SolutionsFin

SolutionsFin - 1 Mathematics 1c Solutions Final Examination...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Mathematics 1c: Solutions, Final Examination Due: Wednesday, June 9, at 10am 1. (a) [7 points] Let f : R 3 R 2 be defined by f ( x,y,z ) = ( e - 2 xy ,x 2 - z 2 - 4 x + sin( x + y + z ) ) and let g : R 2 R be a function such that g (1 , 0) = - 1, and g (1 , 0) = i - 3 j . Calculate the gradient of the composition g f at the point (0 , 0 , 0). Solution. Let f ( x,y,z ) = ( u ( x,y,z ) ,v ( x,y,z )) and h ( x,y,z ) = g ( f ( x,y,z )) . By the chain rule, ∂h ∂x = ∂g ∂u ∂u ∂x + ∂g ∂v ∂v ∂x , ∂h ∂y = ∂g ∂u ∂u ∂y + ∂g ∂v ∂v ∂y , and ∂h ∂z = ∂g ∂u ∂u ∂z + ∂g ∂v ∂v ∂z . Thus, at (0 , 0 , 0) , ∂h ∂x = (1)(0) + ( - 3)( - 4 + 1) = 9 ∂h ∂y = (1)(0) + ( - 3)(1) = - 3 ∂h ∂z = (1)(0) + ( - 3)(1) = - 3 . Therefore the gradient of g f at (0 , 0 , 0) is 9 i - 3 j - 3 k . (b) [5 points] Find the equation of the tangent plane to the level set g f = - 1 at the point (0 , 0 , 0), where g and f are defined in part (a). Solution. The gradient from (a) is orthogonal to the level set g f = - 1, so therefore the tangent plane is given by ( x - 0)(9) + ( y - 0)( - 3) + ( z - 0)( - 3) = 0 . This may be rewritten as 3 x - y - z = 0 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 (c) Consider the function f ( x,y ) = x 2 + 3 xy + y 2 + 16. i. [4 points] Show that f has a minimum at the origin along the x -axis and the y -axis. Solution. Note that f ( x, 0) = x 2 +16. Since this is smallest when x 2 = 0, which only happens at x = 0, the origin is a minimum along the x -axis. Note also that f (0 ,y ) = y 2 +16, which is smallest at y = 0. Thus the origin is a minimum along the y -axis. ii. [4 points] Show that the origin is not a minimum of f by computing the eigenvalues of the second derivative matrix of f evaluated at the origin. Solution. The matrix of second partial derivatives of f is Hf ( x,y ) = 2 f ∂x 2 2 f ∂x∂y 2 f ∂y∂x 2 f ∂y 2 ! = ± 2 3 3 2 ² . The characteristic polynomial of this matrix is det ± 2 - λ 3 3 2 - λ ² = λ 2 - 4 λ +4 - 9 = λ 2 - 4 λ - 5 = ( λ - 5)( λ +1) . Thus the eigenvalues of Hf (0 , 0) are 5 and - 1. Since one of these is positive and the other negative, the origin is a saddle point, not a minimum. 2. Answer each of the following three questions:
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 07/19/2010 for the course MA 1C taught by Professor Ramakrishnan during the Spring '08 term at Caltech.

Page1 / 8

SolutionsFin - 1 Mathematics 1c Solutions Final Examination...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online