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Rolle's theorem on Envelope Equations

  1. Jul 25, 2012 #1
    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..
     

    Attached Files:

  2. jcsd
  3. Jul 27, 2012 #2
    would appreciate if anyone can clear this up..
     
  4. Jul 28, 2012 #3
    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: Jul 28, 2012
  5. Jul 28, 2012 #4
    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!
     
    Last edited: Jul 28, 2012
  6. Jul 28, 2012 #5
    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).
     
  7. Jul 28, 2012 #6
    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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