1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivative when you just have Constraints

  1. Mar 6, 2009 #1
    Suppose that you have a set of real variables {x1,x2,...,xn}. If x1 = f(x2,x3,...,xn) then this represents a constraint on all the variables. In this case, it's possible to find dx1/dxi as long as f is differentiable. But not all possible constraints among the xj are of this form. How might one extend the concept of derivative to the case when the constraint on all the variables is not necessarily that one variable is a function of the others?
  2. jcsd
  3. Mar 6, 2009 #2
    Welcome to the world of http://en.wikipedia.org/wiki/Differential_manifold" [Broken]!

    Suppose you have a differentiable function [tex]g:\mathbb{R}^n\to\mathbb{R}^k[/tex] and your constraints are given by the equation g(x)=0 (this is the general form). If the function g is "nice", for example if 0 is a regular value of g, then the set where g(x)=0 is a differentiable manifold of dimension n-k. As an example, take [tex]g(x)=x_1^2+x_2^2-1[/tex], then the resulting manifold is the unit circle.
    One can then define the derivative of a differentiable function between manifolds abstractly, but to get partial derivatives (and do concrete calculations) you need to choose local coordinates (aka a chart) on the manifold.
    Last edited by a moderator: May 4, 2017
  4. Mar 8, 2009 #3
    Does your example allow you to say that [tex]\frac{dx_1}{dx_2}\big|_{x_1=x_2=\frac{1}{\sqrt(2)}} = -1[/tex]?
  5. Mar 9, 2009 #4
    Yes, [tex]x_1[/tex] defines a real-valued function on the unit circle, and near the point [tex](1/\sqrt{2},1/\sqrt{2})[/tex] one can "parametrize" the circle by [tex]x_2[/tex], i.e. [tex]x_2[/tex] gives differentiable local coordinates. In these coordinates [tex]x_1=\sqrt{1-x_2^2}[/tex] and you can simply compute the usual derivative.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook