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# math for chemistry watch

1. The question is in the attached file. Thanks for any help offered.
Attached Files
2. If z.doc (40.0 KB, 93 views)
3. Are you sure it's not 3x(dz/dx)= 2y(dz/dy)?
4. (Original post by EierVonSatan)
Are you sure it's not 3x(dz/dx)= 2y(dz/dy)?
Err..no, it is partial differential as stated. The f does simply means that z is a function of x^2y^3 and therefore z is a function of x and y right?
5. (Original post by shengoc)
Err..no, it is partial differential as stated. The f does simply means that z is a function of x^2y^3 and therefore z is a function of x and y right?
yes, I know its partial (I'm being lazy) my question was to the order of the derivatives...
6. (Original post by EierVonSatan)
yes, I know its partial (I'm being lazy) my question was to the order of the derivatives...
meaning...
7. (Original post by shengoc)
meaning...
is it or ?

As I can't get z = x2y3 to fit the former.
8. (Original post by EierVonSatan)
Are you sure it's not 3x(dz/dx)= 2y(dz/dy)?
I think it should be this - thiss works.

When it is 3x(pz/py) = / = 2y(pz/px)

However,

3x(pz/px) = 2y(pz/py)

The question itself is wrong.
9. (Original post by DeanK2)
I think it should be this - thiss works.

When it is 3x(pz/py) = / = 2y(pz/px)

However,

3x(pz/px) = 2y(pz/py)

The question itself is wrong.
what is that p you used in your equation anyway?
10. (Original post by shengoc)
what is that p you used in your equation anyway?
p = partial symbol ...I would imagine
11. (Original post by EierVonSatan)
is it or ?

As I can't get z = x2y3 to fit the former.
No, EVS, the question was stated that way, and it is 3x(partial of z wrt y at constant x) = 2y(partial of z wrt x at constant y), just to make sure i didn't type anything wrong in the file i attached.

But yeah, am I allowed to imply that z = x^2 x y^3 ?
12. It seems from the document that all you have been asked to show is the scanned equation holds true for a function z you are free to choose.

It looks like the statemnt if z = (x^2)(y^3) is written by you.

Why don't you try a different function for z?
13. Taking a random example (as the relation should hold for all 'nice' functions) f(u) = u2 - 1 and let u = x^2y^3

knowing that x and y are independant (i.e. z = f(x, y) =/= 0)

Partial (dz/dx) = 4x^3y^6 and partial (dz/dy) = 6x^4y^5

so 2y(4x^3y^6) =/= 3x(6x^4y^5) according to the relation

but 3x(4x^3y^6)= 2y(6x^4y^5) ?
14. (Original post by EierVonSatan)
Taking a random example (as the relation should hold for all 'nice' functions) f(u) = u2 - 1 and let u = x^2y^3

knowing that x and y are independant (i.e. z = f(x, y) =/= 0)

Partial (dz/dx) = 4x^3y^6 and partial (dz/dy) = 6x^4y^5

so 2y(4x^3y^6) =/= 3x(6x^4y^5) according to the relation

but 3x(4x^3y^6)= 2y(6x^4y^5) ?
I see your point, but i was told that what we are asked to show in this question can be done. Hmm, thanks anyway, for all your help.
15. (Original post by EierVonSatan)
Taking a random example (as the relation should hold for all 'nice' functions) f(u) = u2 - 1 and let u = x^2y^3

knowing that x and y are independant (i.e. z = f(x, y) =/= 0)

Partial (dz/dx) = 4x^3y^6 and partial (dz/dy) = 6x^4y^5

so 2y(4x^3y^6) =/= 3x(6x^4y^5) according to the relation

but 3x(4x^3y^6)= 2y(6x^4y^5) ?
Yes, i managed to show what the question ask to show. Apparently, a chain rule works like you mentioned above. If I let z=f(u), u = x^2 y^3

therefore dz/dx = dz/du x du/dx

i did the same for dz/dy,

then make dz/du as subject of formula and equate the two, then i managed to show it. Does this make sense to you?

p/s - I just realised there may be errors in the question, i got 3x(partial z wrt x) in the end instead of what the question stated. hmm, i will leave it that way then, thanks a lot, EVS
16. (Original post by shengoc)
p/s - I just realised there may be errors in the question, i got 3x(partial z wrt x) in the end instead of what the question stated. hmm, i will leave it that way then, thanks a lot, EVS
You do realise that this is what I was asking you from the beginning?
17. (Original post by shengoc)
Yes, i managed to show what the question ask to show. Apparently, a chain rule works like you mentioned above. If I let z=f(u), u = x^2 y^3

therefore dz/dx = dz/du x du/dx

i did the same for dz/dy,

then make dz/du as subject of formula and equate the two, then i managed to show it. Does this make sense to you?

p/s - I just realised there may be errors in the question, i got 3x(partial z wrt x) in the end instead of what the question stated. hmm, i will leave it that way then, thanks a lot, EVS
There are bound to be errors in the question - it's Grout.
18. (Original post by EierVonSatan)
You do realise that this is what I was asking you from the beginning?
no, actually i don't; i didn't have a clue how to do that until the chain rule comes up and only then, i realise that, my apologies for that, EVS.
19. (Original post by cpchem)
There are bound to be errors in the question - it's Grout.
Thanks, yeah, i checked all my workings again, and it seemed to work out logically, if the question was set otherwise. Anyway, thanks for letting me know.
20. (Original post by shengoc)
no, actually i don't; i didn't have a clue how to do that until the chain rule comes up and only then, i realise that, my apologies for that, EVS.

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