Register to reply

Curvature Proof Problem

by MichaelT
Tags: curvature, proof
Share this thread:
MichaelT
#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)?
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
Dick
#2
Dec13-08, 07:08 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
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
#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
#4
Dec14-08, 09:13 AM
Sci Advisor
HW Helper
Thanks
P: 25,228
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


Register to reply

Related Discussions
Proof: Curvature Zero -> Motion along a line Calculus & Beyond Homework 3
Curvature problem Calculus & Beyond Homework 5
Surface curvature problem General Math 0
:frown: Normal curvature integral proof Calculus & Beyond Homework 1
Curvature Circle Proof Calculus & Beyond Homework 0