Find the curvature of r(t) = <1, t^2, e^t>
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Find the curvature of r(t) = <1, t^2, e^t>

[From: ] [author: ] [Date: 11-09-30] [Hit: ]
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I don't know what 'm doing wrong. I'm getting this radical that can probably be simplified, but that would be a hell of a lot of work and I don't think it's right in the first place.

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I am used to using x-hat, y-hat, and z-hat for unit vectors so I will just use x,y,z below in the same way

r(t) = t x + t^2 y + e^t z

r' = x + 2t y + e^t z

r'' = 2 y + e^t z

r' cross r'' = (2t e^t - 2e^t)x - e^t y + 2 z = 2e^t (t - 1)x - e^t y + 2z

|r'| = sqrt{ 1 + 4t^2 + e^{2t} } <---- there is no way to simplify this

|r'|^3 = {1 + 4t^2 + e^{2t} }^{3/2}

curvature = (r' cross r'') / |f'|^3

= {2e^t (t - 1)x - e^t y + 2z } / {1 + 4t^2 + e^{2t} }^{3/2} x...................[Ans.]

If you want the magnitude (not vector) of curvature

take the magnitude

= sqrt{ 4e^t (t - 1)^2 + e^{2t} + 4 } / { 1 + 4t^2 + e^{2t} }^{3/2}

No way to simplify this either
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