21c_02f_1bs

# 21c_02f_1bs - Math 21C First Midterm Version B Oct 21 2002...

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Math 21C First Midterm Version B Oct. 21, 2002 1. dx/dt =4 t + 3 and dy/dt =3 t 2 - 6 t . (a) length = R 4 - 2 p (4 t +3) 2 +(3 t 2 - 6 t ) 2 dt (b) To be horizontal, dy/dt = 0, which means t =0o r t = 2. This gives the two points ( - 1 , 2) and (13 , - 2). 2. (a) Let c = --→ BA = h 1 , 1 , - 1 i and a = --→ BC = h x - 1 , 3 , 3 i . Since a must be perpendic- ular to c , a · c =0. Thus x = 1 and so C is (1 , 3 , 4) and a = h 0 , 3 , 3 i . (b) Let b = -→ AC = h- 1 , 2 , 4 i . The cosine is a · b | a || b | = 6+12 9+9 1+4+16 = 18 18 21 = r 18 21 = r 6 7 . 3. (a) First plane : Since h 1 - 0 , 1 - 0 , 0 - 0 i and h 1 - 0 , 1 - 0 , 2 - 0 i are parallel to the plane, their cross product is a normal. The cross product is h 2 , - 2 , 0 i . Since the origin is in the plane, the equation is 2 x - 2 y = 0 or, equivalently, x - y =0. Second plane : The equation is h 1 , 0 , 2 i·h x - 0 ,y - 0 ,z - 0 i = 0; that is, x +2 z =0. (b) We have the equations x - y = 0 and x +2 z = 0 for the planes. Since both must hold for the intersection, we could take, say, z = t . Then x = - 2 t and y = - 2 t . In other words, h x, y, z i = t h- 2 , - 2 , 1 i
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## This note was uploaded on 06/17/2009 for the course MATH 20C taught by Professor Helton during the Fall '08 term at UCSD.

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