Chain rule in functions of two variables

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Main Question or Discussion Point

Please help me on this. I am trying to make and exercise from an author M.D. Hatton (an english).

Let x = x(r, w) = r. cos (w)
Let y = y(r,w) = r. sen (w)
Let V = V(x,y). So V depends on r and w.

By chain rule (I put "d" for the partial derivative)

dV = dV . dx + dV. dy
-- -- -- -- --
dr dx dr dy dr


So

dV = dV . cos(w) + dV. sen(w) (A)
-- -- --
dr dx dy



And the chain rule also says that

dV = dV . dx + dV. dy
-- -- -- -- --
dw dx dw dy dw


so


dV = - r dV . sen(w) + r. dV. cos(w) (B)
-- -- --
dw dx dy


Now Hatton says: "solving these equation for dV/dx and dV/dy, we find:

dV = dV cos(w) - 1 dV . sen(w). (C)
-- -- -- . --
dx dr r dw


and

dV = dV sen(w) + 1. dV . cos(w). (D)".
-- -- -- --
dy dr r dw

I do not understand. How does (A) produces (C)? How does (B) produces (D)????

Remember that all are partial derivatives. Thanks for the help.

P. Castilla.
 

Answers and Replies

  • #2
Zurtex
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When you are doing that you perhaps want to use the [code] tag. So it comes out like this:

Code:
Please help me on this. I am trying to make and exercise from an author M.D. Hatton (an english).

Let x = x(r, w) = r. cos (w)
Let y = y(r,w) = r. sen (w)
Let V = V(x,y). So V depends on r and w.

By chain rule (I put "d" for the partial derivative)

dV  =  dV . dx  +  dV. dy
--      --    --      --  --
dr      dx    dr      dy  dr


So 

dV  =  dV . cos(w)  +  dV. sen(w)        (A)
--      --                   --  
dr      dx                   dy



And the chain rule also says that

dV  =   dV .  dx  +  dV. dy
--       --    --       --  --
dw      dx    dw     dy  dw


so 


dV  =   - r dV . sen(w) +  r. dV. cos(w)    (B)
--            --                    -- 
dw           dx                    dy


Now Hatton says: "solving these equation for dV/dx  and dV/dy, we find:

dV   =  dV  cos(w)  -  1    dV . sen(w).         (C)
--       --                 -- . --
dx       dr                  r   dw


and 

dV   =  dV  sen(w)  +  1.  dV . cos(w).        (D)".
--       --                 --   --
dy       dr                  r   dw

I do not understand. How does (A) produces (C)? How does (B) produces (D)????

Remember that all are partial derivatives. Thanks for the help.

P. Castilla.
But really it would be much easier to learn latex so you can do this:

[tex]\frac{\partial y}{\partial x}[/tex]

You can learn how to do that here: https://www.physicsforums.com/showthread.php?t=8997


I'll look at your problem later, but what do you mean by sen?
 
  • #3
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By "sen" I mean "sin". I wrote it in spanish, sorry.
 
  • #4
saltydog
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Castilla, that's a bit hard to follow. I tell you what, the prettier the math, the fewer the errors. Or maybe that's just me I don't know. Anyway, Try and learn the math editor here called Latex. Just click on equations and a window will pop up with the commands and also learn the other stuff Zurtex said. Here it is:

[tex]x=f(r,w)[/tex]

[tex]y=g(r,w)[/tex]

[tex]V=h(x,y)[/tex]

Then:

[tex]\frac{\partial V}{\partial r}=\frac{\partial V}{\partial x}\frac{\partial x}{\partial r}+
\frac{\partial V}{\partial y}\frac{\partial y}{\partial r}[/tex]

[tex]\frac{\partial V}{\partial w}=\frac{\partial V}{\partial x}\frac{\partial x}{\partial w}+
\frac{\partial V}{\partial y}\frac{\partial y}{\partial w}[/tex]

Edit: oh yea, just calculate each partial now, form the products above and the sums, and that's it. Need help with that?
 
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  • #5
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Saltydog, when I wrote my first post I didn't know that it will be all mixed up in the screen (er, maybe I am not using "the mot juste", english is not my natal tongue). But Zurtex has rearranged the exercise fairly well.

With the formula that you (Saltydog) have stated, I have obtained two equations, those identified with "A" and "B". My problem is that (using "A" and "B" as inputs) Hatton gets equations "C" and "D". That's the step I don't perceive.

By the way, how do one begins with latex? Which is the first key that one has to press in the keyboard?

Thanks.

P. Castilla.
 
  • #6
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4
The starting and ending tags are (tex) and (/tex), but instead of (), put []. I'll look up the thread where you can learn the code.

Jameson
 
  • #7
saltydog
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Castilla said:
Saltydog, when I wrote my first post I didn't know that it will be all mixed up in the screen (er, maybe I am not using "the mot juste", english is not my natal tongue). But Zurtex has rearranged the exercise fairly well.

With the formula that you (Saltydog) have stated, I have obtained two equations, those identified with "A" and "B". My problem is that (using "A" and "B" as inputs) Hatton gets equations "C" and "D". That's the step I don't perceive.

By the way, how do one begins with latex? Which is the first key that one has to press in the keyboard?

Thanks.

P. Castilla.
Hello Castilla. I stand corrected for suggesting you just jump into LaTex. Sorry. Now just tell me something hard to do like juggle 4 oranges. Go ahead. Anyway go to the General Physics Forum and choose the "Introducing LaTex" thread or just click on my equations. A window will pop up with the commands. You can cut and paste them into your post. Also, for the problem above: We have:

[tex]\frac{\partial V}{\partial r}=\frac{\partial V}{\partial x} Cos(w)+\frac{\partial V}{\partial y} Sin(w)[/tex]

[tex]\frac{\partial V}{\partial w}=-r \frac{\partial V}{\partial x} Sin(w)+r \frac{\partial V}{\partial y} Cos(w)[/tex]

Can you treat these two expressions like you know two linear equations when you have to solve for two unknowns? You mutiply one by some factor, then add or subtract them from one another so that one of the partial types cancels out and then solve for the partial of interest in terms of the remaining terms? Try it. :smile:
 
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  • #8
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I am trying, but it seems I was more stupid than I thought...
 
  • #9
saltydog
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Castilla said:
I am trying, but it seems I was more stupid than I thought...
Easy Castilla, just say this: "Salty, grab two oranges in each hand, toss them up in the air, round and round, end of story, just like that".

I'm kidding alright. :smile: I tell you what, what would happen if you multiply the first equation througout by rCos(w) and the second by Sin(w) and then subtract the second from the first?

Edit: Note change above: Multiply second by Sin(w) too. :yuck:
 
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  • #10
saltydog
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Castilla, you got this alright? Or are you just mad about that orange stunt I pulled? I was just funnin'. Just wait, next time I ask a question in here, either no one will answer or they'll say, you guessed it, "try juggling some oranges Salty". Here's how I worked it:

For the first, I multiplied top by rCos(w) and the bottom by Sin(w) then subtracted bottom from top:

[tex]rCos(w)\frac{\partial V}{\partial r}=\frac{\partial V}{\partial x}rCos^2(w)+\frac{\partial V}{\partial y}(rSin(w)Cos(w))[/tex]

[tex]Sin(w)\frac{\partial V}{\partial w}=-r\frac{\partial V}{\partial x}Sin^2(w)+\frac{\partial V}{\partial y}(rSin(w)Cos(w))[/tex]

Subtracting:

[tex]\frac{\partial v}{\partial x}\{r(Cos^2(w)+Sin^2(w))\}=rCos(w)\frac{\partial v}{\partial r}-Sin(w)\frac{\partial V}{\partial w}[/tex]

Giving:

[tex]\frac{\partial V}{\partial x}=Cos(w)\frac{\partial V}{\partial r}-\frac{Sin(w)}{r}\frac{\partial V}{\partial w}[/tex]


For the second one, multiply top by rSin(w), bottom by Cos(w) and then add:

[tex]rSin(w)\frac{\partial V}{\partial r}=r\frac{\partial V}{\partial x}Sin(w)Cos(w)+r\frac{\partial V}{\partial y}Sin^2(w)[/tex]

[tex]Cos(w)\frac{\partial V}{\partial w}=-r\frac{\partial V}{\partial x}Sin(w)Cos(w)+r\frac{\partial V}{\partial y}Cos^2(w)[/tex]

Adding:

[tex]rSin(w)\frac{\partial V}{\partial r}+Cos(w)\frac{\partial V}{\partial w}=\frac{\partial V}{\partial y}\left[r(Sin^2(w)+Cos^2(w))\right][/tex]

This gives:

[tex]\frac{\partial V}{\partial y}=Sin(w)\frac{\partial V}{\partial r}+\frac{Cos(w)}{r}\frac{\partial V}{\partial w}[/tex]
 
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  • #11
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Excellent, Saltydog.

To prove you I have followed your steps: there is a tiping mistake (error de tipeo, in spanish) in the last but one equation, an "r" is missing in the RHS.

Thanks for your patience.

P.Castilla.
 
  • #12
saltydog
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Castilla said:
Excellent, Saltydog.

To prove you I have followed your steps: there is a tiping mistake (error de tipeo, in spanish) in the last but one equation, an "r" is missing in the RHS.

Thanks for your patience.

P.Castilla.
I don't see the typo Castilla. I do have parenthesis around those two trig functions. Is that what you're talking about? Can you kindly tell me? I do like my math to be drop-on-your-knees perfect in every detail. :smile:
 
  • #13
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0
You are right. Time to check this lousy glasses.

P. Castilla.
 

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