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20esolns6 - Homework 6 Solutions MTH 20E 7.3.1 Tu Tv =(2 2u...

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Homework 6 Solutions MTH 20E 7.3.1 T u = (2 , 2 u, 0) T v = (0 , 1 , 2 v ) n = (2 , 2 u, 0) × (0 , 1 , 2 v ) = (4 uv, - 4 v, 2) If (2 u, u 2 + v, v 2 ) = (0 , 1 , 1), then u = 0 , v = 1; hence, the equation of the tangent plane is ( x - 0 , y - 1 , z - 1) · (0 , - 4 , 2) = 0 i.e., - 4 y + 2 z = - 2 7.3.5 Consider the spherical coordinate system; if we identify u with φ and v with θ , and set ρ = 1, we have the parametrization in question. Hence, this surface is the unit sphere in R 3 . The tangent vectors for this surface are T u = (cos v cos u, sin v cos u, - sin u ) , T v = ( - sin v sin u, cos v sin u, 0) . Then the normal vector for this surface is T u × T v = (cos v sin 2 u, sin v sin 2 u, cos u sin u ) and its length is T u × T v = ( cos v sin 2 u ) 2 + ( sin v sin 2 u ) 2 + (cos u sin u ) 2 = sin u. Thus, the unit normal vector for the unit sphere is (cos v sin u, sin v sin u, cos u ) = ( x, y, z ). 7.4.6 One can parametrize the unit sphere in the manner described in 7.3.5; it remains to place bounds on u and v in order to describe the surface in question. This surface can also be described as the portion of the unit sphere bounded by the cone x 2 + y 2 - z 2 = 0 , z 0 . Recall that in spherical coordinates, fixing φ constant describes a cone em- anating from the origin along the positive z -axis; hence, our parametriza- tion should have bounds 0 v 2 π , 0 u c , where c is the angle 1
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between the positive z -axis and our boundary cone. The value of c can
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