# Rolle's theorem on Envelope Equations

1. Jul 25, 2012

### unscientific

1. The problem statement, all variables and given/known data

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

3. 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 doesnt mean that the derivative approaches 0 as well..

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2. Jul 27, 2012

### unscientific

would appreciate if anyone can clear this up..

3. Jul 28, 2012

### who_

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

$$f(x,y,a_1) = \frac{\partial f(x,y,a_1)}{\partial h} = 0$$

All the stuff beforehand just sets up the context for this definition. Search Wiki for more info on the envelop of curves.

Last edited: Jul 28, 2012
4. Jul 28, 2012

### unscientific

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!

Last edited: Jul 28, 2012
5. Jul 28, 2012

### who_

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).

6. Jul 28, 2012

### unscientific

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.