14.1%20Ex31-32

# 14.1%20Ex31-32 - 460 C H A P T E R 14 C A L C U L U S O F...

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460 CHAPTER 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) 31. The intersection of the plane y = 1 2 with the sphere x 2 + y 2 + z 2 = 1 SOLUTION Substituting y = 1 2 in the equation of the sphere gives: x 2 + ± 1 2 2 + z 2 = 1 x 2 + z 2 = 3 4 This circle in the horizontal plane y = 1 2 has the parametrization x = 3 2 cos t , z = 3 2 sin t . Therefore, the points on the intersection of the plane y = 1 2 and the sphere x 2 + y 2 + z 2 = 1, can be written in the form ³ 3 2 cos t , 1 2 , 3 2 sin t ² , yielding the following parametrization: r ( t ) = * 3 2 cos t , 1 2 , 3 2 sin t + . 32. The intersection of the surfaces z = x 2 y 2 and z = x 2 + xy 1 We solve for x and z in terms of y . Equating the two equations and solving for x gives: x 2 y 2 = x 2 + 1 y 2 = 1 = 1 y 2 Notice that on the intersection curve, y 6= 0(since y = 0gives z = x 2 and z = x 2 1, which are not equal for any z ). Dividing by y we get: x = 1 y 2 y =− y + 1 y We now substitute x in terms of y in the equation z = x 2 y 2 , to obtain: z = Ã 1 y 2 y ! 2 y 2 = ( 1 y 2 ) 2 y 4 y 2 = 1 2 y 2 y 2 2 + 1 y 2 Therefore the points of the intersection can be written in the form
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