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I was studying partial differential equations and came upon the formula:

$$\frac{dy}{dx} = - \frac {\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$ where $f(x,y) = c$ is an implicit relation. The formula was arrived at by the use of total derivative formula.

I encountered a question:

If the curves $f(x,y)=0$ and $\Phi (y,z) = 0$ touch, show that at the point of contact, $ \frac {\partial f}{\partial x} \cdot \frac > {\partial \Phi}{\partial y} = \frac {\partial f}{\partial y} \cdot > \frac {\partial \Phi}{\partial z}$

Through the use of above mentioned formula for implicit differentiation i have arrived at : $$ \frac {dx}{dz} = \frac {\frac {\partial f}{\partial x} \cdot \frac {\partial \Phi}{\partial y}}{\frac {\partial f}{\partial y} \cdot \frac {\partial \Phi}{\partial z}}$$

I am unable to understand why this should be equal to 1. Is it to do with the touching of the two curves. If so how? Can somebody shed some light on this?

P.S. : Sorry for the long question.

Inquistador
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    How is $x$ a function of $z$? –  Nov 16 '16 at 02:10
  • Are you sure it is supposed to be $\Phi(y,z)$ and not $\Phi(x,y)$? Because if $f$ and $\Phi$ don't define curves in the same coordinate plane (e.g. $x,y$) then it's hard to think what you mean by them "touching". – David K Nov 16 '16 at 02:23
  • It is given that $f(x,y) = 0$ and $\phi(y,z) = 0$. This implies that $f$ and $\phi$ are both implicit function in $x,y$ and $$y,z$ respectively. So x and can be written in terms of y and y can be written in terms of z, from definition of implicit functions. So Indirectly x is represented as a function of z.

    @Jack

     – Inquistador Nov 16 '16 at 05:42
  • @DavidK $f$ and $\Phi$ may be represented as curves that are on two different planes the x-y for $f$ and y-z for $\Phi$ though it is possible that $\frac {dx}{dz}$ may not be found directly, it may be found indirectly through ${\frac {dx}{dy}}/{\frac {dz}{dy}}$. That is how i arrived at the equation mentioned above, where i am stuck. – Inquistador Nov 16 '16 at 05:49
  • OK, give an example of two curves and explain how they "touch". Then we can explore whether the proposition is true or not. – David K Nov 16 '16 at 13:32
  • @DavidK $f = x^3 +3y$ and $\Phi = z^3 + 3y$ It can be seen that if the two function touch at any point their corresponding values will be equal. So $f = \Phi$ putting the equation we would arrive at $x=z$ that is for every value $k$ where $x=z=k$ we would have $f=\Phi$. There are some function generators available on the internet or you can use matlab to plot the functions to visualize them better. – Inquistador Nov 16 '16 at 14:04
  • The graphs of $f=x^3+3y$ and $\Phi=z^3+3y$ intersect at the point $(0,0,0)$, that is, $x=y=z=0$. At the point of intersection, the two curves are perpendicular to each other; also at that point $\frac{\partial f}{\partial x}=\frac{\partial\Phi}{\partial z}=0$, so both sides of the formula you're supposed to prove are zero but the right side of your equation for $\frac{dx}{dz}$ is indeterminate. I think it might be more interesting to try an example where none of the partial derivatives is zero at the intersection of the curves. – David K Nov 16 '16 at 15:41
  • I'm assuming, by the way, that $f=x^3+3y$ is restricted to the $x,y$ plane and $\Phi=z^3+3y$ is restricted to the $y,z$ plane. Otherwise they'd be formulas of surfaces, not of curves. Or do we actually want surfaces after all, not curves? – David K Nov 16 '16 at 15:45

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