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cain_herod_sample_problems

cain_herod_sample_problems - Solutions to some of the...

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Solutions to some of the problems in Chapters 15 and 16 of Cain & Herod. 1 Section 15.1 15.1, Problem 1 We use x = u and y = v as parameters. The surface can then be parametrized by r ( u, v ) = u, v, u + 2 v 2 , u, v 0 . 15.1, Problem 2 If we let x = 2 x , y = y and z = 2 2 z , the equation becomes ( x ) 2 + ( y ) 2 + ( z ) 2 = 4 2 , which is a sphere centered at the origin with radius 4 in the x , y , z coordinates. We can use spherical coordinates to parametrize the sphere: x = 4 cos θ sin φ, y = 4 sin θ sin φ, z = 4 cos φ, 0 θ 2 π, 0 φ π. Reverting back to the x, y, z coordinates, we have thus x = 2 cos θ sin φ, y = 4 sin θ sin φ, z = 2 cos φ, 0 θ < 2 π, 0 φ π. or, letting u = θ and v = φ , r ( u, v ) = 2 cos u sin v, 4 sin u sin v, 2 cos v , 0 u 2 π, 0 v π. 15.1, Problem 3 If x 2 + y 2 = 1, we can write, using polar coordinates that x = cos θ and y = sin θ for some θ with 0 θ < 2 π . Using u = θ and v = z as parameters, the cylinder can thus be parametrized by r ( u, v ) = cos u, sin u, v , 0 u 2 π, -∞ < v < . 15.1, Problem 4 If the surface is parametrized by r ( u, v ) = u cos v, u sin v, u , 0 v 2 π, - 1 u 1 , 1

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we have x 2 + y 2 = u 2 cos 2 v + u 2 sin 2 v = u 2 = z 2 . The surface is thus the part of the hyperboloid x 2 + y 2 = z 2 between the planes z = - 1 and z = 1. 15.1, Problem 5 If the surface is parametrized by r ( u, v ) = u cos v, u sin v, u 2 , 0 v 2 π, 1 u 2 , we have x 2 + y 2 = u 2 cos 2 v + u 2 sin 2 v = u 2 = z . The surface is thus the part of the paraboloid z = x 2 + y 2 between the planes z = 1 and z = 2. 15.1, Problem 6 The equation for the sphere of radius 3 centered at the point (1 , 2 , 3) is ( x - 1) 2 + ( y - 2) 2 + ( z - 3) 2 = 3 2 . If we let x = x - 1, y = y - 2 and z = z - 3, we get the equation ( x ) 2 + ( y ) 2 + ( z ) 2 = 3 2 , which is a sphere centered at the origin with radius 3 in the x , y , z coordinates. We can use spherical coordinates to parametrize the sphere: x = 3 cos θ sin φ, y = 3 sin θ sin φ, z = 3 cos φ, 0 θ 2 π, 0 φ π. Reverting back to the x, y, z coordinates, we have thus x = 1 + 3 cos θ sin φ, y = 2 + 3 sin θ sin φ, z = 3 + 3 cos φ, 0 θ 2 π, 0 φ π. or, letting u = θ and v = φ , r ( u, v ) = 1 + 3 cos u sin v, 2 + 3 sin u sin v, 3 + 3 cos v , 0 u 2 π, 0 v π. 15.1, Problem 7 We use the parametrization x = a cos θ sin φ, y = a sin θ sin φ, z = a cos φ, 0 θ 2 π, 0 φ π. or r ( φ, θ ) = a cos θ sin φ, a sin θ sin φ, a cos φ , 0 θ 2 π, 0 φ π. We have r φ = a cos θ cos φ, a sin θ cos φ, - a sin φ , and r θ = - a sin θ sin φ, a cos θ sin φ, 0 . Thus, r φ × r θ = a 2 cos θ sin 2 φ, a 2 sin θ sin 2 φ, a 2 cos φ sin φ = a sin φ r ( φ, θ ) . 2
This implies that the normal vector r φ × r θ to the sphere at a point P on the surface has the same direction has the direction vector of that point, -→ OP . In particular, if P = ( a 3 , a 3 , - a 3 ), the normal vector will have the same direction as a 3 , a 3 , - a 3 or as 1 , 1 , - 1 . The normal line at P as thus vector equation s ( t ) = a 3 , a 3 , - a 3 + t 1 , 1 , - 1 , -∞ < t < , or, more simply, s ( t ) = t 1 , 1 , - 1 , -∞ < t < .

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