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Solutions7

# Solutions7 - 1 Mathematics 1c Solutions Homework Set 7 Due...

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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 = 3 cos θ sin φ, y = 2 sin θ sin φ, z = cos φ for θ in [0 , 2 π ] and φ in [0 , π ] . Solution. Here, T θ = ( - 3 sin θ sin φ, 2 cos θ sin φ, 0) and T φ = (3 cos θ cos φ, 2 sin θ cos φ, - sin φ ) . Thus, T θ × T φ = ( - 2 cos θ sin 2 φ, - 3 sin θ sin 2 φ, - 6 sin φ cos φ ) and k T θ × T φ k = sin φ (5 sin 2 θ sin 2 φ + 32 cos 2 φ + 4) 1 / 2 . Hence a unit normal vector is n = T θ × T φ k T θ × T φ k = 1 sin φ p 5 sin 2 θ sin 2 φ + 32 cos 2 φ + 4 × ( - 2 cos θ sin 2 φ, - 3 sin θ sin 2 φ, - 6 sin φ cos φ ) . 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 0 , y 0 , 0) , where x 2 0 + y 2 0 = 25 . (d) Show that the pair of lines ( x 0 , y 0 , 0) + t ( - y 0 , x 0 , 5) and ( x 0 , y 0 , 0) + t ( y 0 , - x 0 , 5) lie in the surface and as well as in the tangent plane found in part (c) .

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2 Solution. This solution uses hyperbolic functions! (a) x = 5 cosh u cos θ y = 5 cosh u sin θ z = 5 sinh 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 θ, cosh u sin θ, - sinh u ) . (c) Since the normal vector of the tangent plane is parallel to the gradient f ( x 0 , y 0 , 0) = (2 x 0 , 2 y 0 , 0), an equation of the plane is ( x 0 , y 0 , 0) · ( x - x 0 , y - y 0 , z - 0) = 0 , i.e., x 0 ( x - x 0 ) + y 0 ( y - y 0 ) = 0 . (d) One simply substitutes into the equation of the surface and the tangent plane and verifies that they are satisfied.
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Solutions7 - 1 Mathematics 1c Solutions Homework Set 7 Due...

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