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Rolle's theorem on Envelope Equations |
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| Jul25-12, 10:17 AM | #1 |
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Rolle's theorem on Envelope Equations
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.. |
| Jul27-12, 03:10 AM | #2 |
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would appreciate if anyone can clear this up..
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| Jul28-12, 01:40 AM | #3 |
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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. |
| Jul28-12, 05:43 AM | #4 |
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Rolle's theorem on Envelope Equations
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! |
| Jul28-12, 06:14 AM | #5 |
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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).
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| Jul28-12, 09:02 AM | #6 |
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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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