# Calculus questions

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• Aug 22nd 2006, 03:30 AM
Glaysher
Calculus 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, 03:38 AM
Glaysher
And this one. For (b) replace lower limit 0 with $\frac{1}{cos \theta}$ in answer
• Aug 22nd 2006, 08:36 AM
ThePerfectHacker
Quote:

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|
$

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.
• Aug 22nd 2006, 08:58 AM
ThePerfectHacker
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,
$f_x=\frac{x}{\sqrt{x^2+y^2+z^2}}$

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

$f_z=\frac{z}{\sqrt{x^2+y^2+z^2}}$
• Aug 22nd 2006, 08:58 AM
CaptainBlack
Quote:

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|
$

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
• Aug 22nd 2006, 09:13 AM
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.
• Aug 22nd 2006, 09:22 AM
Glaysher
Yes it did occur to me later that I had made a mistake by missing out the minus. Thanks.
• Aug 22nd 2006, 09:29 AM
Glaysher
Quote:

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|
$

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.
• Aug 22nd 2006, 05:38 PM
sarahisme
Quote:

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))
• Aug 22nd 2006, 05:39 PM
ThePerfectHacker
Quote:

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.
• Aug 23rd 2006, 02:17 AM
sarahisme
Quote:

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

(c)
http://img215.imageshack.us/img215/903/picture6ax5.png
• Aug 23rd 2006, 03:51 AM
Glaysher
Quote:

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.
• Aug 23rd 2006, 07:45 AM
ThePerfectHacker
Quote:

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

(c)
http://img215.imageshack.us/img215/903/picture6ax5.png

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.
• Aug 23rd 2006, 06:43 PM
sarahisme
Quote:

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))

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, 07:38 PM
ThePerfectHacker
Quote:

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

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

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