Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

7 Pages

235_tt2_s10_soln

Course: MATH 235/237, Spring 2010
School: Waterloo
Rating:

Word Count: 1164

Document Preview

235 Math 1. Short Answer Problems Term Test 2 Solutions [2] a) Let B = {v1 , . . . , vk } be an orthonormal basis for a subspace S of an inner product space V . Dene projS and perpS . Solution: Let v V , then projS (v ) and perpS (v ) are the unique vectors such that v = projS (v ) + perpS (v ) where projS (v ) S and perpS (v ) S . In particular, projS (v ) =&lt; v, v1 &gt; v1 + + &lt; v, vk...

Register Now

Unformatted Document Excerpt

Coursehero >> Canada >> Waterloo >> MATH 235/237

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
235 Math 1. Short Answer Problems Term Test 2 Solutions [2] a) Let B = {v1 , . . . , vk } be an orthonormal basis for a subspace S of an inner product space V . Dene projS and perpS . Solution: Let v V , then projS (v ) and perpS (v ) are the unique vectors such that v = projS (v ) + perpS (v ) where projS (v ) S and perpS (v ) S . In particular, projS (v ) =< v, v1 > v1 + + < v, vk > vk and perpS = v projS (v ). [2] b) Consider the quadratic form Q(x) = 4x2 5x2 + 2x2 3x1 x2 + 4x1 x3 + 2x2 x3 . 1 2 3 Write down the symmetric matrix A such that Q(x) = xT Ax. 4 3/2 2 Solution: A = 3/2 5 1 2 1 2 10 00 , 01 1 1 01 00 [2] c) Find a basis for the orthogonal compliment of the subspace S = Span of M (2, 2) under the standard inner product. Solution: Observe that dim S = dim M (2, 2) dim S = 4 2 = 2 and that v1 = and v2 = 1 0 are orthogonal to the basis vectors of S . Hence, 11 S = Span{v1 , v2 }. [2] d) Let Q(x1 , x2 ) = ax2 + 2bx1 x2 + cx2 where ac b2 < 0. Prove that Q is indenite. 1 2 Solution: Let A = ab have eigenvalues 1 and 2 . Then we have 1 2 = det A = bc ac b2 < 0 and so one eigenvalue is positive and the other is negative. Hence Q is indenite. [2] e) Use the Triangularization theorem to prove that every symmetric matrix is orthogonally diagonalizable. Solution: Let A be a symmetric matrix. Then A has all real eigenvalues, so by the triangularization theorem, we have that there exists an orthogonal matrix P such that P T AP = T where T is upper triangular. But, then T T = (P T AP )T = P T AT P = P T AP = T. Hence, T is symmetric and T must be diagonal. Hence, P orthogonally diagonalizes A. 1 2 222 [6] 2. Let A = 2 2 2. Find an orthogonal matrix P that diagonalizes A and 222 the corresponding diagonal matrix. Solution: Clearly the eigenvalues are 0, 0 and 6. For = 0 we get 22 111 2 2 0 0 0 . 22 000 1 0 . Applying the Gram-Schmidt and w2 = 1 1 1 1 1 1 and v2 = 1. procedure, we get orthonormal eigenvectors v1 = 2 6 0 2 4 2 2 1 0 1 1 1 2 4 2 0 1 1. So, v3 = 1. For = 6 we get A 6I = 3 2 2 4 00 0 1 1/ 2 1/6 1/3 000 Thus, we take P = 1/ 2 1/ 6 1/3 and get D = 0 0 0. 006 0 2 / 6 1/ 3 2 A 0I = 2 2 1 1 Hence, we get eigenvectors w1 = 0 3 [4] 3. Find a and b to obtain the best tting equation of the form y = a + bx for the following data: x 1 0 1 2 y 4 3 1 2 4 3 1 111 Solution: Let AT = and y = . Then we get 1 1 0 1 2 2 a = (AT A)1 AT y 42 26 1 = 4 1 1 1 1 3 1 0 1 2 1 2 6 5/2 = 7 2 = Thus, the line of best t is y = 1 6 2 20 2 4 5 2 2x. [4] 4. Find an orthonormal basis for P2 using the inner product p(x), q (x) = p(0)q (0) + p(1)q (1) + p(2)q (2), by applying the Gram-Schmidt procedure to the basis {1, x, x2 }. Solution: Let w1 = 1, w2 = x and w3 = x2 . Let v1 = w1 . v 2 = w2 w2 , v1 3 v1 = x (1) = x 1 2 v1 3 w3 , v1 w3 , v2 v 3 = w3 v1 v2 2 v1 v2 2 5 4 1 = x2 (1) (x 1) = x2 2x + 3 2 3 = 2/3 we get an orthonormal basis is x2 2x+ 1 1 3 , x1 , 3 2 2/3 Then, since x2 2x + 1 3 . 4 1 3 1 [7] 5. Let A = 0 1 2 . Find an orthogonal matrix P and upper triangular matrix T 021 T such that P AP = T . 1 0. Thus, Solution: Observe that 1 is an eigenvalue of A with unit eigenvector e1 = 0 3 we can extend {e1 } to the standard basis for R and so we take P = I . Thus P T AP = 12 1 bT where A1 = . 21 0 A1 We have C () = det(A1 I ) = 2 2 3 = ( 3)( + 1). For = 3 we get A1 3I = 2 2 1 1 so v1 = 2 2 00 1 2 1 . 1 1 2 We to need extend {v1 } to an orthonormal basis for R2 , so we take v2 = 1 . 1 1 0 0 2 Thus, we take P1 = 0 1/2 1/ and we get that 0 1/ 2 1/ 2 0 0 1 0 0 1 3 1 1 1 1 2 T 2 2 P1 AP1 = 0 1/2 1/ 0 1 2 0 1/2 1/ = 0 3 0 . 021 0 0 1 0 1/ 2 1/ 2 0 1/ 2 1/ 2 5 6. Let A = 53 and Q(x) = xT Ax. 3 3 [4] a) Find an orthogonal matrix P and diagonal matrix D such that P T AP = D. Solution: We have C () = 5 3 = 2 2 24 = ( 6)( + 4). 3 3 1 3 1 3 , so v1 = 3 9 00 3 . 1 Thus, the eigenvalues of A are 6 and 4. For = 6 we get A 6I = For = 4 we get A + 4I = 1 10 93 31 1 , so v2 = 1 . 10 3 31 00 3/10 1/ 10 60 Hence, we can take P = to get D = . 0 4 1/ 10 3/ 10 [1] b) Classify Q(x) as positive denite, negative denite or indenite. Solution: Since A has a positive eigenvalues and a negative eigenvalue we have that Q(x) is indenite. [2] c) Write Q so that it has no cross terms and give the change of variables which brings it into this form. Solution: Using the change of variables 3/10 1/ 10 x = Py = 1/ 10 3/ 10 2 2 we get Q(x) = 6y1 4y2 . y1 (3/10)y1 + (1/10)y2 , = y2 (1/ 10)y1 (3/ 10)y2 6 [3] 7. Let A be a positive denite symmetric matrix. Show that A = BB T for some matrix B with orthogonal columns. Solution: Let 1 , . . . , n be the eigenvalues of A which are all positive since A is positive denite. Let D be the diagonal matrix whose diagonal entries are the square roots of the eigenvalues of A. Since A is symmetric, there exists an orthogonal matrix P such that P T AP = D2 = DDT A = P DDT P T = P D(P D)T . Thus, we let B = P D. Hence, the columns of B are just scalar multiples of columns of P and hence the columns of B are orthogonal since the columns of P are orthogonal. [3] 8. Let A be an m n matrix with n > m and let x Rn . Prove that if AAT is invertible, then the point u Null(A) which is nearest to x is u = x AT (AAT )1 Ax. Solution: By the approximation theorem, the vector u Null(A) which is closest to x is projNull(A) (x). Let v = perpNull(A) (x) so that x = u + v . Then, v Null(A) = Col(AT ). Hence, v = AT t for some vector t Rm . In particular, t is the vector in Col(AT ) which is nearest to x. Hence, by the method of least squares, we have that (AT )T (AT )t = (AT )T x AAT t = Ax, so, t = (AAT )1 Ax. Thus, we have u = x v = x AT (AAT )1 Ax as required. 7 [3] 9. Let A be an n n positive denite symmetric matrix and C be an invertible n n matrix. Prove that B = C T AC is also a positive denite symmetric matrix. Solution: Since A is symmetric, B T = (C T AC )T = C T AT C T T = C T AC = B Therefore, B is symmetric. Since A, is positive denite, xT Bx = xT C T ACx = (Cx)T A(Cx) 0 for all x Rn , and xT Bx = (Cx)T A(Cx) = 0 implies that Cx = 0, and hence x = 0 since C is invertible. Thus, B is positive denite. [3] 10. Let A be a symmetric matrix and let L : Rn Rn be the linear map given by L(x) = Ax. Show that there exists an orthogonal basis B = {v1 , . . . , vn } for Rn such that L is a linear combination of the projections P1 , . . . , Pn where Pi : Rn Rn is the projection Pi (x) = projvi (x) for 1 i n. Solution: Since A is symmetric there exists an orthogonal matrix P = v1 vn and diagonal matrix D such that A = P DP T where the vi are eigenvectors of A corresponding to eigenvalues i for 1 i n. Hence, L(x) = Ax = P DP T x = (P D)(P T x) = 1 v 1 n v n v1 x . . . vn x = 1 (v1 x)v1 + + n (vn x)vn = 1 projv1 (x) + + n projvn (x)
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

Waterloo - MATH - 235/237
Math 235Assignment 0Due: Not To Be Submitted1. Determine projv x and perpv x where a) v = (2, 3, 2) and x = (4, 1, 3). b) v = (1, 2, 1, 3) and x = (2, 1, 2, 1). 2. Prove algebraically that projv (x) and perpv x are orthogonal. 3. Solve the system z1 (1
Waterloo - MATH - 235/237
Math 235Assignment 1Due: Wednesday, May 12th1. Let A be an m n matrix and B be an n p matrix. a) Prove that rank(AB ) rank(A). b) Prove that rank(AB ) rank(B ). c) Prove that if B is invertible, then rank(AB ) = rank(A). 2. Let T : V W be a linear mapp
Waterloo - MATH - 235/237
Math 235Assignment 1 Solutions1. Let A be an m n matrix and B be an n p matrix. a) Prove that rank(AB ) rank(A). Solution: Since the rank of a matrix is equal to the dimension of its column space, we consider the column space of A and AB . Observe that
Waterloo - MATH - 235/237
Waterloo - MATH - 235/237
Math 235Assignment 2 Solutions1. For each of the following linear transformations, determine a geometrically natural basis B and determine the matrix of the transformation with respect to B . a) perp(2,1,2) Solution: Pick v1 = (2, 1, 2). We want to pick
Waterloo - MATH - 235/237
Math 235Assignment 3Due: Wednesday, May 26th1. For each of the following pairs of vector spaces, dene an explicit isomorphism to establish that the spaces are isomorphic. Prove that your map is an isomorphism. a) P3 and R4 . b) The vector space P = cfw
Waterloo - MATH - 235/237
Math 235Assignment 3 Solutions1. For each of the following pairs of vector spaces, dene an explicit isomorphism to establish that the spaces are isomorphic. Prove that your map is an isomorphism. a) P3 and R4 . Solution: We dene L : P3 R4 by L(a3 x3 + a
Waterloo - MATH - 235/237
Math 235Assignment 4Due: Wednesday, Jun 2nd1. Prove that the product of two orthogonal matrices is an orthogonal matrix. 2. Prove that if R is an orthogonal matrix, then det R = 1. Give an example of a matrix A that has det A = 1, but is not orthogonal
Waterloo - MATH - 235/237
Math 235Assignment 4 Solutions1. Prove that the product of two orthogonal matrices is an orthogonal matrix. Solution: Let P and Q be orthogonal matrices. Then we have (P Q)T (P Q) = QT P T P Q = QT Q = I, since P T P = I and QT Q = I . Thus P Q is also
Waterloo - MATH - 235/237
Waterloo - MATH - 235/237
Math 235Assignment 5 Solutionsa) Use the Gram-Schmidt process to produce an orthonormal basis for S . 2 1 1 0 1 0 Solution: Denote the given basis by z1 = , z2 = , z3 = . Let w1 = z1 . 1 1 1 1 1 1, 1 2 1 1 1 0 1 3 z2 w1 Then, we get w2 = z2 projw1 (z2 )
Waterloo - MATH - 235/237
Waterloo - MATH - 235/237
Math 235Assignment 6 Solutions1. Show that the following are equivalent for a symmetric matrix A: (1) A is orthogonal (2) A2 = I (3) All the eigenvalues of A are 1 Solution: (1) (2) (2) (3) If A is orthogonal then I = AAT = AA, since A is symmetric. Av
Waterloo - MATH - 235/237
Math 235Assignment 7Due: Wednesday, June 30th1. For each quadratic form Q(x), determine the corresponding symmetric matrix A. By diagonalizing A, Write Q so that it has no cross terms and give the change of variables which brings it into this form. Cla
Waterloo - MATH - 235/237
Math 235Assignment 7 Solutions1. For each quadratic form Q(x), determine the corresponding symmetric matrix A. By diagonalizing A, Write Q so that it has no cross terms and give the change of variables which brings it into this form. Classify each quadr
Waterloo - MATH - 235/237
Waterloo - MATH - 235/237
Math 235Assignment 8Due: Wednesday, July 14th1. Sketch the graph of 9x2 + 4xy + 6y 2 = 21 showing both the original and new axes. Solution: The corresponding symmetric matrix is polynomial is C ( ) = 9 2 = 2 15 + 50 = ( 10)( 5). 2 6 A 1 I = 1 2 1 2 . 2
Waterloo - MATH - 235/237
Math 235Assignment 9Due: Wednesday, July 21st1. Suppose that a real 2 2 matrix A has 2 + i as an eigenvalue with a corresponding 1+i eigenvector . Determine A. i 0 2 1 2. Determine a real canonical form of A = 2 2 1 and give a change of basis matrix 0
Waterloo - MATH - 235/237
Math 235Assignment 9 Solutions1. Suppose that a real 2 2 matrix A has 2 + i as an eigenvalue with a corresponding 1+i eigenvector . Determine A. i Solution: Since A is real, we know that A has real canonical form B = brought into this form by P = 11 . W
Waterloo - MATH - 235/237
Math 235Assignment 10 Not To Be Submitted 1+i 1i 1. Consider C3 with its standard inner product. Let z = 2 i , w = 2 3i. 1 + i 1 a) Evaluate z , w and w, 2iz . b) Find a vector in spancfw_z, w that is orthogonal to z . c) Write the formula for the proj
Waterloo - MATH - 235/237
Math 235Assignment 10 Solutions 1+i 1i 1. Consider C3 with its standard inner product. Let z = 2 i , w = 2 3i. 1 + i 1 a) Evaluate z , w and w, 2iz . Solution: We have z , w = (1 + i)(1 + i) + (2 i)(2 + 3i) + (1 + i)(1) = 2i 1 + 8i + 1 i = 9i w, 2iz =
Waterloo - MATH - 235/237
Math 235 - Final Exam Fall 2009NOTE: The questions on this exam does not exactly reect which questions will be on this terms exam. That is, some questions asked on this exam may not be asked on our exam and there may be some questions on our exam not ask
Waterloo - MATH - 235/237
Math 235Final F09 AnswersNOTE: These are only answers to the problems and not full solutions! On the nal exam you will be expected to show all steps used to obtain your answer. 1. Short Answer Problems 3 i i a) A = . 2 1 b) A is Hermitian since A = A, a
Waterloo - MATH - 235/237
Math 235 - Final Exam Spring 2009NOTE: The questions on this exam does not exactly reect which questions will be on this terms exam. That is, some questions asked on this exam may not be asked on our exam and there may be some questions on our exam not a
Waterloo - MATH - 235/237
Math 235Final S09 AnswersNOTE: These are only answers to the problems and not full solutions! On the nal exam you will be expected to show all steps used to obtain your answer. 1. a) A basis for the nullspace is cfw_x, hence the nullity of L is 1. Thus,
Waterloo - MATH - 235/237
Math 235Final Exam InformationThursday August 5, 9:00 AM - 11:30 AMLOCATION: PAC 1, 2, 3Material Covered: Entire Course, with an emphasis on material after term test 2. Information: - Surfaces in R3 are not covered. - Fourier Series are not covered. -
Waterloo - MATH - 235/237
Math 235 1. Short Answer ProblemsSample Term Test 1 - 1a) Give the denition of an inner product , on a vector space V . b) Let B = cfw_v1 , . . . , vn be orthonormal in an inner product space V and let v V such that v = a1 v1 + + an vn . Prove that ai
Waterloo - MATH - 235/237
Math 235Sample Term Test 1 - 1 AnswersNOTE: - Only answers are provided here (and some proofs). On the test you must provide full and complete solutions to receive full marks. 1. Short Answer Problems a) Give the denition of an inner product , on a vect
Waterloo - MATH - 235/237
Math 235 1. Short Answer ProblemsSample Term Test 1 - 2 1 0 0 1 a) Write a basis for the rowspace, columnspace and nullspace of A = 0 0 1 1 . 000 0 b) Let B = cfw_v1 , . . . , vn be orthonormal in an inner product space V and let v = a1 v1 + + an vn .
Waterloo - MATH - 235/237
Math 235Sample Term Test 1 - 2 AnswersNOTE: - Only answers are provided here (and some proofs). On the test you must provide full and complete solutions to receive full marks. 1. Short Answer Problems 1 0 0 1 a) Let A = 0 0 1 1 . Write a basis for the R
Waterloo - MATH - 235/237
Math 235 1. Short Answer ProblemsSample Term Test 2 - 1a) Let S be a subspace of an inner product space V . What is the denition of S . b) State the Principal Axis Theorem. c) Determine the matrix for the quadratic form Q(x, y, z ) = 3x2 y 2 + z 2 2xy +
Waterloo - MATH - 235/237
Math 235Sample Term Test 2 - 1 AnswersNOTE: - Only answers are provided here (and some proofs). On the test you must provide full and complete solutions to receive full marks. 1. Short Answer Problems a) Let S be a subspace of an inner product space V .
Waterloo - MATH - 235/237
Math 235 1. Short Answer ProblemsSample Term Test 2 - 2a) State the Principal Axis Theorem. b) Let A be an m n matrix. Prove that AT A is symmetric. c) State the denition of a quadratic form Q(x) on Rn being negative denite. d) Consider the quadratic fo
Waterloo - MATH - 235/237
Math 235Sample Term Test 2 - 2 AnswersNOTE: - Only answers are provided here (and some proofs). On the test you must provide full and complete solutions to receive full marks. 1. Short Answer Problems a) State the Principal Axis Theorem. Solution: A mat
Waterloo - MATH - 235/237
Math 235Midterm InformationTuesday, June 8th, 4:30 - 6:20 p.mMaterial Covered: Sections 4-5, 4-6, 4-7, 7-4 (not including Fourier series), 7-1. You need to know: - All denitions and statements of theorems. - How to nd a basis of the rowspace, column sp
Waterloo - MATH - 235/237
Math 235Term Test 2 InformationTuesday, July 6th, 4:30 - 6:20 p.mRoom Assignments: MC 4059: A - G MC 4061: H - Lin MC 4045: Liu - P MC 4020: Q - Wang MC 4021 Wardell - Z Material Covered: Sections 7-2, 7-3, Triangularization, 8-1, 8-2. You need to know
Waterloo - MATH - 235/237
Math 235Tutorial: Term Test 1 Review1: State the denition of: a) One-to-one b) Onto c) An orthogonal matrix (what are 2 other equivalent denitions?) d) An inner product 1 0 1/2 1/2 1/2 1/2 , , 2 1 1/2 1/2 1/2 1/2 0 2 T under the inner product A, B = tr
Waterloo - MATH - 235/237
SOSMATH235MIDTERM2REVIEWPACKAGE HelloMATH235students,mynameisTaiCaiandIamtheSOStutorthistermforMATH235.This packageisdesignedtosupplementyourstudyingforthesecondmidtermonNovember16,2010. Wheneverpossible,Ihaveincludedexamplesthatarenotfromclassoryourtextb
Waterloo - MATH - 235/237
Math 235 1. Short Answer ProblemsTerm Test 1 Solutions[1] a) State the denition of the rank of a linear mapping L : V W . Solution: rank(L) = dim Range(L).[2] b) Let B = cfw_v1 , . . . , vn be a basis for a vector space V and let L : V W be an isomorp
Waterloo - MATH - 235/237
Math 235 31 5 2 1. Let A = 2 1 32 4 7 1 5 2 3 3 2Assignment 1Due: Wednesday, Sept 22nd 0 1 0 0 1 1 0 0 01 0 2 . 1 1 00 3 1 0 4 , then the RREF of A is R = 0 7 1 0a) Find rank(A) and dim(Null(A). b) Find a basis for Row(A). c) Find a basis for Null(A).
Waterloo - MATH - 235/237
Math 235 31 5 2 1. Let A = 2 1 32 4 7 1 5 2 3 3 2Assignment 1 Solutions 3 1 0 4 , then the RREF of A is R = 0 7 1 0 0 1 0 0 1 1 0 0 01 0 2 . 1 1 00a) Find rank(A) and dim(Null(A). Solution: rank(A) = 3 and dim(Null(A) = 5 3 = 2 b) Find a basis for Row(A
Waterloo - MATH - 235/237
Waterloo - MATH - 235/237
Math 235Assignment 2 Solutions1. For each of the following linear transformations, determine a geometrically natural basis B and determine the matrix of the transformation with respect to B . a) The projection proj(3,2) : R2 R2 onto the line x = t Solut
Waterloo - MATH - 235/237
Math 235Assignment 3Due: Wednesday, Oct 6th1. For each of the following pairs of vector spaces, dene an explicit isomorphism to establish that the spaces are isomorphic. Prove that your map is an isomorphism. a) M (2, 2) and P3 . b) The vector space P
Waterloo - MATH - 235/237
Math 235Assignment 3 Solutions1. For each of the following pairs of vector spaces, dene an explicit isomorphism to establish that the spaces are isomorphic. Prove that your map is an isomorphism. a) M (2, 2) and P3 . Solution: We dene L : M (2, 2) P3 by
Waterloo - MATH - 235/237
Math 235Assignment 4Due: Wednesday, Oct 13th1. Prove that the product of two orthogonal matrices is an orthogonal matrix. 2. Observe that the dot product of two vectors x, y Rn can be written as x y = xT y. Use this fact to prove that if an n n matrix
Waterloo - MATH - 235/237
Math 235Assignment 4 Solutions1. Prove that the product of two orthogonal matrices is an orthogonal matrix. Solution: Let P and Q be orthogonal matrices. Then we have (P Q)T (P Q) = QT P T P Q = QT Q = I, since P T P = I and QT Q = I . Thus P Q is also
Waterloo - MATH - 235/237
Waterloo - MATH - 235/237
Math 235Assignment 5 Solutions1. On M (2, 2) dene the inner product &lt; A, B &gt;= tr(B T A) and let S = Span 10 01 1 1 , , 01 10 01 .a) a) Use the Gram-Schmidt procedure to produce an orthonormal basis for S . 10 and 01 are already orthogonal. Then 2 1 1 v
Waterloo - MATH - 235/237
Waterloo - MATH - 235/237
Math 235 1 1. Let A = 2 4 diagonalizes A 2 2 2 andAssignment 6 Solutions 4 2. Find an orthogonal matrix P that 1 the corresponding diagonal matrix.Solution: The characteristic polynomails is 1 2 4 1 2 4 2 2 = det 2 2 2 C () = det(A I ) = det 2 4 2 1 3 0
Waterloo - MATH - 235/237
Math 235Assignment 7Due: Wednesday, Nov 10th1. For each quadratic form Q(x), determine the corresponding symmetric matrix A. By diagonalizing A, Write Q so that it has no cross terms and give the change of variables which brings it into this form. Clas
Waterloo - MATH - 235/237
Math 235Assignment 7 Solutions1. For each quadratic form Q(x), determine the corresponding symmetric matrix A. By diagonalizing A, Write Q so that it has no cross terms and give the change of variables which brings it into this form. Classify each quadr
Waterloo - MATH - 235/237
MATH237: EXAM-AID SOSNIALL W. MACGILLIVRAY (EDITOR); VINCENT CHAN (WRITER)1.October 31st, 2010 ScalarFunctions(1.1 1.2)DEFINITION 1.1. Suppose A and B are sets. A function f is a rule that determines how a subset of A is associated with a subset of B
Waterloo - MATH - 235/237
Here are answers to most of the problems. I have, of course, not included ones in which the question asks to prove or verify something. Also, for now, I have not had a chance to include pictures. Math 237 Problem Set 1 AnswersA1. a) Range: z R. b) Range:
Waterloo - MATH - 235/237
Here are answers to most of the problems. I have, of course, not included ones in which the question asks to prove or verify something. Also, for now, I have not had a chance to include pictures. Math 237 Problem Set 2 AnswersA3. a) L(a,b,c) (x, y, z ) =
Waterloo - MATH - 235/237
Here are answers to most of the problems. I have, of course, not included ones in which the question asks to prove or verify something. Also, for now, I have not had a chance to include pictures. Math 237 A1.dw (2) dtProblem Set 3 Answers = 76.A2. a) I
Waterloo - MATH - 235/237
Here are answers to most of the problems. I have, of course, not included ones in which the question asks to prove or verify something. Also, for now, I have not had a chance to include pictures. Math 237 A1. a) Hf (2, 3) = 9 6 . 6 4 Problem Set 4 Answers
Waterloo - MATH - 235/237
Math 237Problem Set 5 Answers12 , 33A1. i) (0, 1), (1, 1) and (0, 0) are all saddle points.is a local min.ii) (0, 0) is a saddle point, (1, 1/2) is a local max. iii) (0, 1/ 3) local min, (0, 1/ 3) local max, (1, 0) are saddle points. iv) (0, k ), k Z
Waterloo - MATH - 235/237
Math 237 A1. a) r = 8, =4 . 3 3 . 4Problem Set 6 Answers b) r = 2, = . 6 d) r = 5, = arctan(1/2). b) (x, y ) = (3 3/2, 3/2). d) (x, y ) = ( 3, 1). b) Area= . 4c) r = 2, = A2. a) (x, y ) = (1, 3). c) (x, y ) = (3/2, 3 3/2). A3. a) Area= . 4B1. a) Are
Waterloo - MATH - 235/237
Math 237 A1. i)Problem Set 7 Answers ii) iii)A2.A3. (4.980, 0.6224). A4. a) A5. 3e e e 3e c) 2e2 0 . 4 4 2(e 1) 2(e + 1)04 11uv 2 1 , 2 (u + v ) .A6. a) F 1 (u, v ) = ln b) DF = A7. i) ii)ex 1 1/(u v ) 1(u v ) , DF 1 = . ex 1 1/2 1/ 2 y = 2x. = k o