Property of Implicit Function - Calculus

Question

The differentiable fuction $z=z(x,y)$ is given implicitly by equation $f(\frac{x}{y},z)=0$, where $f(u,v)$ is supposed to be differentiable and $\frac{\partial f}{\partial v}(u,v)\neq0$. Verify that $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0.$$ This is the exercise 6 from Guidorizzi's "A Course in Calculus", Vol. 2, Section 12.2 - Derivative of functions defined implicitly. Implicit Functions Theorem.

What I've done:

Let $F(x,y,z)$ be defined by $F(x,y,z)=f(\frac{x}{y},z)$. Then we have that $z=z(x,y)$ is implicitly given by $F(x,y,z(x,y))=0$. By hypothesis, $z$ is differentiable, thus follows from the Implicit Functions Theorem that $$\frac{\partial z}{\partial x}=-\frac{\frac{\partial F}{\partial x}(x,y,z(x,y))}{\frac{\partial F}{\partial z}(x,y,z(x,y))}=-\frac{\frac{\partial f}{\partial x}(\frac{x}{y},z)}{\frac{\partial f}{\partial z}(\frac{x}{y},z)}$$ and $$\frac{\partial z}{\partial y}=-\frac{\frac{\partial F}{\partial y}(x,y,z(x,y))}{\frac{\partial F}{\partial z}(x,y,z(x,y))}=-\frac{\frac{\partial f}{\partial y}(\frac{x}{y},z)}{\frac{\partial f}{\partial z}(\frac{x}{y},z)}.$$ But at this point, I'm a little bit confused about how to compute $-\frac{\frac{\partial f}{\partial x}(\frac{x}{y},z)}{\frac{\partial f}{\partial z}(\frac{x}{y},z)}$ and $-\frac{\frac{\partial f}{\partial y}(\frac{x}{y},z)}{\frac{\partial f}{\partial z}(\frac{x}{y},z)}$. Can anybody explain me how to proceed? Thanks!

Answer 1

$f(u,v)=0$ where $u=\frac{x}{y}$ and $v=z(x,y)$ $$du=\frac{1}{y}dx-\frac{x}{y^2}dy$$ $$dv=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$ Since $f(u,v)=0$, we have : $$\frac{\partial f}{\partial u}du+\frac{\partial f}{\partial v}dv=0$$ $$\frac{\partial f}{\partial u}\big(\frac{1}{y}dx-\frac{x}{y^2}dy\big)+\frac{\partial f}{\partial v}\big(\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy\big)=0$$ $$\big(\frac{1}{y}\frac{\partial f}{\partial u}+\frac{\partial z}{\partial x}\frac{\partial f}{\partial v}\big)dx+ \big(-\frac{x}{y^2}\frac{\partial f}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial f}{\partial v}\big)dy=0$$ which implies : $$\frac{1}{y}\frac{\partial f}{\partial u}+\frac{\partial z}{\partial x}\frac{\partial f}{\partial v}=0$$ and $$-\frac{x}{y^2}\frac{\partial f}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial f}{\partial v}=0$$ Then, combining the two last equations, (i.e. multiply the first by $x$ , multiply the second by $y$ and add both), leads to : $$\big(x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}\big)\frac{\partial f}{\partial v}=0$$ Since $\frac{\partial f}{\partial v}$ is not uniformaly nul : $$x\frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=0$$

Answer 2

We need the partial derivative of $f$ relative to $x$. Thus, consider $y$ and $z$ to be constants and derivate $f(u(x),v(x))$ relative to $x$, with $u(x)=\frac{x}{y}$ and $v(x)=z$. We have, by the Chain Rule $$\frac{\partial f}{\partial x}(u(x),v(x))=\frac{\partial f}{\partial u}(u(x),v(x))\cdot\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}(u(x),v(x))\cdot\frac{\partial v}{\partial x}$$ $$=\frac{\partial f}{\partial u}(u(x),v(x))\cdot\frac{1}{y}+\frac{\partial f}{\partial v}(u(x),v(x))\cdot0$$ $$\therefore \frac{\partial f}{\partial x}(\frac{x}{y},z)=\frac{1}{y}\cdot \frac{\partial f}{\partial u}(\frac{x}{y},z).$$ Following with the same idea but now derivating with respect to $y$, we have $$\frac{\partial f}{\partial y}(u(x),v(x))=\frac{\partial f}{\partial u}(u(x),v(x))\cdot\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}(u(x),v(x))\cdot\frac{\partial v}{\partial y}$$ $$=\frac{\partial f}{\partial u}(u(x),v(x))\cdot\frac{-x}{y^2}+\frac{\partial f}{\partial v}(u(x),v(x))\cdot0$$ $$\therefore \frac{\partial f}{\partial y}(\frac{x}{y},z)=-\frac{x}{y^2}\cdot \frac{\partial f}{\partial u}(\frac{x}{y},z).$$ From the Implicit Function Theorem, we have $$\frac{\partial z}{\partial x}=-\frac{\frac{\partial F}{\partial x}(x,y,z(x,y))}{\frac{\partial F}{\partial z}(x,y,z(x,y))}=-\frac{\frac{\partial f}{\partial x}(\frac{x}{y},z)}{\frac{\partial f}{\partial z}(\frac{x}{y},z)}$$ and $$\frac{\partial z}{\partial y}=-\frac{\frac{\partial F}{\partial y}(x,y,z(x,y))}{\frac{\partial F}{\partial z}(x,y,z(x,y))}=-\frac{\frac{\partial f}{\partial y}(\frac{x}{y},z)}{\frac{\partial f}{\partial z}(\frac{x}{y},z)}.$$ Thus $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=-\frac{1}{\frac{\partial f}{\partial z}(\frac{x}{y},z)}\left[ x\cdot\frac{1}{y}\frac{\partial f}{\partial u}(\frac{x}{y},z) + y\cdot \left( \frac{-x}{y^2}\right)\frac{\partial f}{\partial u}(\frac{x}{y},z)\right]=0.$$

Answer 3

I won't go into the calculations, but give an intuitive argument why $$x{\partial z\over\partial x}+ y{\partial z\over \partial y}\equiv 0\tag{1}$$ should hold:

Along rays through the origin the quotient ${x\over y}$ is constant. Since $z=\psi(x,y)$ is determined from an equation of the form $f\bigl({x\over y},z\bigr)=0$ it follows that $\psi$ is constant along rays through the origin. This implies that at each point $(x,y)\ne(0,0)$ the gradient $\nabla\psi$ is orthogonal to the direction of this ray through $(x,y)$, i.e., orthogonal to $(x,y)$. That's what $(1)$ says.