Derive the formula for gradient using chain rule

[Happiness]

Consider a surface defined by the equation ##g(x, y, z)=0##. The intersection between this surface and the plane ##z=c## produces a curve that can be plotted on an x-y plane. Find the gradient of this curve.

By chain rule,

##\frac{\partial y}{\partial x}=\frac{\partial y}{\partial g}\frac{\partial g}{\partial x}##

Using the reciprocity relation ##\frac{\partial y}{\partial g}=\Big(\frac{\partial g}{\partial y}\Big)^{-1}##, we have

##\frac{\partial y}{\partial x}=\frac{(\frac{\partial g}{\partial x})}{(\frac{\partial g}{\partial y})}##

This differs from the correct answer by a negative sign. What's wrong with this method?

 

[andrewkirk]

Greater care is needed when handling partial derivatives. They cannot be treated in general the same as total derivatives. In particular it is not necessarily the case that

##\frac{\partial y}{\partial x}=\frac{\partial y}{\partial g}\frac{\partial g}{\partial x}##

Indeed, in this case the ##\frac{\partial y}{\partial x}## that is being calculated is with the values of ##g## and ##z## held constant, so ##\frac{\partial y}{\partial g}## is undefined or infinite, while ##\frac{\partial g}{\partial x}## is zero.

 

[Happiness]

The following website calculates the gradient as follows:

Differentiate ##g## wrt ##x## while holding ##z## constant.

Then we will have ##\frac{\partial y}{\partial x}=-\frac{(\frac{\partial g}{\partial x})}{(\frac{\partial g}{\partial y})}##. Why ain't ##\frac{\partial g}{\partial x}## and ##\frac{\partial g}{\partial y}## zero in this case?

In what situations must the two partials on the RHS of the chain rule have the same variables held constant? Like you mentioned for post #1, ##(\frac{\partial y}{\partial g})_{gz}(\frac{\partial g}{\partial x})_{gz}##. But why is it not required in the attachment above, where ##(\frac{\partial g}{\partial y})_{xz}(\frac{\partial y}{\partial x})_{gz}##?

Source:
http://www.sjsu.edu/faculty/watkins/envelopetheo.htm

 

[andrewkirk]

I find the presentation on that web page unnecessarily confusing. It attempts to use the formula for a total derivative in a case where all derivatives are partial. Sometimes one can get away with this, but sometimes one can't, so it is best avoided.

A more correct derivation would be as follows:

Say the point ##(x_0,y_0)## is on the curve. Let us parametrise the curve through that point by function ##\gamma:\mathbb R\to\mathbb R^2## and wlog set ##\gamma(0)=(x_0,y_0)##. Then we have ##g(\gamma_1(t).\gamma_2(t),c)=0##, where ##\gamma_1## and ##\gamma_2## are the component functions of ##\gamma##.

We then apply ##\frac d{dt}## to this - ie taking a total derivative - to get:

$$0=\frac d{dt}0=\frac d{dt}g(\gamma_1(t).\gamma_2(t),c)=\frac{\partial g}{\partial x}\frac {dx}{dt}+
\frac{\partial g}{\partial y}\frac {dy}{dt}+\frac{\partial g}{\partial c}\frac {dc}{dt}
=\frac{\partial g}{\partial x}\frac {dx}{dt}+
\frac{\partial g}{\partial y}\frac {dy}{dt}+\frac{\partial g}{\partial c}\cdot 0$$

Rearranging, we get

$$-\frac{\partial g}{\partial x}/\frac{\partial g}{\partial y}=\frac{dy}{dt}/\frac{dx}{dt}=\frac{dy}{dx}$$
as required. Note that the transformation in the final step is allowed because we are dealing with total derivatives, not partial derivatives.