Derivative when you just have Constraints

In summary, the concept of derivative can be extended to cases where constraints on variables are not necessarily of the form x1 = f(x2,x3,...,xn). This can be done by defining a differentiable function g:\mathbb{R}^n\to\mathbb{R}^k and using the equation g(x)=0 to represent constraints. The derivative can be calculated using local coordinates, such as in the case of a differentiable manifold.
  • #1
mXSCNT
315
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?
 
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  • #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.
 
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  • #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]?
 
  • #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.
 

1. What is a constraint in terms of derivatives?

A constraint is a condition or restriction that must be satisfied in order for a derivative to exist or be valid. It can be a mathematical equation, an inequality, or a physical limitation.

2. How do constraints affect the calculation of derivatives?

Constraints limit the possible values of the independent variables, which in turn affects the values of the dependent variable and the resulting derivative. In some cases, constraints can make it impossible to take a derivative at certain points.

3. Can you give an example of a constraint affecting a derivative?

Say we have a function f(x,y) = x^2 + y^2, and we want to find the derivative with respect to x. If we have a constraint that x + y = 5, then this restricts the possible values of x and y, and therefore the values of the derivative will change depending on which point on the constraint line we are looking at. For example, at the point (2,3), the derivative would be 4x + 2y = 14.

4. How do we incorporate constraints into the derivative equation?

We can use Lagrange multipliers to incorporate constraints into the derivative equation. This involves adding the constraint equation multiplied by a constant (the Lagrange multiplier) to the original function before taking the derivative. The resulting equation can then be solved for the desired derivative.

5. What are some real-life applications of using derivatives with constraints?

Derivatives with constraints are used in a variety of fields, such as economics, physics, and engineering. For example, in economics, derivatives with constraints can be used to maximize profits while adhering to budget or resource constraints. In physics, they can be used to calculate optimal trajectories for objects subject to certain physical limitations. In engineering, they can be used to optimize designs while considering constraints such as material strength or cost.

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