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MATH1813 Problem Set 4 Ans

MATH1813 Problem Set 4 Ans - University of Hong Kong Fall...

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University of Hong Kong Fall 2011 Math 1813 Mathematical Methods for Actuarial Science Instructor: Siye Wu Solution 4 I. In cylindrical polar coordinates: Z = { ( r cos θ, r sin θ, z ) T R 3 : 0 6 θ < 2 π, r = 2 , z R } S = { ( r cos θ, r sin θ, z ) T R 3 : 0 6 θ < 2 π, r > 0 , z R , r 2 + z 2 = 9 } C = { ( r cos θ, r sin θ, z ) T R 3 : 0 6 θ < 2 π, r > 0 , z R , r 2 = z 2 } In spherical polar coordinates: Z = { ( r sin θ cos φ, r sin θ sin φ, r cos θ ) T R 3 : 0 < θ < π, 0 6 φ < 2 π, r sin θ = 2 } S = { ( r sin θ cos φ, r sin θ sin φ, r cos θ ) T R 3 : 0 6 θ 6 π, 0 6 φ < 2 π, r = 3 } C = { ( r sin θ cos φ, r sin θ sin φ, r cos θ ) T R 3 : θ = π 4 or 3 π 4 , 0 6 φ < 2 π, r > 0 } II. We have J f ( x, y ) = - y sin( xy ) - x sin( xy ) cos( x + y ) cos( x + y ) , so J f (0 , π ) = 0 0 - 1 - 1 . (a) One has J F (2 , 3) = J f (0 , π ) J g (2 , 3) = 0 0 - 1 - 1 3 7 - 2 0 = 0 0 - 1 - 7 . (b) Note that F (2 , 3) = f (0 , π ) = (1 , 0) T . The linear approximation L ( s, t ) at point ( a, b ) T is L ( s, t ) = F ( a, b ) + J F ( a, b ) s - a t - b . So L (2 . 01 , 2 . 98) = F (2 , 3) + J F (2 , 3) 0 . 01 - 0 . 02 =(1 , 0) T + 0 0 - 1 - 7 0 . 01 - 0 . 02 = (1 , 0) T + (0 , 0 . 13) T = (1 , 0 . 13) T .

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