midterm2_solution_problems1thru3

midterm2_solution_problems1thru3 - 1 Let C be a curve...

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Unformatted text preview: 1 Let C be a curve defined by the position function r ( t ) = < sin 2 t, t, cos 2 t > . (a) Calculate the equation of the Normal plane at the point (0 , , 1). Its easier than you think. Solution : The Normal plane is spanned by N and B , so it is perpendicular to T . The point (0 , , 1) corresponds t = . r ( t ) = < 2 cos 2 t, 1 ,- 2 sin 2 t > T = r ( t ) | r ( t ) | = 1 5 < 2 cos 2 t, 1 ,- 2 sin 2 t > T ( ) = < 2 5 , 1 5 , > Then the equation of the Normal plane at (0 , , 1) is 2 5 ( x- 0) + 1 5 ( y- ) + 0( z- 1) = 0 which can be simplified as 2 x + y- = 0 (b) Calculate the equation of the Osculating plane at the point (0 , , 1). Solution : The Normal plane is spanned by T and N , so it is perpendicular to B . T = 1 5 <- 4 sin 2 t, ,- 4 cos 2 t > N = T | T | = <- sin 2 t, ,- cos 2 t > N ( ) = < , ,- 1 > B = T N = <- 1 5 , 2 5 , > Then the equation of the Osculating plane at (0 , , 1) is- 1 5 ( x...
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This note was uploaded on 05/15/2008 for the course MATH 32A taught by Professor Gangliu during the Spring '08 term at UCLA.

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midterm2_solution_problems1thru3 - 1 Let C be a curve...

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