2005Fall20Final-sol

# 2005Fall20Final-sol - Name ID Solutions to Final...

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Unformatted text preview: Name: ID#: Solutions to Final Examination Math 20 Introduction to Multivariable Calculus and Linear Algebra January 23, 2006 Rules: • This is a three-hour exam. • Calculators are not allowed. • Unless otherwise stated, show all of your work. Full credit may not be given for an answer alone. • You may use the backs of the pages or the extra pages for scratch work. Do not unstaple or remove pages as they can be lost in the grading process. • Please do not put your name on any page besides the first page. If you like, you may put your ID number on the top of each page you write on. Hints: • Read the entire exam to scan for obvious typos or questions you might have. • Budget your time so that you don’t run out. • Problems may stretch across several pages. • Relax and do well! Students who, for whatever reason, submit work not their own will ordinarily be required to withdraw from the College. —Handbook for Students Summary Data Problem 1 2 3 4 5 6 7 8 9 Total Maximum Possible 15 15 16 8 10 9 10 9 8 100 Mean 14.90 14.57 11.05 4.29 8.29 7.05 9.38 7.52 7.29 84.33 Median 15 15 14 4 10 9 10 9 8 83 Mode 15 15 15 8 10 9 10 9 8 83 Standard Deviation 0.29 0.90 4.97 2.90 2.58 3.39 1.43 2.06 1.64 12.49 Correlation with total-0.0563 0.6751 0.8548 0.3789 0.4882 0.7300 0.6620 0.4463 0.5798 1.0000 Part I. Techniques 1 1 1. (15 Points) For n × n matrices A and B , define [ A,B ] = AB- BA. Notice AB = BA if and only if [ A,B ] = 0 . For X = 1 Y = 1 H = 1- 1 Find (i) [ H,X ] Solution. We have HX = 1- 1 1 = 1 XH = 1 1- 1 =- 1 = ⇒ [ H,X ] = 2 = 2 X. N (ii) [ H,Y ] Solution. Similar to the above we have [ H,Y ] =- 2 Y . N (iii) [ X,Y ] Solution. [ X,Y ] = H . N / 15 1 2 2 2. (15 Points) Let f x y = 4 xy- 2 x 4- y 2 (a) Find ∂f ∂x x y and ∂f ∂y x y . Solution. We have ∂f ∂x x y = 4 y- 8 x 3 = 4( y- 2 x 3 ) ∂f ∂y x y = 4 x- 2 y = 2(2 x- y ) . N (b) Find the critical points of f . Solution. We need both partial derivatives to be zero. From the second equation above, we see that we must have y = 2 x . Hence 2 x = 2 x 3 and the possibilities are x = 0 or x = ± 1. Hence the critical points are , 1 2 ,- 1- 2 , N (c) For each critical point, decide if it’s a local maximum or a local minimum. Solution. We have ∂ 2 f ∂x 2 =- 24 x 2 ∂ 2 f ∂x∂y = 4 ∂ 2 f ∂y ∂x = 4 ∂ 2 f ∂y 2 =- 2 The determinant of the hessian matrix is 48 x 2- 16, which is positive if x = ± 1. Also at these points the top-left entry is negative. Thus 1 2 and- 1- 2 are local maxima. The point is a saddle point....
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2005Fall20Final-sol - Name ID Solutions to Final...

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