November 4th, 2016, 07:44 AM  #1 
Member Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2  Simple chainrule exercise, help? Equation: 2*Z'x + 3*Z'y = 0 Simplify through variable change: u = x, v = 3x  2y I'm very confused here. First, I don't know the function at hand, how am I suppose to solve it? Secondly, am I suppose to replace x with u and then solve y = (v3x)/2 and replace y with that?! Very confusing exercise, all help is appreciated. 
November 4th, 2016, 10:12 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
I presume your "Z'x" indicates the partial derivative with respect to x. What function, that you don't know, are you talking about? If you mean z, this is an equation involving the "unknown" z with variables x and y and you are to convert it to an equivalent equation involving the "unknown" z with variable u and v. When you have two variable, here x and y, and each depends on two other variables, x(u,v) and y(u,v) then $\displaystyle \frac{\partial z}{\partial x}= \frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+ \frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$ and $\displaystyle \frac{\partial z}{\partial y}= \frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+ \frac{\partial z}{\partial v}\frac{\partial v}{\partial y}$ $\displaystyle 2\frac{\partial z}{\partial x}+ 3\frac{\partial z}{\partial y}= 2\left(\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+ \frac{\partial z}{\partial v}\frac{\partial v}{\partial x}\right)+ 3\left(\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+ \frac{\partial z}{\partial v}\frac{\partial v}{\partial y}\right)= 0$ Here, here u= x so $\displaystyle \frac{\partial u}{\partial x}= 1$ and $\displaystyle \frac{\partial v}{\partial y}= 0$. v= 3x 2y so $\displaystyle \frac{\partial v}{\partial x}= 3$ and $\displaystyle \frac{\partial v}{\partial y}= 2$ So $\displaystyle \frac{\partial z}{\partial x}= \frac{\partial z}{\partial u}+ 3\frac{\partial z}{\partial v}$ and $\displaystyle \frac{\partial z}{\partial y}= 2\frac{\partial z}{\partial v}$. Then $\displaystyle 2\frac{\partial z}{\partial x}+ 3\frac{\partial z}{\partial y}= 2\left(\frac{\partial z}{\partial u}+ 3\frac{\partial z}{\partial v}\right)+ 3\left(2\frac{\partial z}{\partial v}\right)= 0$. 
November 4th, 2016, 10:36 AM  #3 
Member Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2 
This is what I don't get, lets say they meant z = z(x(u,v), y(u,v)) Shouldn't dz/dx = dz/dx? I mean, where asking: How much does z change when x changes? We don't ask anything about any u,v. They do change x but that's not what we're asking for. Also: du/dx what is that? x is dependent on u and v, u isnt dependent on x so where you get this du/dx from? I just fail to see the connection here.. :/ 
November 4th, 2016, 01:49 PM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
In your original post you wrote "u = x, v = 3x  2y" so u and v are certainly dependent on x and y (u isn't actually "dependent" of on y so the partial derivative of y is $\displaystyle \frac{\partial u}{\partial y}= 0$).

November 4th, 2016, 05:19 PM  #5 
Member Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2 
Oh so the function z should be thought of as: z = z(u(x,y), v(x,y)) ? 
November 5th, 2016, 04:35 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
Yes, that is what the notation means.

November 5th, 2016, 07:41 AM  #7 
Member Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2 
Thanks a ton man! Finally understood this abit better! 

Tags 
chainrule, exercise, simple 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
kind of silly/simple question with probably a simple answer.  GumDrop  Math  4  October 4th, 2016 05:34 PM 
Numerical methodschainrule  bhmk  Differential Equations  1  October 10th, 2014 04:20 PM 
Simple exercise on determine some dimensions.  beesee  Physics  4  August 23rd, 2014 03:10 PM 
A simple exercise in the number field  Simple Minded  Real Analysis  1  December 12th, 2013 09:57 AM 
Simple, not so simple question about areas of triangles  jkh1919  Algebra  1  November 20th, 2011 09:14 AM 