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Curvature Question

  • Thread starter shards5
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  • #1
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Homework Statement


Find the curvature K(t) of the curve r(t) = (-4sin(t)) i + (-4sin(t)) j + (5cos(t)) k.


Homework Equations


K(t) = |r'(t) x r"(t)| / |r'(t)|3

The Attempt at a Solution


r'(t) = (-4cos(t))i + (-4cos(t))j + (-5sin(t))k
r"(t) = (4sin(t))i + 4sin(t))j + (-5cos(t))k
|r'(t)| = sqrt(16cos2(t) + 16cos2(t) + 25sin2(t))
r'(t) x r"(t) = [20cos(t)+20sin(t)]i - [20cos(t) + 20sin(t)]k +0j
|r'(t) x r"(t)| = sqrt([20cos(t)+20sin(t)]^2 + [-20cos(t) - 20sin(t)]^2)
Answer should be:
sqrt([20cos(t)+20sin(t)]^2 + [-20cos(t) - 20sin(t)]^2)/[sqrt(16cos2(t) + 16cos2(t) + 25sin2(t))]3
But it isn't, so I am confused as to what I am doing wrong.
 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
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think you added a minus... though i didn't check the original cross product
r'(t) x r"(t) = [20cos(t)+20sin(t)]i - [20cos(t) + 20sin(t)]k +0j
so shouldn't it become
|r'(t) x r"(t)| = sqrt([20cos(t)+20sin(t)]^2 + [20cos(t) + 20sin(t)]^2)
|r'(t) x r"(t)| = sqrt(2[20cos(t)+20sin(t)]^2)
 
  • #3
38
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I don't think that would be it since the cross product formula says that for the middle number it is subtraction. Besides, it is being squared, even if it is negative the answer which results will be positive.
 
  • #4
lanedance
Homework Helper
3,304
2
r(t) = (-4sin(t)) i + (-4sin(t)) j + (5cos(t)) k.
r'(t) = (-4cos(t)) i + (-4cos(t)) j + (-5sin(t)) k.
r''(t) = (4sin(t)) i + (4sin(t)) j + (-5cos(t)) k.

ok redoing the cross product you get
|r'(t) x r"(t)| = i(20c^2 + 20s^2) + j(-20s^2 - 20c^2) + k(-16cs +16cs)

which becomes
|r'(t) x r"(t)| = i(1) + j(-1) + k(0)
 

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