Final Exam (2018-19)

MATH2822 Mathematical Methods for Actuarial Science II Examination Questions Answer All Nine Questions 1. (a) (3 points) Show that the distance from a point P ( x 0 , y 0 , z 0 ) in space to the plane Ax + By + Cz = D is | Ax 0 + By 0 + Cz 0 - D | √ A 2 + B 2 + C 2 . (b) Given two lines L 1 : x = 2 t, y = 4 + 3 t, z = - 2 - t, and L 2 : x = - 1 - t, y = 3 + 2 t, z = 10 + 4 t. i. (3 points) Find the plane containing L 1 and parallel to L 2 . ii. (1 point) Find the distance between the lines L 1 and L 2 . Solution. (a) Take any point Q ( x, y, z ) on the plane Ax + By + Cz = D . The distance from P to the plane is the absolute value of the scalar component of -- * QP in the direction of A i + B j + C k . The value is (( x - x 0 ) i + ( y - y 0 ) j + ( z - z 0 ) k ) · A i + B j + C k √ A 2 + B 2 + C 2 = Ax + By + Cz - Ax 0 - By 0 - Cz 0 √ A 2 + B 2 + C 2 = D - Ax 0 - By 0 - Cz 0 √ A 2 + B 2 + C 2 . (b) i. The plane is parallel to both L 1 and L 2 and is therefore orthogonal to i j k 2 3 - 1 - 1 2 4 = 14 i - 7 j + 7 k = 7(2 i - j + k ) . As it contains the point (0 , 4 , - 2), an equation is given by 2 x - y + z = - 4 - 2 = - 6 . ii. The required distance is the distance from ( - 1 , 3 , 10) to the plane obtained in i., i.e., - 2 - 3 + 10 - ( - 6) √ 6 = 11 √ 6 . 1

2. Let f ( x, y ) = xy ln( x 2 + y 2 ) when ( x, y ) 6 = (0 , 0) and f (0 , 0) = 0. (a) (3 points) Show that f is continuous at (0 , 0). (b) (3 points) Determine whether f is differentiable at (0 , 0). Solution. (a) Note that 0 ≤ xy ln( x 2 + y 2 ) ≤ p x 2 + y 2 p x 2 + y 2 ln( x 2 + y 2 ) = ( x 2 + y 2 ) ln( x 2 + y 2 ) . (Or, using G.M. ≤ A.M., | xy | ≤ ( x 2 + y 2 ) / 2.) By L’Hospital’s rule, lim s → 0 + s ln s = 0. It follows that lim ( x,y ) → (0 , 0) ( x 2 + y 2 ) ln( x 2 + y 2 ) = 0. We then get from the sandwich principle that lim ( x,y ) → (0 , 0) xy ln( x 2 + y 2 ) = 0 = f (0 , 0). Therefore f is continuous at (0 , 0). (b) Along the x - and y -axes, f ≡ 0 and hence f x (0 , 0) = f y (0 , 0) = 0. We have f ( x, y ) - f (0 , 0) - xf x (0 , 0) - yf y (0 , 0) p x 2 + y 2 = xy ln( x 2 + y 2 ) p x 2 + y 2 ≤ p x 2 + y 2 ln( x 2 + y 2 ) . By L’Hospital’s rule again, lim s → 0 + √ s ln s = 0. So, lim ( x,y ) → (0 , 0) f ( x, y ) - f (0 , 0) - xf x (0 , 0) - yf y (0 , 0) p x 2 + y 2 = 0 , and f is differentiable at (0 , 0). 2

4. (4 points) Find the greatest and smallest values that the function f ( x, y, z ) = x 2 + 2 y 2 + 6 z takes on the ellipsoid x 2 + y 2 + 3 z 2 = 3 . Solution. We have to find the maximum and minimum of the function f ( x, y, z ) = x 2 + 2 y 2 + 6 z subject to the constraint g ( x, y, z ) = x 2 + y 2 + 3 z 2 - 3 = 0 . Using the Lagrange multiplier method, the gradient equation ∇ f = λ ∇ g is 2 x i + 4 y j + 6 k = λ (2 x i + 2 y j + 6 z k ) , or 2 x = 2 λx (1) 4 y = 2 λy (2) 6 = 6 λz (3) From (1), x = 0 or λ = 1, and from (2), y = 0 or λ = 2. If x = 0 and y = 0, then g = 0 yields z = ± 1. If x = 0 and λ = 2, then (3) yields z = 1 / 2 and then g = 0 yields y = ± 3 / 2. If λ = 1, then y = 0 and from (3), z = 1. From g = 0, we get x = 0. Evaluating f at the 4 points (0 , ± 3 / 2 , 1 / 2) and (0 , 0 , ± 1), we get f (0 , ± 3 / 2 , 1 / 2) = 15 / 2 , f (0 , 0 , 1) = 6 and f (0 , 0 , - 1) = - 6 . The greatest and smallest values of f on the ellipsoid are 15/2 and - 6 respectively. 4

5. (4 points) Find the volume of the region bounded below by the plane z = 0, laterally by the circular cylinder x 2 + y 2 = 4 x , and above by the paraboloid z = x 2 + y 2 . Solution. In cylindrical coordinates, the circular cylinder and the paraboloid are given by r = 4 cos θ , for θ ∈ [ - π/ 2 , π/ 2], and z = r 2 respectively. Hence the volume is Z π/ 2 - π/ 2 Z 4 cos θ 0 Z r 2 0 r dz dr dθ = Z π/ 2 - π/ 2 Z 4 cos θ 0 r 3 dr dθ = Z π/ 2 - π/ 2 r 4 4 r =4 cos θ r =0 dθ = 64 Z π/ 2 - π/ 2 cos 4 θ dθ = 16 Z π/ 2 - π/ 2 (2 cos 2 θ ) 2 dθ = 16 Z π/ 2 - π/ 2 (1 + cos 2 θ ) 2 dθ = Z π/ 2 - π/ 2 [16 + 32 cos 2 θ + 8(2 cos 2 2 θ )] dθ = Z π/ 2 - π/ 2 (24 + 32 cos 2 θ + 8 cos 4 θ ) dθ = 24 π. 5

7. (4 points) Evaluate Z 1 1 / 2 Z 4 x 1 /x e √ xy x dy dx + Z 2 1 Z 4 /x x e √ xy x dy dx. Solution. It is not hard to see that Z 1 1 / 2 Z 4 x 1 /x e √ xy x dy dx + Z 2 1 Z 4 /x x e √ xy x dy dx = ZZ R e √ xy x dA, where R is the region in the first quadrant bounded by the hyperbolas xy = 1, xy = 4 and the lines y = x , y = 4 x . Introduce the transformation u 2 = xy and v 2 = y/x . Then x = u/v and y = uv . The corresponding uv -region is bounded by the lines u = 1, u = 2, v = 1 and v = 2. The Jacobian of the transformation is ∂ ( x, y ) ∂ ( u, v ) = 1 v - u v 2 v u = 2 u v . Hence ZZ R e √ xy x dA = Z 2 1 Z 2 1 e u u/v · 2 u v dv du = 2( e 2 - e ) . 7

8. (a) (3 points) Let S = { X 1 , X 2 , . . . , X k } , where X 1 =       a 11 a 12 . . . a 1 n       , X 2 =       a 21 a 22 . . . a 2 n       , . . . , X k =       a k 1 a k 2 . . . a kn       are vectors in R n . Assume that k ≤ n and define S ⊥ = [ x 1 , . . . , x n ] t ∈ R n : a i 1 x 1 + · · · + a in x n = 0 for all i = 1 , . . . , k . (In the definition above, [ x 1 , . . . , x n ] t denotes the transpose of the row vector [ x 1 , . . . , x n ]. It is therefore a column vector.) Identifying S ⊥ as the null space of a certain matrix, or otherwise, show that S ⊥ is a subspace of R n and that dim S ⊥ ≥ n - k . (b) Let S = { X 1 , X 2 , X 3 } , where X 1 =      0 1 0 2      , X 2 =      1 - 1 1 - 3      , X 3 =      1 1 1 1      , are vectors in R 4 . i. (2 points) Using the left-to-right algorithm, or otherwise, find a basis for h X 1 , X 2 , X 3 i consisting (only) of vectors in S . ii. (3 points) Find a basis for S ⊥ . Solution. (a) Let A =       a 11 a 12 · · · a 1 n a 21 a 22 · · · a 2 n . . . . . . . . . a k 1 a k 2 · · · a kn       . Then S ⊥ = N ( A ), the null space of A , and is therefore a subspace of R n . Since the row space of A , R ( A ), is spanned by k vectors, rank A = dim R ( A ) ≤ k . By the rank-nullity theorem, dim S ⊥ = nullity A = n - rank A ≥ n - k. 8

9. Let A =    1 0 - 2 0 5 0 - 2 0 4    . (a) (2 points) Find all eigenvalues of A . (b) (5 points) Find a nonsingular matrix P such that P - 1 AP is a diagonal matrix. Find also P - 1 . (c) (2 points) Find explicitly a matrix B , which is not a scalar multiple of A , such that B 2 + 4 B = A. (There are more than one such B , you just need to exhibit one of them.) Solution . (a) The characteristic polynomial of A is 1 - λ 0 - 2 0 5 - λ 0 - 2 0 4 - λ = (1 - λ )(5 - λ )(4 - λ ) - ( - 2)(5 - λ )( - 2) = (5 - λ )( λ 2 - 5 λ + 4 - 4) = - λ (5 - λ ) 2 Hence the eigenvalues of A are 0 and 5 (twice). (b) For the eigenvalue 0, we have A - 0 I 3 =    1 0 - 2 0 5 0 - 2 0 4    elementary row operations → · · · →    1 0 - 2 0 1 0 0 0 0    , and hence [2 , 0 , 1] t is a corresponding eigenvector. For the eigenvalue 5, A - 5 I 3 =    - 4 0 - 2 0 0 0 - 2 0 - 1    elementary row operationss → · · · →    2 0 1 0 0 0 0 0 0    , we get two linearly independent eigenvectors: [0 , 1 , 0] t and [1 , 0 , - 2] t . 10

Let P =    2 0 1 0 1 0 1 0 - 2    . Then P - 1 = 1 5    2 0 1 0 5 0 1 0 - 2    and P - 1 AP =    0 0 0 0 5 0 0 0 5    . (c) The equation x 2 + 4 x = 5 has two solutions in 1 and - 5. Let D = diag [0 , 1 , - 5]. Then D 2 + 4 D = diag [0 , 5 , 5]. We have ( PDP - 1 ) 2 + 4 PDP - 1 = P ( D 2 + 4 D ) P - 1 = P diag [0 , 5 , 5] P - 1 = A. Therefore, let B = P diag [0 , 1 , - 5] P - 1 =    - 1 0 2 0 1 0 2 0 - 4    . Then B is not a scalar multiple of A , and it satisfies B 2 + 4 B = A . Verification (not needed in the answer): B 2 + 4 B =    5 0 - 10 0 1 0 - 10 0 20    +    - 4 0 8 0 4 0 8 0 - 16    =    1 0 - 2 0 5 0 - 2 0 4    = A. 11