My Math Forum Simple chainrule exercise, help?

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 November 4th, 2016, 06:44 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 = (v-3x)/2 and replace y with that?! Very confusing exercise, all help is appreciated.
 November 4th, 2016, 09:12 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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 $\displaystyle 2\frac{\partial z}{\partial x}+ 3\frac{\partial z}{\partial y}= 2\left(\frac{\partial z}{\partial u}+ 3\frac{\partial z}{\partial v}\right)+ 3\left(-2\frac{\partial z}{\partial v}\right)= 0$. Thanks from Addez123
 November 4th, 2016, 09:36 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, 12:49 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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 $\displaystyle \frac{\partial u}{\partial y}= 0$). Thanks from Addez123
 November 4th, 2016, 04:19 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, 03:35 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 Yes, that is what the notation means. Thanks from Addez123
 November 5th, 2016, 06:41 AM #7 Member   Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2 Thanks a ton man! Finally understood this abit better!

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