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Unformatted text preview: Ma 116  Calculus II Hw4 (Solutions) Due: March 25, 2011 Pledge and Sign : An important component of these written exercises is the quality of your presentation, including: legibility, organization of the solution, and clearly stated reasoning where appropriate. Points will be deducted for sloppy work or insufficient explanations. 1. The sphere S of radius 4 centered at the origin is given by the equation x 2 + y 2 + z 2 = 16 , the circular cylinder C of radius 2 with the axis passing through the point (2 , , 0) and orthogonal to the xyplane is given by the equation ( x 2) 2 + y 2 = 4 . Let be the curve of intersection of S and C (see the picture below). (a) Show that can be described by the parametric equations x = 2(1 + cos t ) , y = 2sin t, z = 4sin parenleftbigg t 2 parenrightbigg , t ( 2 , 2 ) . (b) Find all points of intersection of with the plane x = 2. (c) Find the parametric equation of the tangent line to at the point Q (2 , 2 , 2 2). Solution : (a) is described by the system braceleftbigg x 2 + y 2 + z 2 = 16 ( x 2) 2 + y 2 = 4 so, the parametrization x = 2(1 + cos t ) , y = 2sin t, z = 4sin parenleftbigg t 2 parenrightbigg , t ( 2 , 2 ) describes the curve if the expressions for x, y and z solve the system. Hence, we plug them in and check. We have x 2 + y 2 + z 2 = (2(1 + cos t )) 2 + (2sin t ) 2 + parenleftbigg 4sin parenleftbigg t 2 parenrightbiggparenrightbigg 2 = 4 + 8cos t + 4cos 2 t + 4sin 2 t + 16sin 2 parenleftbigg t 2 parenrightbigg = 4 + 8cos t + 4 + 16 1 cos t 2 = 4 + 8cos t + 4 + 8 8cos t = 16 , and ( x 2) 2 + y 2 = (2(1 + cos t ) 2) 2 + (2sin t ) 2 = (2cos t ) 2 + (2sin t ) 2 = 4 . (b) x = 2 implies (2 2) 2 + y 2 = 4 and y 2 = 4, so, y = 2. In both cases y = 2 , y = 2, from x 2 + y 2 + z 2 = 16 we get z 2 = 8 and z = 2 2. Hence, we get four points (2 , 2 , 2...
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This note was uploaded on 10/09/2011 for the course MATHEMATIC MA116 taught by Professor Duboski during the Fall '11 term at Stevens.
 Fall '11
 Duboski
 Calculus

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