20Chw2 - x-9 2 = cos t and y +4 3 = sin t , or in other...

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MATH 20C graded homework 2. (18+2 points) 12.5 # 26 (6 points): Find the equation of the plane containing the point P = ( - 1 , 0 , 1) and r ( t ) = h t + 1 , 2 t, 3 t - 1 i . Solution: Since the plane contains the line r ( t ), and r (0) = h 2 , 0 , - 2 i and r (1) = h 2 , 2 , 2 i , then Q = (2 , 0 , - 2) and R = (2 , 2 , 2) are two points on the plane. The vectors v = ~ PQ = h 2 , 0 , - 2 i and w = ~ PR = h 3 , 2 , 1 i lie on the plane, so a normal vector to the plane is given by n = v × w = h 4 , - 8 , 4 i . Therefore, the equation of the plane is 4( x + 1) - 8( y - 0) + 4( z - 1) = 0 , or x - 2 y + z = 0 . 13.1 # 30 (6 points): Find a parametrization of the ellipse ( x 2 ) 2 + ( y 3 ) 2 = 1 in the xy -plane, translated to have center (9 , - 4 , 0) . Solution: The ellipse in the xy -plane with center at (9 , - 4 , 0) has equation ± x - 9 2 ² 2 + ± y + 4 3 ² 2 = 1 . Since cos 2 t + sin 2 t = 1, we can parametrize by setting
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Unformatted text preview: x-9 2 = cos t and y +4 3 = sin t , or in other words, r ( t ) = h 2 cos t + 9 , 3 sin t-4 , i is a parametrization of the translated ellipse. Note: Also see Section 13.2 Example 3 in your text for a dierent viewpoint. 13.2 # 28 (6 points): Evaluate d dt r ( g ( t )) using the chain rule, where r ( t ) = h 4 sin 2 t, 6 cos 2 t i and g ( t ) = t 2 . Solution : Since r ( t ) = h 8 cos 2 t,-12 sin 2 t i , then d dt r ( g ( t )) = g ( t ) r ( g ( t )) = 2 t h 8 cos 2 t 2 ,-12 sin 2 t 2 i = h 16 t cos 2 t 2 ,-24 t sin 2 t 2 i ....
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This note was uploaded on 03/28/2011 for the course MATH 20C taught by Professor Helton during the Winter '08 term at UCSD.

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