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# Multivariable Calculus 18.02, Lecture Notes for Weeks 7-9 18.02

Massachusetts Institute of Technology (MIT)

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This 4 page Class Notes was uploaded by Zach Hauk on Saturday April 23, 2016. The Class Notes belongs to 18.02 at Massachusetts Institute of Technology (MIT) taught by Prof. Denis Auroux in Spring 2016. Since its upload, it has received 21 views. For similar materials see Multivariable Calculus in Mathematics (M) at Massachusetts Institute of Technology (MIT).

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Date Created: 04/23/16

Multivariable Calculus: Notes 3 Zach Hauk April 24, 2016 Second Derivative Test : ▯ Given a function in 3-space f(x;y), we can evaluate local minima, maxima, and saddle points. Minimum: point with locally lowest value Maximum: point with locally highest value Saddle Point: critical point which is neither a minimum nor a maximum, and is not degenerate. ex. f(x;y) = x ▯ y , saddle point at (0;0) 20 0 5 ▯20 0 ▯4 ▯2 0 2 4 ▯5 Degenerate Point: critical point which is not a local minimum or maximum point or a saddle point. ▯ By evaluating the critical points of f and using the second derivative test, we can evaluate the minimas, maximas, and saddle points of f ▯ We can ▯nd the critical points of f by ▯nding solutions to the homogeneous equations 1 @f @f (x0;y0) = 0; (x 0y 0 = 0 (1) @x @y ▯ The solutions of x0and y 0escribe the coordinates of a "turning point", known as a critical point, in the equation. ▯ After evaluating the critical points, we can test whether each critical point is a minima, maxima, saddle point, or is degenerate using the second derivative test. 1. De▯ne the variables A;B;C as @ f @ f @ f @ f A = ; B = = ; C = (2) @x 2 @x@y @y@x @y2 2. Plug in each critical point into the expression AC ▯ B 2 (3) 3. Use the following inequalities to determine the type of critical point. AC ▯ B > 0: A > 0: Local minimum, A < 0: Local maximum AC ▯ B < 0: Saddle point AC ▯ B = 0: Higher order derivative test required Example: Find the local minima, maxima, and saddle points of the function f(x;y) = x + 2y ▯ 3x in 3-space Critical Points: @f = 0; @f = 0 (4) @x @y 3x ▯ 6x 0 0; 4y 0 0 (5) x0= 0;2; y 0 0 (6) (x0;y 0 = (0;0); (2;0) (7) Second Derivative Test: @ f @ f @ f @ f A = ; B = = ; C = (8) @x 2 @x@y @y@x @y2 A = 6x 0 6; B = 0; C = 4 (9) 2 f or AC ▯ B = (0;0) 2 2 AC ▯ B = 24x ▯ 20 (10) 2 AC ▯ B = ▯24 < 0 (11) AC ▯ B < 0: saddle point at (0;0) 2 f or AC ▯ B = (0;0) AC ▯ B = 24 > 0; A = 12 > 0 (12) 2 AC ▯ B > 0. A > 0: local minimum at (2;0) Solution: Saddle point at (0;0), local minimum at (2;0). Hessian Matrices : 2 ▯ The second derivative test’s determining expression AC▯B bears a resemblance to the equation of the determinant of a 2-by-2 matrix. ▯ ▯ ▯▯ ▯▯ 2 ▯ ▯▯= ▯▯ ▯ ▯ , AC ▯ B (13) ▯ By reversing the determinant operation, we can ▯nd a matrix whose determi- nant will tell us what type of critical point we have. This matrix is called the Hessian Matrix. ▯ ▯ ▯A B ▯ AC ▯ B =2 ▯ ▯ = jHj (14) B C ▯ since A = @xx, B = @ xy= @ yxC = @ yy (where @ abis the partial derivative with respect to variables a and b), we get the more generalized form ▯ ▯ @xx @xy H = (15) @ yx @yy ▯ We can expand this to any number of dimensions using the simple rule H i;j= @xi;j (16) where x ns the axis in dimension n. Extra The bordered Hessian Matrix of f and g is de▯ned as ▯ ▯ 0 Og T (17) Og H(f) where O is the gradient operation. 3 Exercises: ▯ Find the Hessian matrix and it’s determinant for each function. x 1. f(x;y) = e y 2. f(x;y) = lnx + lny 3. f(x;y) = x + y 2 4. f(x;y) = e + x y 5. f(x;y) = cosx ▯ siny 4

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