2 cos u sin v 2 sin u and φ v 2 sin u sin v 2 sin u

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2 cos u sin v, - 2 sin u ) and φ v = ( - 2 sin u sin v, 2 sin u cos v, 0 ) giving φ u × φ v = ( 4 sin 2 u cos v, 4 sin 2 u sin v, 4 sin u cos u ) . A normal vector at (1 , 0 , 2) is φ u × φ v parenleftbigg π 4 , 3 π 4 parenrightbigg = ( - 2 , 2 , 2 ) . The tangent plane is given by ( - 2 , 2 , 2 ) · ( ( x, y, z ) - (1 , 0 , 2) ) = 0. which simplifies to x - y - 2 z = - 1. (b) The gradient is f ( x, y, z ) = 2 ( x - 2 , y +1 , z ) and f (1 , 0 , 2) = ( - 2 , 2 , 2 2 ) . The tangent plane is given by 0 = f (1 , 0 , 2) · ( x - (1 , 0 , 2) ) = ( - 2 , 2 , 2 2 ) · ( ( x, y, z ) - (1 , 0 , 2) ) which also simplifies to x - y - 2 z = - 1.
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MATB42H Solutions # 6 page 5 (c) The partial derivatives are g x = - ( x - 2) radicalbig 4 - ( x - 2) 2 - ( y + 1) 2 and g y = - ( y + 1) radicalbig 4 - ( x - 2) 2 - ( y + 1) 2 so g x (1 , 0 , 2) = 1 2 and g y (1 , 0 , 2) = - 1 2 . The tangent plane is given by z = g (1 , 0 , 2)+ g x (1 , 0 , 2) ( x - 1)+ g y (1 , 0 , 2) ( y - 0) = 2 + 1 2 ( x - 1 ) - 1 2 y which we can rewrite as x - y - 2 z = - 1. 9. (a) We are given Φ ( u, v ) = ( u cos v, u sin v, u ) , 0 u 4, 0 v 2 π . Now φ u = parenleftbigg cos v 2 u , sin v 2 u , 1 parenrightbigg , φ v = ( - u sin v, u cos v, 0 ) and φ u × φ v = ( - u cos v, - u sin v, 1 2 ) , so bardbl φ u × φ v bardbl = radicalbigg u cos 2 + u sin 2 + 1 4 = radicalbigg u + 1 4 . The surface area is integraldisplay Φ dS = integraldisplay D bardbl φ u × φ v bardbl dA = integraldisplay 2 π 0 integraldisplay 4 0 radicalbigg u + 1 4 du dv = 0 2 0 2 0 2 4 x y z 1 2 integraldisplay 2 π 0 integraldisplay 4 0 4 u + 1 du dv = parenleftbigg 2 π 2 parenrightbiggparenleftbigg 2 3 parenrightbiggparenleftbigg 1 4 parenrightbiggparenleftbigg 4 u + 1 parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle 4 0 = π 6 parenleftbigg (17) 3 / 2 - 1 parenrightbigg = π 6 ( 17 17 - 1 ) . (b) We are given Φ ( u, v ) = ( sin u cos v, sin u sin v, cos u ) , 0 u π 3 , 0 v 2 π . This is a familar parametrization, for which we know that bardbl φ u × φ v bardbl = | sin u | = sin u . The sur- face area is integraldisplay Φ dS = integraldisplay D bardbl φ u × φ v bardbl dA = integraldisplay 2 π 0 integraldisplay π/ 3 0 sin u du dv = ( 2 π ) bracketleftbigg - cos u bracketrightbigg π/ 3 0 = ( 2 π ) parenleftbigg - 1 2 + 1 parenrightbigg = π . x y z LParen1 0,0,1 RParen1
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MATB42H Solutions # 6 page 6 (c) We are given Φ ( s, t ) = ( s cos t, s sin t, 2 t ) , 0 s 1, 0 t 2 π . Now φ s = ( cos t, sin t, 0 ) , φ t = ( - s sin t, s cos t, 2 ) and φ s × φ t = ( 2 sin t, - 2 cos t, s ) so bardbl φ s × φ t bardbl = radicalbig 4 sin 2 t + 4 cos 2 t + s 2 = 4 + s 2 . The surface area is integraldisplay Φ dS = integraldisplay D bardbl φ s × φ t bardbl dA = integraldisplay 2 π 0 integraldisplay 1 0 4 + s 2 ds dt = ( 2 π ) bracketleftBig s 2 4 + s 2 + 2 ln | s + 4 + s 2 |
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