You are Here: Home >< Maths

# Differentiating a Vector Field

Announcements Posted on
Would YOU be put off a uni with a high crime rate? First 50 to have their say get a £5 Amazon voucher! 27-10-2016
1. Hi,

Just need a bit of help with differentiating a vector field.

I need to find

I am unsure what to differentiate with respect to as i need to have a vector as the outcome. Do I differentiate w.r.t x for i, w.r.t y for j, and w.r.t to z for k?

Any help is appreciated.
2. If is a vector then is a matrix, not a vector, where (for instance) .

If you were differentiating the -component w.r.t. and the -component w.r.t. and so on then what you'd have would be and not .
3. (Original post by nuodai)
If is a vector then is a matrix, not a vector, where (for instance) .

If you were differentiating the -component w.r.t. and the -component w.r.t. and so on then what you'd have would be and not .
Okay, thank you. Thats a big help because the question is to find the values a,b,c such that F is soleneidal. i.e.
4. (Original post by HappyHammer15)
Okay, thank you. Thats a big help because the question is to find the values a,b,c such that F is soleneidal. i.e.
Right. Bear in mind that is the sum of the respective derivatives of the components, so it's not a vector. (I didn't make this clear in my last post.)
5. (Original post by nuodai)
Right. Bear in mind that is the sum of the respective derivatives of the components, so it's not a vector. (I didn't make this clear in my last post.)
Okay thank you, however, in this example i do not get a variable c in the . Thank you for your help thus far.
6. (Original post by HappyHammer15)
Okay thank you, however, in this example i do not get a variable c in the . Thank you for your help thus far.
So?
7. (Original post by around)
So?
I get a=-0.5 and b = 2. Is c an arbitrary constant then?
8. (Original post by nuodai)
Right. Bear in mind that is the sum of the respective derivatives of the components, so it's not a vector. (I didn't make this clear in my last post.)
To add to this: remember that the notation is suggestive and is formally like a dot product. i.e. think of and and where in your example and so on.

Then, you form the 'dot product'

where you interpret the 'product' of, say with in the obvious way i.e. you take the partial derivative of w.r.t.
9. (Original post by HappyHammer15)
I get a=-0.5 and b = 2. Is c an arbitrary constant then?
Exactly, if it doesn't contribute anything then you can take any value.

e.g. Find all pairs (a,b) such that a = 0.
10. (Original post by Jake22)
To add to this: remember that the notation is suggestive and is formally like a dot product. i.e. think of and and where in your example and so on.

Then, you form the 'dot product'

where you interpret the 'product' of, say with in the obvious way i.e. you take the partial derivative of w.r.t.

If i do it in this way i get a very complicated solution that is quite difficult to reduce. Will the way in which i first did it suffice?, i.e differentiating w.r.t x, differentiating w.r.t y and so on.
11. (Original post by HappyHammer15)
If i do it in this way i get a very complicated solution that is quite difficult to reduce. Will the way in which i first did it suffice?, i.e differentiating w.r.t x, differentiating w.r.t y and so on.
I never wrote that you should calculate it in any other way... I said to do exactly that i.e. differentiate the i component wrt to x, the j component wrt y etc. and then sum them up. All I wrote was just a convinient way of remembering what the operation gives you. You shouldn't get anything different at all.

The point is to think of as a vector and as a kind of dot product of vectors whose formula in components you are (or at least should be) very familiar with. The reason that it isn't a 'true' dot product is that is a vector in a different vector space to i.e. the dual space.

Note: there was a mistake in my initial post which I have now corrected - I put commas instead of + signs on the right hand side but this shouldn't have affected how you would calculate.
12. (Original post by Jake22)
I never wrote that you should calculate it in any other way... I said to do exactly that i.e. differentiate the i component wrt to x, the j component wrt y etc. and then sum them up. All I wrote was just a convinient way of remembering what the operation gives you. You shouldn't get anything different at all.

Note: there was a mistake in my initial post which I have now corrected - I put commas instead of + signs on the right hand side.
Okay, i see now. I was reading and i was trying to compute
This seems like such a silly mistake to make now. Thanks for all of your help though.

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: April 18, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### University open days

Is it worth going? Find out here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams