

Originally Posted by sarahisme

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.

**ThePerfectHacker** · August 23rd 2006, 06:45 AM

Originally Posted by sarahisme

but arn't we saying that we are integrating r from -1/cos(theta) to 1/cos(theta)? in which case, can we do that really?

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

(c)

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.
---
I think in some instances the radius can be negative.

**sarahisme** · August 23rd 2006, 05:43 PM

Originally Posted by ThePerfectHacker

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.
---
I think in some instances the radius can be negative.

when you say bowtie region, do you mean this? (the graph of x*cos(x))

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

**ThePerfectHacker** · August 23rd 2006, 06:38 PM

Originally Posted by sarahisme

when you say bowtie region, do you mean this? (the graph of x*cos(x))

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

No,
$|x|\geq |y|$
$|x|\leq 1$

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