0
$\begingroup$

Not sure how to interpret this question or where to start.
$\text{Assuming that the equation}$ $$F(x,y,z) = 0$$ $\text{defines} z \text{implicitly as a differentiable function of} \: x \: \text{and} \: y \: \text{and that}$ $$F_{zx} = F_{xz}$$ $\text{show that}$ $$\frac{\partial ^2 z}{\partial x^2} = \frac{-(F_{z})^2 F_{xx} + 2F_{z}F_{x}F_{xz} - (F_{x})^2 F_{zz}}{(F_{z})^3}.$$

I have no idea how to use the given equation to imply that. I know how to implicity differentiate functions but when they actually give a function...
Also, from their definition, it means that this is true right?
$z \equiv z(x,y)$
(We haven't learn the implicit function theorem btw).

$\endgroup$
1
$\begingroup$

A heuristic approach: from $F(x,y,z)=0$, you differentiate in terms of $x$ and $y$ repsectively (just treat $z=z(x,y)$) to get \begin{align} F_x+F_zz_x &= 0 \\ F_y+F_zz_y &=0 \end{align} and you solve $(z_x,z_y)=-\frac1F_z(F_x,F_y)$.

Then you differentiate $F_x+F_zz_x=0$ wrt $x$ again to obtain $$F_{xx} + F_{zx}z_x + z_x(F_{xz}+F_{zz}z_x)+F_zz_{xx}=0$$ and you get $z_{xx}$. It turns out $z_y$ is unused, but in general even if you only need the second partial derivative in a particular defining variable, you will still have to differentiate in all the defining variables as all first order partial derivatives can arise later on.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks. How did you get your two results in the beginning? I used chain rule on $F$ and obtained $$F_{x} = F_{z} z_{x} + F_{y} y_{x}$$. Is it true that $y_{x} = 0$ (as is with any other combination of x,y,z)? Since we can think of it as partial derivative of y w.r.t x. $\endgroup$ – Twenty-six colours Apr 28 '17 at 8:47
  • 2
    $\begingroup$ since you let $(x,y)$ define $z$, then $(x,y)$ themselves should be independent variables and $y_x=0$. $\endgroup$ – Vim Apr 28 '17 at 8:48
  • $\begingroup$ I see, the sign should be the opposite then right? (Might've just been a typo in your paragraph) $\endgroup$ – Twenty-six colours Apr 28 '17 at 8:49
  • 2
    $\begingroup$ which sign? I don't think anything is wrong in my first order derivatives. It's just the first entry of $$[F_x,F_y,F_z]\begin{bmatrix}\frac{\partial x}{\partial x} & \frac{\partial x}{\partial y}\\ \frac{\partial y}{\partial x}& \frac{\partial y}{\partial y}\\ \frac{\partial z}{\partial x}& \frac{\partial z}{\partial y}\end{bmatrix}$$ Actually I wonder how you got yours. Also, it seems that the denominator in the answer should be $F_z^3$ instead of $F_z^2$, could you confirm it again? $\endgroup$ – Vim Apr 28 '17 at 8:53
  • $\begingroup$ My apologies, you are correct about the denominator.. My first order derivative for x is found by chain rule. Am I using it incorrectly? $\frac{\partial F}{\partial x} = \frac{\partial F}{\partial x}\frac{\partial z}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial y}{\partial x}$. $\endgroup$ – Twenty-six colours Apr 28 '17 at 8:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.