Curvature Proof Problem


by MichaelT
Tags: curvature, proof
MichaelT
MichaelT is offline
#1
Dec13-08, 06:08 PM
P: 25
1. The problem statement, all variables and given/known data
If c is given in terms of some other parameter t and c'(t) is never zero, show that
k = ||c'(t) x c"(t)||/||c'(t)||3

The first two parts of this problem involved a path parametrized by arc length, but this part says nothing about that, so I assume that this path is not parametrized by arc length.

2. Relevant equations
I have found that
T'(t) = c"(t) = [||c'(t)||2c"(t) - c'(t)(c'(t) dot c"(t))]/||c'(t)||3

I am having trouble figuring out how to relate the curvature to the equation
k = ||c'(t) x c"(t)||/||c'(t)||3

3. The attempt at a solution
I can express the equation as the components so
F(x,y,z) = [(z"y'-y"z')2+(x"z'-x'z")2(y"x'-x'y")2]1/2/(x'2+y'2+z'2)3/2

Where should I go from here, and is the above equation useful at all? Do I need to find
||c"(t)|| as well (and is that even possible)?
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Dick
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#2
Dec13-08, 07:08 PM
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P: 25,176
What you have the numerator there is (c'(t).c'(t))*c''(t)-c'(t)(c'(t).c''(t)). There is a vector identity that tells you that that is the same as c'(t)x(c''(t)xc'(t)). When you take the norm of that note that c'(t) and c''(t)xc'(t) are orthogonal. So the sin(theta) in the cross product is one. Is that enough of a hint?
MichaelT
MichaelT is offline
#3
Dec14-08, 12:37 AM
P: 25
I am definitely getting there with this hint.

so I take the norm of c'(t) X (c"(t) X c'(t)) and get

||c'(t) X (c"(t) X c'(t))|| = ||c'(t)||* ||c"(t) X c'(t)||sin(theta) where theta = pi/2

so ||c'(t) X (c"(t) X c'(t))|| = ||c'(t)||* ||c"(t) X c'(t)||

Now I need to relate this to k = ||c"(t)|| correct? I have taken the norm of the numerator, but not of the denominator ||c'(t)||^3

Thanks for your help!

Dick
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#4
Dec14-08, 09:13 AM
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P: 25,176

Curvature Proof Problem


So far you've only got dT/dt. To get the curvature you want dT/ds, where s is arclength. The wikipedia calls it Lagranges formula http://en.wikipedia.org/wiki/Cross_product


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