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Unformatted text preview: Your Name Final Exam The exam starts at 3:10pm and lasts until 5:30pm, at which point you must turn in your work. You cannot use calculators, any books or other supplementary materials. However, you are allowed to have a formula sheet the size of which is a half of a regular sheet of paper. You can write on both sides of it. If you need to ask a question, please raise your hand and I will address your question. If cheating is noticed draconian measures will be taken. Please write legibly. Sketches should be done in a clear and neat way. The main features of graphs should be specified. Your explanations should be brief and clear. Please don’t forget to write your name above each of the problems. Good luck! 1 Your Name Problem 1 The surface S is given in Cartesian coordinates by x 5 + y 5 + z 5 + 1 = ( x + y + z + 1) 5 . Find the parametric equations of the line of intersection of the tangent plane at the origin and the tangent plane at the point (1 , , 0). Solution Let F ( x, y, z ) = x 5 + y 5 + z 5 ( x + y + z +1) 5 +1 then the surface S is given by F ( x, y, z ) = 0. We have F x = 5 x 4 5( x + y + z + 1) 4 , F y = 5 y 4 5( x + y + z + 1) 4 , F z = 5 z 4 5( x + y + z + 1) 4 . The equation of the tangent plane at the origin to the surface S is 5 x 5 y 5 z = 0 , or, simplifying, we have x + y + z = 0 . The equation of the tangent plane at the point (1 , , 0) to the surface S is 75( x 1) 80 y 80 z = 0 , or, simplifying, we have 15 x + 16 y + 16 z 15 = 0 . We denote the normal vector to the first plane by ~n 1 = (1 , 1 , 1) and the normal vector to the second plane by ~n 2 = (15 , 16 , 16). The cross product of these two normal vectors we denote by ~v . It equals to ~v = (0 , 1 , 1). The point with the position vector ~ r = ( 15 , 15 , 0) belongs to both planes. Therefore, the vector equation of the line of intersection is ~ r ( t ) = ( 15 , 15 , 0) + t (0 , 1 , 1) and the parametric equations are x = 15 , y = 15 t, z = t, where t ∈ R . 2 Your Name Problem 2 Sketch the curve given in polar coordinates r 2 = sin(3 θ ) and find the area that it encloses. Solution The sketch is given in Fig. 1. –1 –0.8 –0.6 –0.4 –0.2 0.2 0.4 0.6 –0.8 –0.6 –0.4 –0.2 0.2 0.4 0.6 0.8 Figure 1: r 2 = sin(3 θ ) Using the symmetry of the curve we obtain the area that it encloses is A = 3 · 1 2 Z π/ 3 r 2 dθ = 3 2 Z π/ 3 sin(3 θ ) dθ = 1 . 3 Problem 3 Find critical points of the function f ( x, y ) = 2 x 3 + xy 2 + 5 x 2 + y 2 and define whether they are local maxima, minima or saddle points....
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 Fall '08
 Lanzara
 Physics, Work, Dρ, fyy, kzdxdydz

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