# sdfsf - e t-t 2-t c with r(0 =(1 0 c = r =(0 1 0 so that c...

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MATH 241H - EXAM 1 - Solutions - September 20, 2004 1. (a) neither since b 6 = c a and a · b = 19 6 = 0. (b) perpendicular since a · c = 0. 2. The vector from (2 , 5 , - 3) to (3 , 1 , - 1) is L = (1 , - 4 , 2) so we can take (2 , 5 , - 3) + t (1 , - 4 , 2), giving parametric equations x = 2 + t, y = 5 - 4 t, z = - 3 + 2 t . (This is not the only correct answer) 3. The plane contains the point P 0 = (3 , - 3 , 1). The line contains the point P 1 = (1 , - 1 , 0) and is parallel to the vector L = (3 , 1 , 2). Thus the vectors P 1 P 0 = (2 , - 2 , 1) and L are in the plane so we can take N = (2 , - 2 , 1) × (3 , 1 , 2) = ( - 5 , - 1 , 8). Using (1 , - 1 , 0) as the point we get equation - 5( x - 1) - ( y + 1) + 8 z = 0 which simpliﬁes to 5 x + y - 8 z = 4. 4. lim t 0 F ( t ) = j - 3 k since the expression for t 6 = 0 has this as limit, and the limit does not care about the value of F (0). 5. a ( t ) = ( e t , - 2 , 0) so v ( t ) = ( e t , - 2 t, 0) + c with v (0) = (1 , 0 , 0) + c = v 0 = (1 , 0 , - 1). Thus c = (0 , 0 , - 1), so we have v ( t ) = ( e t , - 2 t, - 1). Now r ( t ) = (
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Unformatted text preview: e t ,-t 2 ,-t ) + c with r (0) = (1 , , 0) + c = r = (0 , 1 , 0) so that c = (-1 , 1 , 0). Thus r ( t ) = ( e t-1 , 1-t 2 ,-t ). That is, r ( t ) = ( e t-1) i + (1-t 2 ) j-t k and v ( t ) = e t i-2 t j-k . 6. (a) v ( t ) = (2 t, t 2 ,-2) so k v ( t ) k = √ 4 t 2 + t 4 + 4 = t 2 + 2. Thus L = Z 2 1 ( t 2 + 2) dt = t 3 / 3 + 2 t | 2 1 = 8 / 3 + 4-1 / 3-2 = 13 / 3 . (b) a T = d/dt ( t 2 + 2) = 2 t . Now a ( t ) = (2 , 2 t, 0) so a N = √ 4 + 4 t 2-4 t 2 = 2. (c) v (1) = (2 , 1 ,-2) and a (1) = (2 , 2 , 0) so v (1) × a (1) = (4 ,-4 , 2). Now k v (1) × a (1) k = √ 36 = 6 and k v (1) k = √ 9 = 3 so κ (1) = 6 / 27 = 2 / 9....
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