Rolle's theorem on Envelope Equations

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Homework Help Overview

This discussion revolves around the application of Rolle's theorem in the context of finding equations of envelopes by eliminating an independent variable, specifically in relation to a function f(x,y,a). Participants are exploring the implications of differentiability and the conditions under which the partial derivatives equal zero.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the reasoning behind the equality of partial derivatives as h approaches zero. Some participants question the necessity of differentiability for the application of Rolle's theorem. Others discuss the definition of the envelope of a family of curves and the conditions that must be satisfied for points on the envelope curve.

Discussion Status

Participants are actively engaging with the concepts, with some providing insights into the definitions and conditions necessary for the envelope equations. There is a recognition of the need for further clarification on the relationships between the functions and their derivatives, though no consensus has been reached on the implications of these relationships.

Contextual Notes

There are discussions around assumptions related to differentiability and the specific forms of the equations involved, including the distinction between the family of curves and their envelope. Some participants express uncertainty about the implications of their findings and the definitions being used.

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Homework Statement



This section describes using Rolle's theorem to find equation of envelopes by eliminating the independent variable, in this case a1.

The Attempt at a Solution



By Rolle's theorem for some 0<k<1,

∂f(x,y,a+kh)/∂a = [ f(x,y,a) - f(x,y,a+h) ]/h = 0

As h→0, P→P1

∴lim(h→0) ∂f(x,y,a+kh)/∂a = ∂f(x,y,a)/∂a

I understand why ∂f(x,y,a+kh)/∂a = 0, but why must ∂f(x,y,a)/∂a = 0 too?

Just because ∂f(x,y,a+kh)/∂a = 0, as you let h→0 it doesn't mean that the derivative approaches 0 as well..
 

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would appreciate if anyone can clear this up..
 
You need the assumption that f is differentiable. Also, they are not deducing that the partial derivative is 0 from the limit of h going to 0. They are defining the envelope of the family of curves f(x,y,a) as all the pts (x,y) such that there exists an a such that

[tex]f(x,y,a_1) = \frac{\partial f(x,y,a_1)}{\partial h} = 0[/tex]

All the stuff beforehand just sets up the context for this definition. Search Wiki for more info on the envelop of curves.
 
Last edited:
I have given some thought about this problem and have come up with some conclusions:

for a family of curves g(x,y,ck) = 0, where a particular value of ck will give you such a particular curve:

Equation
For all values of ck within range, a certain point (xk,yk) from the member family will lie both on the family curve and the envelope, and tangential to each other.

Therefore the resulting envelope curve depends on c, x and y. But since every single point on the envelope curve is due to a point on a member-family curve, the envelope must therefore satfisfy some function:

h(x,y,c) = 0 (Yes the 'k' beneath the c was intentionally left out)

albeit definitely a different form from g(x,y,ck).

Derivative
Every point (x,y) is determined for a particular value of c; in a way x(c), y(c). So for

∂h/∂c = [ h(x,y,c+Δc) - h(x,y,c) ]/Δc.

But if you keep (x,y) fixed, the numerator = 0 as it doesn't matter what value of c is given.

Hence, (∂h/∂c) = 0

Would appreciate any inputs!
 
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That pretty much sums it up. But be careful: its not the case that for every c_k there is some (x_k,y_k), but that for every (x,y) on the envelope curve there exists a value x such that the envelope curve is tangent to g at (x,y,c).
 
Update:

I thought some more about it, and came up with new conclusions:

g(x,y,c) = 0 and (∂g/∂c) = 0

should be the solutions, instead of h because it is very difficult to find h(x,y,c).

g(x,y,c) = 0 alone is meaningless, as it simply shows the family of curves. Now, when you combine this result with the constraint (∂g/∂c) = 0 voila' it gives you the envelope equation!

Each expression on its own is meaningless, only when you combine it do you get the envelope equations.
 

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