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s3 - University of Toronto Scarborough Department of...

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University of Toronto Scarborough Department of Computer & Mathematical Sciences MAT B42H 2011/2012 Solutions #3 1. We are given that the path followed is γ ( t ) = p e t , 1 t , cos p π t 6 PP so the tangent is γ ( t ) = p e t , 1 t 2 , π 6 sin p π t 6 PP . The particle leaves the curve at γ (1) and fol- lows the direction γ (1); i.e., it travels along the line γ (1) + s γ (1) = p 1 e , 1 , cos π 6 P + s p 1 e , 1 , π 6 sin p π 6 PP , s 0. When t = 3, s = 2, and the particle is lo- cated at p 1 e , 1 , cos π 6 P +2 p 1 e , 1 , π 6 sin p π 6 PP = p 1 e , 1 , 3 2 P + p 2 e , 2 , π 6 P = p 1 e , 1 , 3 3 π 6 P . 2. (a) This is just a straight line with direction vector ( 1 , 1 , 1 , 1) (1 , 2 , 3 , 4) = ( 2 , 3 , 4 , 5). For 0 t 1, γ ( t ) = (1 , 2 , 3 , 4) + t ( 2 , 3 , 4 , 5) = (1 2 t, 2 + 3 t, 3 4 t, 4 + 5 t ) traces out this line segment. (b) The curve { ( x, y ) R 2 | 5 x 2 + 9 y 2 = 4 } is just an ellipse. To parametrize we put x = 2 5 cos t and y = 2 3 sin t . Now 5 x 2 + 9 y 2 = 5 p 4 5 cos 2 t P + 9 p 4 9 sin 2 t P = 4(cos 2 t + sin 2 t ) = 4. For 0 t 2 π , γ ( t ) = p 2 5 cos t, 2 3 sin t P traces out the ellipse in the counterclockwise direction. (c) To parametrize the curve of intersection of the surfaces z = 1 + 3 x and 10 x 2 + y 2 z 2 +6 = 0 we substitute the ±rst into the second giving 0 = 10 x 2 + y 2 ( 1+ 3 x ) 2 + 6 = 10 x 2 + y 2 1 6 x 9 x 2 + 6 = ( x 2 6 x ) + y 2 + 5. Completing the square we have ( x 2 3 ) 2 + y 2 + 5 9 = 0 or ( x 2 3 ) 2 + y 2 = 4. The projection of this curve in the xy –plane is a circle of radius 2 centered at (3 , 0). We put x

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s3 - University of Toronto Scarborough Department of...

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