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# hw4sol - Math 53 Homework 4 Solutions 13.1 # 14: Since x =...

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Unformatted text preview: Math 53 Homework 4 Solutions 13.1 # 14: Since x = cos t and y = cos t , the curve lies in the vertical plane y = x . Moreover, y 2 + z 2 = cos 2 t + sin 2 t = 1, so the curve lies on the cylinder y 2 + z 2 = 1 (or also on the cylinder x 2 + z 2 = 1). It is therefore the ellipse where the cylinder and the plane intersect. x y z 13.1 # 22: x = e t cos 10 t , y = e t sin 10 t , z = e t : then x 2 + y 2 = e 2 t cos 2 10 t + e 2 t sin 2 10 t = e 2 t = z 2 . So the curve lies on the cone z 2 = x 2 + y 2 . Also, z is always positive. So the graph must be I. 13.1 # 23: x = cos t , y = sin t , z = sin 5 t : then x 2 + y 2 = cos 2 t + sin 2 t = 1, so the curve lies on the circular cylinder of radius 1 centered on the z-axis. Each of x, y, z is periodic, and the curve repeats itself after t reaches 2 ; moreover z oscillates between 1 and 1 (5 times over the interval 0 t 2 ). So the graph must be V. 13.2 # 6: (a),(c) x ( t ) = e t , y ( t ) = e t ; note that eliminating t gives y = 1 /x . x y (1 , 1) vector r (0) r (0) (b) r ( t ) = e t e t . 13.2 # 9: differentiating each component (and using the product rule), r ( t ) = ( sin t + t cos t, 2 t, cos 2 t 2 t sin 2 t ) . 13.2 # 25: differentiating vector r ( t ) = ( e t cos t, e t sin t, e t ) , we get r ( t ) = ( e t (cos t + sin t ) , e t (cos t sin t ) , e t ) . The point (1 , , 1) corresponds to t = 0, and the tangent vector there is r (0) = ( 1 , 1 , 1 ) . Thus, the tangent line is directed along the vector ( 1 , 1 , 1 ) , and parametric equations are x = 1 t , y = t , z = 1 t . 1 13.2 # 31: Note that vector r 1 ( t ) = ( 1 , 2 t, 3 t 2 ) while vector r 2 ( t ) = ( cos t, 2 cos 2 t, 1 ) . Since both curves pass through the origin at t = 0, the tangent vectors there are respectively r 1 (0) = ( 1 , , ) and r 2 (0) = ( 1 , 2 , 1 )...
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## hw4sol - Math 53 Homework 4 Solutions 13.1 # 14: Since x =...

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