- #1

E'lir Kramer

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Hello all, and thanks again to all the help I've been getting with this book. This is a two part problem in Advanced Calculus of Several Variables, C. H. Edwards Jr. I have the first part and the second part should be easy, but I find I'm stumped.

Since the second part builds on the solution of the first part, I'll give both problem statements and the proof of my first part.

II.1.2a: Let [itex]f : \Re \to \Re^{n}[/itex] and [itex]g : \Re \to \Re^{n}[/itex] be two differentiable curves with f'(t) ≠ 0 and g'(t) ≠ 0 for all all [itex] t \in \Re [/itex]. Suppose the two points [itex] p = f(s_{0})[/itex] and [itex] q = g(t_{0}) [/itex] are closer than any other pair of points on the two curves. Then prove that the vector p - q is orthogonal to both velocity vectors [itex]f'(s_{0}), g'(t_{0})[/itex]. Hint: the point [itex] (s_{0}, t_{0}) [/itex] must be a critical point for the function [itex]\rho : \Re^{2} -> \Re [/itex] defined by [itex] \rho(s, t) = \left \|

f(s) - g(t)

\right\|^{2}[/itex]

From an earlier problem I have proved three lemmas to my satisfaction:

Lemma 1: [itex]<a,a> - <b,b> = <a+b, a-b>[/itex]

Lemma 2: If [itex] n(t) = <c, f(t)> [/itex] for some constant c, then [itex] n'(t) = <c, f'(t)>[/itex]

Lemma 3: if [itex] n(t) = <f(t), f(t) > = \left \| f(t) \right \|^{2}[/itex], then [itex] n'(t) = 2<f'(t), f(t)> [/itex].

Now the problem statement formally is that we must show that [itex] <p-q, f'(t)> = 0[/itex], [itex] <p-q, g'(t)> = 0[/itex].

This is just a matter of algebra and the observation that [itex]\rho`(s_{0}, t_{0}) = 0 [/itex] with respect to either partial derivative. We get this from the hint, but the hint was unnecessary.

[itex] \rho = <f(s), f(s)> + <g(t), g(t)> - 2<f(s), g(t)>[/itex]

Applying lemmas 2 and 3, and differentiating first wrt s:

[itex] \frac{d\rho}{ds}(s, t) = 2<f'(s), f(s)> - 2<f'(s), g'(t)> \\

= 2<f(s) - g(t), f'(s)> \\

\frac{d\rho}{dt}(s, t) = 2<g'(t), g(t)> - 2<f'(s), g'(t)> \\

= 2<f(s) - g(t), g'(t)>[/itex]

When we plug in [itex](s_{0}, t_{0})[/itex] to these equations, we know that they are equal to 0 by observation that [itex](s_{0}, t_{0})[/itex] is a critical point. At the same time, the expression for [itex]f(s) - g(t)[/itex], which appears in both equations, turns into [itex]f(s_{0}) - g(t_{0}) = p - q[/itex] and the equations become [itex] 0 = < p - q, g'(t_{0}) >[/itex] and [itex] 0 = < p - q, f'(s_{0})> [/itex], QED.

Now the part I can't get is II.1.2b:

So far I just have some aimless restating of facts that I know. All of them are available as results of part A above, except for two:

[itex] f'(s) = (1, 2, -1) \\

g'(t) = (1, 1, 2)[/itex] for all s and t.

Somehow I've got to use these equations in reverse and solve for [itex]s_{0}[/itex] and [itex]t_{0}[/itex]. With that observation I have two equations of two unknowns:

[itex] <f(s_{0}) - g(t_{0}), (1, 2, -1)> = 0 = <f(s_{0}) - g(t_{0}), (1, 1, 2)>[/itex]. But now I am truly stuck. I need to get these variables out of the functions, but g(t) is not linear, so I can't easily construct an inverse function for it.

Since the second part builds on the solution of the first part, I'll give both problem statements and the proof of my first part.

II.1.2a: Let [itex]f : \Re \to \Re^{n}[/itex] and [itex]g : \Re \to \Re^{n}[/itex] be two differentiable curves with f'(t) ≠ 0 and g'(t) ≠ 0 for all all [itex] t \in \Re [/itex]. Suppose the two points [itex] p = f(s_{0})[/itex] and [itex] q = g(t_{0}) [/itex] are closer than any other pair of points on the two curves. Then prove that the vector p - q is orthogonal to both velocity vectors [itex]f'(s_{0}), g'(t_{0})[/itex]. Hint: the point [itex] (s_{0}, t_{0}) [/itex] must be a critical point for the function [itex]\rho : \Re^{2} -> \Re [/itex] defined by [itex] \rho(s, t) = \left \|

f(s) - g(t)

\right\|^{2}[/itex]

From an earlier problem I have proved three lemmas to my satisfaction:

Lemma 1: [itex]<a,a> - <b,b> = <a+b, a-b>[/itex]

Lemma 2: If [itex] n(t) = <c, f(t)> [/itex] for some constant c, then [itex] n'(t) = <c, f'(t)>[/itex]

Lemma 3: if [itex] n(t) = <f(t), f(t) > = \left \| f(t) \right \|^{2}[/itex], then [itex] n'(t) = 2<f'(t), f(t)> [/itex].

Now the problem statement formally is that we must show that [itex] <p-q, f'(t)> = 0[/itex], [itex] <p-q, g'(t)> = 0[/itex].

This is just a matter of algebra and the observation that [itex]\rho`(s_{0}, t_{0}) = 0 [/itex] with respect to either partial derivative. We get this from the hint, but the hint was unnecessary.

[itex] \rho = <f(s), f(s)> + <g(t), g(t)> - 2<f(s), g(t)>[/itex]

Applying lemmas 2 and 3, and differentiating first wrt s:

[itex] \frac{d\rho}{ds}(s, t) = 2<f'(s), f(s)> - 2<f'(s), g'(t)> \\

= 2<f(s) - g(t), f'(s)> \\

\frac{d\rho}{dt}(s, t) = 2<g'(t), g(t)> - 2<f'(s), g'(t)> \\

= 2<f(s) - g(t), g'(t)>[/itex]

When we plug in [itex](s_{0}, t_{0})[/itex] to these equations, we know that they are equal to 0 by observation that [itex](s_{0}, t_{0})[/itex] is a critical point. At the same time, the expression for [itex]f(s) - g(t)[/itex], which appears in both equations, turns into [itex]f(s_{0}) - g(t_{0}) = p - q[/itex] and the equations become [itex] 0 = < p - q, g'(t_{0}) >[/itex] and [itex] 0 = < p - q, f'(s_{0})> [/itex], QED.

Now the part I can't get is II.1.2b:

**Apply the result of (a) to find the closest pair of points on the "skew" straight lines in**[itex] \Re^{3} [/itex]**defined by**[itex]f(s) = (s, 2s, -s)[/itex]**and**[itex]g(t) = (t+1, t-2, 2t+3)[/itex].So far I just have some aimless restating of facts that I know. All of them are available as results of part A above, except for two:

[itex] f'(s) = (1, 2, -1) \\

g'(t) = (1, 1, 2)[/itex] for all s and t.

Somehow I've got to use these equations in reverse and solve for [itex]s_{0}[/itex] and [itex]t_{0}[/itex]. With that observation I have two equations of two unknowns:

[itex] <f(s_{0}) - g(t_{0}), (1, 2, -1)> = 0 = <f(s_{0}) - g(t_{0}), (1, 1, 2)>[/itex]. But now I am truly stuck. I need to get these variables out of the functions, but g(t) is not linear, so I can't easily construct an inverse function for it.

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