# directional derivatives.......i think?

## Recommended Posts

Hello,

I am confused by what this question is asking, and i was wondering if anyone could suggest the best way to approach it?

Thanks

sarah

##### Share on other sites

The object $\nabla$ is a vector $\nabla = {\mathbf i} \frac{\partial}{\partial x}+ {\mathbf j} \frac{\partial}{\partial x}+{\mathbf k} \frac{\partial}{\partial x}$

Multiply out the vectors into coordinate form until you have only normal derivative and see what you get.

For example $|\vec{v} \times \vec{r}|$ is a scalar quantity (just a number), so $\nabla |\vec{v} \times \vec{r}|$ is ${\mathbf i} \frac{\partial |\vec{v} \times \vec{r}|}{\partial x}+ {\mathbf j} \frac{\partial |\vec{v} \times \vec{r}|}{\partial x}+{\mathbf k} \frac{\partial |\vec{v} \times \vec{r}|}{\partial x}$

Now put in the form of $|\vec{v} \times \vec{r}|$ and see what ou get.

##### Share on other sites

The object $\nabla$ is a vector $\nabla = {\mathbf i} \frac{\partial}{\partial x}+ {\mathbf j} \frac{\partial}{\partial x}+{\mathbf k} \frac{\partial}{\partial x}$

Multiply out the vectors into coordinate form until you have only normal derivative and see what you get.

For example $|\vec{v} \times \vec{r}|$ is a scalar quantity (just a number)' date=' so [math'] \nabla |\vec{v} \times \vec{r}|[/math] is ${\mathbf i} \frac{\partial |\vec{v} \times \vec{r}|}{\partial x}+ {\mathbf j} \frac{\partial |\vec{v} \times \vec{r}|}{\partial x}+{\mathbf k} \frac{\partial |\vec{v} \times \vec{r}|}{\partial x}$

Now put in the form of $|\vec{v} \times \vec{r}|$ and see what ou get.

hmm thats kind of what i was thinking but then isnt the answer for part (a) just 0?

also i am unsurewhat it means when it says "f® is a function of r = |r|"?

##### Share on other sites

Are you familiar with the summation convention?

So $\vec{v} \times \vec{r} = \epsilon_{ijk} v_i r_j \vec{e}_k$

Then $\frac{\partial}{\partial r_m} \epsilon_{ijk} v_i r_j \vec{e}_k = \epsilon_{ijk} v_i \delta_{mj} \vec{e}_k = \epsilon_{imk} v_i \vec{e}_k$

Intorduce auxilliary variables to make your life easier. So, e.g. $\vec{a} = \vec{v} \times \vec{r}$ allows you to write

$\nabla |\vec{v} \times \vec{r}| = \nabla |\vec{a}| = \nabla \sqrt{\vec{a} \cdot \vec{a}} = \frac{1}{|\vec{a}|} \vec{a} \cdot \nabla \vec{a}$

and carry on from there.

$f®$ really means $f(\sqrt{\vec{r} \cdot \vec{r}})$. i.e. f only depends on the magnitude of the vector.

##### Share on other sites

ok hold on.

i don't understand how you did this step...

$\nabla \sqrt{\vec{a} \cdot \vec{a}} = \frac{1}{|\vec{a}|} \vec{a} \cdot \nabla \vec{a}$

also, once there i can't quite see where to carry on too..... i think i am missing a key thing about this question (although i don't know what it is!, lol )

p.s.

my reasoning behind the answer to (a) being 0 is this....

if $|\vec{v} \times \vec{r}|$ is a scalar quantity (just a number), so $\nabla |\vec{v} \times \vec{r}|$ is ${\mathbf i} \frac{\partial |\vec{v} \times \vec{r}|}{\partial x}+ {\mathbf j} \frac{\partial |\vec{v} \times \vec{r}|}{\partial x}+{\mathbf k} \frac{\partial |\vec{v} \times \vec{r}|}{\partial x}$

then isnt the derivative of a scalar 0?

##### Share on other sites

what do you think of this as an answer for part (b)...?

$\nabla \times (f®\vec{v}) = \frac{f'®}{\sqrt{x^2+y^2+z^2}} [(yc - zb)i - (xc - za)j + (xb - ya)k]$

##### Share on other sites

For your first question, I think my previous notation was a bit misleading.

$\nabla \sqrt{\vec{a} \cdot \vec{a}} = \frac{1}{2} (\vec{a} \cdot \vec{a})^{-1/2} \nabla (\vec{a} \cdot \vec{a}) = \frac{1}{2 |\vec{a}|} \partial_i (a_j a_j) \vec{e}_i = \frac{1}{|\vec{a}|} a_j \vec{e}_i \partial_i a_j$

Is that clearer?

my reasoning behind the answer to (a) being 0 is this....

Your reasoning is a little odd. A scalar quantity can still change with position, so it can still have a gradiant. For example, temperature is a scalar, but the rate of change of temperature depends on where you are and the direction you are heading.

I agree that the expression for (a) is zero, but your reasoning is a bit squiffy. (You should be able to put the bit from this post together with the bit I did before (the third line of my second post) to get the answer.)

For (b) I got $f'® \frac{1}{r} ( \vec{r} \times \vec{v})$ which seems a little different from your answer. Edit: Actually, I take that back - I think it is the same as your answer, just more compactly written.

The best way to do all of these is via the summation convention, where for example $\vec{a} \cdot \vec{b} = a_ib_i$ with an implicit summation over i, and $\vec{a} \times \vec{b} = \epsilon_{ijk} a_ib_j \vec{e}_k$ where $\epsilon_{ijk}$ is totally anti-symmetric in interchange of indices.

##### Share on other sites

then isnt the derivative of a scalar 0?

you are confusing two different uses of the word scalar.

In ordinary analysis, when one talks of scalars one means constants, fixed unvarying elements of the real numbers (or complex numbers). Thus in this sense, if k is a scalar (a real number) then the derivative of kx is k, and the derivative of k is zero.

However, in this context a scalar means a one dimensional vector. It is not a constant, just something that takes values in the basefield. Think of it as a 'scalar quantity' rather than a 'scalar', and to think of those old 'scalars' whose derivative vanishes as 'constants', if it helps.

|vxr| is a scalar quantity: it is an element of R, but it is not a constant: it changes with r, which is a function of (x,y,z).

##### Share on other sites

I am not sure if I have confused you with using the summation convention or not.

Let me do (b) for you via the summation convention so you can really get what I am meaning. (I am choosing (b) since you have already got it right.)

$\nabla \times (f® \vec{v}) = ( \nabla f® ) \times \vec{v}$

but $\nabla f® = \vec{e}_i \partial_i f® = \vec{e}_i \frac{\partial r}{\partial x_i} \frac{\partial}{\partial r} f® = \vec{e}_i \frac{\partial \sqrt{x_j x_j}}{\partial x_i} f'®$$= \vec{e}_i \frac{1}{2} (x_jx_j)^{-1/2} \frac{\partial (x_j x_j)}{\partial x_i} f'®= \vec{e}_i \frac{1}{2r} 2 x_i f'® = \frac{\vec{r}}{r} f'®$

So $\nabla \times (f® \vec{v}) = \frac{1}{r} f'® \; \vec{r} \times \vec{v}$

This gives a much neater expression in terms of vectors rather than components.

##### Share on other sites

you are confusing two different uses of the word scalar.

In ordinary analysis' date=' when one talks of scalars one means constants, fixed unvarying elements of the real numbers (or complex numbers). Thus in this sense, if k is a scalar (a real number) then the derivative of kx is k, and the derivative of k is zero.

However, in this context a scalar means a one dimensional vector. It is not a constant, just something that takes values in the basefield. Think of it as a 'scalar quantity' rather than a 'scalar', and to think of those old 'scalars' whose derivative vanishes as 'constants', if it helps.

|vxr| is a scalar quantity: it is an element of R, but it is not a constant: it changes with r, which is a function of (x,y,z).[/quote']

ah ok, i see, yes that makes more sense now.

##### Share on other sites

I am not sure if I have confused you with using the summation convention or not.

Let me do (b) for you via the summation convention so you can really get what I am meaning. (I am choosing (b) since you have already got it right.)

$\nabla \times (f® \vec{v}) = ( \nabla f® ) \times \vec{v}$

but $\nabla f® = \vec{e}_i \partial_i f® = \vec{e}_i \frac{\partial r}{\partial x_i} \frac{\partial}{\partial r} f® = \vec{e}_i \frac{\partial \sqrt{x_j x_j}}{\partial x_i} f'®$$= \vec{e}_i \frac{1}{2} (x_jx_j)^{-1/2} \frac{\partial (x_j x_j)}{\partial x_i} f'®= \vec{e}_i \frac{1}{2r} 2 x_i f'® = \frac{\vec{r}}{r} f'®$

So $\nabla \times (f® \vec{v}) = \frac{1}{r} f'® \; \vec{r} \times \vec{v}$

This gives a much neater expression in terms of vectors rather than components.

i sort of understand the summation notation, but i had never seen it before you showed me so its still a little difficult to grasp. Thank you very much for showing me it though.

##### Share on other sites

Stick with it. Summation convention absolutely throws everyone the first time they see it (and boasting 'oh I got it straight away replies' are not wanted), but once you've got used to it you'll not understand how you survived without it and you'll have no problem using it. I mean, I can still remember that

$\epsilon_{ijk}\epsilon_{ipq}= \delta_{jp}\delta_{kq} - \delta_{jq}\delta_{kp}$

and I haven't used it for 8 years or more. (It may seem odd that I say you will not understand how you got by without it whilst saying I've not used it for 9 years, but I chose pure maths where summation convention is rarely used. For instance it is common to let v_i be an eigenvector with eigenvalue k_i and we want Av_i=k_iv_i, and we definitely don't want to sum that.)

##### Share on other sites

Stick with it. Summation convention absolutely throws everyone the first time they see it (and boasting 'oh I got it straight away replies' are not wanted)' date=' but once you've got used to it you'll not understand how you survived without it and you'll have no problem using it. I mean, I can still remember that

[math']\epsilon_{ijk}\epsilon_{ipq}= \delta_{jp}\delta_{kq} - \delta_{jq}\delta_{kp}[/math]

and I haven't used it for 8 years or more. (It may seem odd that I say you will not understand how you got by without it whilst saying I've not used it for 9 years, but I chose pure maths where summation convention is rarely used. For instance it is common to let v_i be an eigenvector with eigenvalue k_i and we want Av_i=k_iv_i, and we definitely don't want to sum that.)

yeah i can see it could prove quite useful in the near future for me, i will keep it in mind.

Thanks for all the help matt & severian!

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account