Solutions7

Solutions7 - 1 Mathematics 1c: Solutions, Homework Set 7...

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Unformatted text preview: 1 Mathematics 1c: Solutions, Homework Set 7 Due: Monday, May 24 at 10am. 1. (10 Points) Section 7.3, Exercise 6 . Find an expression for a unit vector normal to the surface x = 3cos sin , y = 2sin sin , z = cos for in [0 , 2 ] and in [0 , ] . Solution. Here, T = (- 3sin sin , 2cos sin , 0) and T = (3cos cos , 2sin cos ,- sin ) . Thus, T T = (- 2cos sin 2 ,- 3sin sin 2 ,- 6sin cos ) and k T T k = sin (5sin 2 sin 2 + 32cos 2 + 4) 1 / 2 . Hence a unit normal vector is n = T T k T T k = 1 sin p 5sin 2 sin 2 + 32cos 2 + 4 (- 2cos sin 2 ,- 3sin sin 2 ,- 6sin cos ) . Since x 2 9 + y 2 4 + z 2 = 1 , the surface is an ellipsoid. 2. (15 Points) Section 7.3, Exercise 15 (a) Find a parameterization for the hyperboloid x 2 + y 2- z 2 = 25 . (b) Find an expression for a unit normal to this surface. (c) Find an equation for the plane tangent to the surface at ( x ,y , 0) , where x 2 + y 2 = 25 . (d) Show that the pair of lines ( x ,y , 0) + t (- y ,x , 5) and ( x ,y , 0) + t ( y ,- x , 5) lie in the surface and as well as in the tangent plane found in part (c) . 2 Solution. This solution uses hyperbolic functions! (a) x = 5cosh u cos y = 5cosh u sin z = 5sinh u (b) Since f ( x,y,z ) = x 2 + y 2- z 2 = 25, the unit normal is n = f k f k , f = (2 x, 2 y,- 2 z ) . Thus, n = 2( x,y,- z ) 2 p x 2 + y 2 + z 2 = 1 p cosh(2 u ) (cosh u cos , cosh u sin ,- sinh u ) . (c) Since the normal vector of the tangent plane is parallel to the gradient f ( x ,y , 0) = (2 x , 2 y , 0), an equation of the plane is ( x ,y , 0) ( x- x ,y- y ,z- 0) = 0 , i.e., x ( x- x ) + y ( y- y ) = 0 ....
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Solutions7 - 1 Mathematics 1c: Solutions, Homework Set 7...

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