- #1
E'lir Kramer
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Advanced Calculus of Several Variables, Edwards, problem II.4.1: Find the shortest distance from the point (1, 0) to a point of the parabola [itex]y^{2} = 4x[/itex].
This is the Lagrange multipliers chapter. There might be another way to solve this, but the only way I'm interested in right now is the Lagrange multipliers method.
The first tricky thing is that he has no supplied us with the function f that we are trying to minimize.
We need an equation for the distance between some point (x,y) and the point (1,0). [itex] d(x,y) = \sqrt{(x-1)^{2} + y^{2}} [/itex]. For convenience we can minimize [itex]d^{2}(x, y) = (x-1)^{2} + y^{2}[/itex] and get the same answer.
Now, this is a tricky part that I may have gotten wrong, since this author has given only a brief overview of the gradient function. [itex]\nabla d^{2}(x, y) = (2x-2, 2y) [/itex].
Now the parabola function is actually the constraint on this distance function. However, I am having a hard time with the form of this. Is the correct constraint [itex] 0 = 4x - y^{2} [/itex]? The thing is, that's not a function, it's just an equation. I don't understand how I'm suppose to take the gradient of it. Now I can imagine a function [itex] g : \Re^{2} \to \Re [/itex] such that [itex] g(x, y) = 4x - y^{2}[/itex]. And I know the parabola are all the points {[itex] { p \in \Re^{2} : g(p) = 0 } [/itex]}. Perhaps I just didn't understand the proof of the Langrangian method, but I am not sure why my constraint has to be the zero set of some function.
If [itex] \nabla g = (4, -2y) [/itex] (is it?), then I have the Lagrangian equality:
[itex] (2x-2, 2y) = \lambda (4, -2y) [/itex].
The three equations to solve are:
[itex] 2x-2 = 4\lambda \\
2y = -2\lambda y \\
0 = 4x - y^{2}
[/itex]
Is this correct? If so, it's unfortunate, because I can't solve it.
I tried first solving for [itex] \lambda = \frac{2x-2}{4} [/itex]
and [itex] \lambda = -1 [/itex] if y ≠ 0. I think this is justified because if y = 0, then [itex] \lambda = 0 [/itex].
Then [itex] -1 = \frac{2x-2}{4} [/itex], so [itex] x = -1 [/itex].
But then we have [itex] 0 = -4 - y^{2} [/itex], so [itex] -4 = y^{2} [/itex]. So y is an imaginary number? That's impossible. There must be some real solution to this problem.
This is the Lagrange multipliers chapter. There might be another way to solve this, but the only way I'm interested in right now is the Lagrange multipliers method.
The first tricky thing is that he has no supplied us with the function f that we are trying to minimize.
We need an equation for the distance between some point (x,y) and the point (1,0). [itex] d(x,y) = \sqrt{(x-1)^{2} + y^{2}} [/itex]. For convenience we can minimize [itex]d^{2}(x, y) = (x-1)^{2} + y^{2}[/itex] and get the same answer.
Now, this is a tricky part that I may have gotten wrong, since this author has given only a brief overview of the gradient function. [itex]\nabla d^{2}(x, y) = (2x-2, 2y) [/itex].
Now the parabola function is actually the constraint on this distance function. However, I am having a hard time with the form of this. Is the correct constraint [itex] 0 = 4x - y^{2} [/itex]? The thing is, that's not a function, it's just an equation. I don't understand how I'm suppose to take the gradient of it. Now I can imagine a function [itex] g : \Re^{2} \to \Re [/itex] such that [itex] g(x, y) = 4x - y^{2}[/itex]. And I know the parabola are all the points {[itex] { p \in \Re^{2} : g(p) = 0 } [/itex]}. Perhaps I just didn't understand the proof of the Langrangian method, but I am not sure why my constraint has to be the zero set of some function.
If [itex] \nabla g = (4, -2y) [/itex] (is it?), then I have the Lagrangian equality:
[itex] (2x-2, 2y) = \lambda (4, -2y) [/itex].
The three equations to solve are:
[itex] 2x-2 = 4\lambda \\
2y = -2\lambda y \\
0 = 4x - y^{2}
[/itex]
Is this correct? If so, it's unfortunate, because I can't solve it.
I tried first solving for [itex] \lambda = \frac{2x-2}{4} [/itex]
and [itex] \lambda = -1 [/itex] if y ≠ 0. I think this is justified because if y = 0, then [itex] \lambda = 0 [/itex].
Then [itex] -1 = \frac{2x-2}{4} [/itex], so [itex] x = -1 [/itex].
But then we have [itex] 0 = -4 - y^{2} [/itex], so [itex] -4 = y^{2} [/itex]. So y is an imaginary number? That's impossible. There must be some real solution to this problem.
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