# D1 2 d2 2 d3 2 therefore t t 0 for all t then

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Unformatted text preview: 3 t) k. Then r ′ (t) = d1 i + d2 j + d3 k = constant vector, r ′ (t) = (d1 )2 + (d2 )2 + (d3 )2 = constant number, 60 (8) T (t) = r ′ (t) = r ′ (t) d1 i + d2 j + d3 k = constant vector. (d1 )2 + (d2 )2 + (d3 )2 Therefore T ′ (t) = 0 for all t. Then T ′ (t) = 0 and hence κ(t) = T ′ (t) = r ′ (t) Curvature for a plane curve formula becomes For a plane curve r(t) = x(t) i + y (t) j , the curvature κ= where x′ = dx , y′ dt = dy , x′′ dt = d2 x dt2 0 = 0. (d1 )2 + (d2 )2 + (d3 )2 |x′ y ′′ − y ′x′′ | and y ′′ = d2 y . dt2 In fact, r = x i + y j, r ′ = x′ i + y ′ j , r T= r r ′ ′ = (9) 3/2 (x′ )2 + (y ′)2 ′ = (x′ )2 + (y ′ )2 . Then x′ i + y ′ j = x′ (x′ )2 + (y ′ )2 ((x′ )2 + (y ′)2 )1/2 1 −2 i + y ′ (x′ )2 + (y ′)2 −1 2 j. By the formula (f g )′ = f ′ g + f g ′, we get T′ = 1 x′′ ((x′ )2 + (y ′)2 )−1/2 − x′ ((x′ )2 + (y ′)2 )−3/2 (2x′ x′′ + 2y ′y...
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## This note was uploaded on 10/02/2012 for the course CALCULUS 2433 taught by Professor Shanyuji during the Fall '12 term at University of Houston.

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