- #1
sardonic
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Homework Statement
If f(x,y,z) = 0, then you can think of z as a function of x and y, or z(x,y). y can also be thought of as a function of x and z, or y(z,x)
Therefore:
[tex] dz= \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y} dy [/tex]
and
[tex] dy= \frac{\partial y}{\partial x}dx + \frac{\partial y}{\partial z} dz [/tex]
Show that
[tex] 1= \frac{\partial z}{\partial y} \frac{\partial y}{\partial z} [/tex]
and then
[tex] -1= \frac{\partial x}{\partial z} \frac{\partial y}{\partial x}\frac{\partial z}{\partial y} [/tex]
Homework Equations
The Attempt at a Solution
Substituting [itex] dy [/itex] into the [itex] dz [/itex] equation you get
[tex] dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y} \frac{\partial y}{\partial x}dx + \frac{\partial y}{\partial z}\frac{\partial z}{\partial y}dz [/tex]
This can be rearranged to show
[tex] dz (1-\frac{\partial z}{\partial y}\frac{\partial y}{\partial z}) = dx(\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x}) [/tex]
and then
[tex](1-\frac{\partial z}{\partial y}\frac{\partial y}{\partial z}) = \frac{dx}{dz}(\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x}) [/tex]
In order to show that [itex]\frac{\partial z}{\partial y}\frac{\partial y}{\partial z} = 1 [/itex], I only need to show that [itex] (\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x})=0 [/itex], but I'm not sure how to do that. As for the second equation, I'm not sure where to get a [itex] \frac{\partial x}{\partial z} [/itex] into the equation in the first place.
