Unformatted text preview: k < 1. For k > 1, as k increases, the hyperbolae move further from the asymptotes. 3. The point (1 , 1 , 1) corresponds to t = 1. For a normal vector to the normal plane we can take r (1) = h 1 , 2 , 3 i , so the equation is ( x1) + 2( y1) + 3( z1) = 0. As a normal vector for the osculating plane we can take r (1) × r 00 (1). This is because B ( t ) is perpendicular to both T ( t ) and N ( t ), and the plane of T ( t ) and N ( t ) is the same as the plane of r ( t ) and r 00 ( t ). Thus, the normal vector we want is h 1 , 2 , 3 i × h , 2 , 6 i = h 6 ,6 , 2 i , so we can write the equation of the osculating plane as 3( x1)3( y1) + ( z1) = 0. 4. The endpoints correspond to t = 0 and t = 2 π , so the arc length is given by Z 2 π  r ( t )  dt = Z 2 π p sin 2 t + cos 2 t + 1 dt = 2 π √ 2 ....
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 Fall '08
 Keeran
 Calculus, Linear Algebra, Geometry, Determinant, Scalar, Vector Space, normal vector

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