2t r t 1 cos 62 then 1 1 t t 1 cos2 t 2 1 2

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Unformatted text preview: y ′ = R cos t, x′′ = −R sin t, y ′′ = −R sin t. By the formula (9), κ= Another solution: |R2 cos2 t + R2 sin2 t| R2 1 = 3= . 2 cos2 r + R2 sin2 t)3/2 (R R R Write r (t) = R cos t i + R sin t j. Then r ′ (t) = −R sin t i + R cos t j, r ′ (t) = T (t) = (−R sin t)2 + (R cos t)2 = R, −R sin t i + R cos t j r ′ (t) = = −sin t i + cos t j. ′ (t) r R Then T ′ (t) = cos ti − sin tj, and hence κ(t) = [Example] T ′ (t) = (cos t)2 + (sin t)2 = 1 T ′(t) 1 =. r ′ (t) R Consider a curve C : x = x(t) = t, y = y (t) = sin t, 0 ≤ t ≤ 2π. Write r(t) = t i + sin t j. Then r ′ (t) = i + cos t j, T (t) = r ′ (t) = √ 1 + cos2 t, i + cos t j r ′ (t) =√ = (1 + cos2 t)−1/2 i + cos t(1 + cos2 t)−1/2 j. 2t r ′ (t) 1 + cos 62 Then 1 1 T ′ (t) = − (1 + cos2 t)− 2 −1 · 2 cos t(−sin t)i 2 1 1 1 + (−sin t)(1 + cos2 t)1 2 + cos t · − (1 + cos2 t)− 2 −1 (2 cos t)(−sin t) j 2 sin t sin t · cos t = 3 i− 3j (1 + cos2 t) 2 (1 + cos2 t) 2 and hence ′ T (t) = κ(t) = r ′ (t) 1 3 (1+cos2 t) 2 √ sin2 tcos2 t + sin2 t (1 + cos2 t) Another formula for curvature = 1 2 |sin t| 3 (1 + cos2 t) 2 . Let r(t) be a parametrization of a curve. Let v (t) := r ′ (t), a(t)...
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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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