- #1
Refraction
- 21
- 0
Homework Statement
If T is implicitly defined via the relationship f(x, y, z, T) = 0 to be a differentiable function of x, y and z, show that the first partial derivative of T with respect to z can be found using:
[tex]
\frac{\partial T}{\partial z} = -\frac{\partial f}{\partial z} / \frac{\partial f}{\partial T}
[/tex]
Homework Equations
This is a similar example from my textbook, for f(x, y, z) = 0, where z is assumed to be a function of x only.
[tex]
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial x} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial x}
[/tex]
[tex]
0 = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial x}[/tex] as x and y are independent variables, meaning the derivative of y with respect to x is 0.
[tex]
\frac{\partial z}{\partial x} = -\frac{\partial f}{\partial x} / \frac{\partial f}{\partial z}
[/tex]
The Attempt at a Solution
I tried using something similar for the question and it happened to work out, but I'd like to actually understand it properly. It's mostly the implicit part that I'm unsure of for this question. Here's what I did for it:
[tex]
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial z} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial z} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial z} + \frac{\partial f}{\partial T}\frac{\partial T}{\partial z}
[/tex]
[tex]
0 = \frac{\partial f}{\partial z} + \frac{\partial f}{\partial T}\frac{\partial T}{\partial z}[/tex] as x and y are not dependent on z, derivatives with respect to z must be 0 (not too sure about this part either)
[tex]
\frac{\partial T}{\partial z} = -\frac{\partial f}{\partial z} / \frac{\partial f}{\partial T}
[/tex]
It gets the right answer but I'm still a bit confused by it. I was also having trouble figuring out what the dependency tree for this relationship would look like.
For the textbook example, it mentions that z is assumed to be a function of x only; I'm not sure if that applies to applies to this question in a similar way or if it's for both x and y (or maybe something completely different).
Last edited: