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Curvature problem |
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| Jun17-08, 08:09 AM | #1 |
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Curvature problem
1. The problem statement, all variables and given/known data
A Curve C is given by the polar equation r=f(theta). Show that the curvature K at the point (r, theta) is K=|2(r')^2 - rr'' + r^2| -------------------- [(r')^2 + r^2]^(3/2) *Represent the curve by r(theta) = r<cos theta, sin theta> 2. Relevant equations I have so far taken the first and second derivatives of x= r cos Theta and y=r sin theta and I know that the formula below is probably involved but i don't know how K(t) = ||r'(t) X r''(t)|| --------------- ||r'(t)||^3 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution |
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| Jun17-08, 09:04 AM | #2 |
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Write your second formula K(theta)=|f'(theta) X f''(theta)|/|f'(theta)|^3 where f(theta)=r(theta)*[cos(theta),sin(theta)]. Now put your derivatives of f(theta) in. Your presentation above has too many r's in it. Is that what's confusing?
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| Jun17-08, 10:50 AM | #3 |
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yea, i dnt know exactly what i'm doing when im plugging things in.. like f(theta) is a vector?.. f(theta) = r cos theta + r sin theta... or... I just don't understand :( but i get the math, just not what im subsituting..
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| Jun17-08, 10:53 AM | #4 |
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Curvature problem
f(theta) IS a vector [r(theta)*cos(theta),r(theta)*sin(theta)]. Those are the x and y components. Consider theta a parameter of the curve, like t.
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| Jun17-08, 11:35 AM | #5 |
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i got that the actual K should be 1/ (r^2 + 1)^(1/2) which i cant seem to relate to the K given.. when i subsitute it, it's too messy
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| Jun17-08, 02:35 PM | #6 |
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