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  • King Fahd University of Petroleum & Minerals

    Department of Mathematics and Statistics

    MATH 302

    FINAL EXAM (172)

    Sunday, May 6, 2018 Allowed Time: 3 Hours

    Name: ___________________________________________________________

    ID Number: _______________________ Serial Number: _______________

    Section Number: _____________ Instructorโ€™s Name: _________________

    Instructions:

    1. Write neatly and legibly. You may lose points for messy work.

    2. Show all your work. No points for answers without justification.

    3. Calculators and Mobiles are not allowed.

    4. Make sure that you have 10 different problems (11 pages).

    Problem No. Points Maximum Points

    1 14

    2 11

    3 11

    4 14

    5 16

    6 14

    7 11

    8 14

    9 20

    10 15 Total: 140

  • M A T H 3 0 2 - T e r m 1 7 2 F I N A L E X A M P a g e | 2

    Q1. Consider the 2ร—2 complex matrix

    ๐ด = ( ๐‘Ž ๐‘ โˆ’ ๐‘Ž 0 ๐‘

    ).

    (a) Find the eigenvalues of ๐ด. (b) For each eigenvalue of ๐ด, determine the eigenvectors. (c) Diagonalize the matrix ๐ด. (d) Using the result of the diagonalization, compute and simplify ๐ด๐‘˜ for each positive integer ๐‘˜.

    Solution:

    (a) Since ๐ด is an upper triangular matrix, eigenvalues are diagonal entries.

    Hence ๐‘Ž, ๐‘ are eigenvalues of ๐ด.

    (b) The corresponding eigenvectors are

    ๐พ1 = ( 1 0

    ) ๐‘Ž๐‘›๐‘‘ ๐พ2 = ( 1 1

    ).

    (c) Thus, a matrix P that diagonalizes A is given by

    ๐‘ƒ = ( 1 1 0 1

    )

    and the diagonal matrix D is

    ๐ท = ๐โˆ’1๐€๐ = ( ๐‘Ž 0 0 ๐‘

    ).

    (d) We have

    ๐ด๐‘˜ = ๐‘ƒ๐ท๐‘˜๐‘ƒโˆ’1 = ( 1 1 0 1

    ) ( ๐‘Ž 0 0 ๐‘

    ) ๐‘˜

    ( 1 โˆ’1 0 1

    )

    = ( 1 1 0 1

    ) (๐‘Ž ๐‘˜ 0

    0 ๐‘๐‘˜ ) (

    1 โˆ’1 0 1

    ) = (๐‘Ž ๐‘˜ ๐‘๐‘˜ โˆ’ ๐‘Ž๐‘˜

    0 ๐‘๐‘˜ ).

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  • M A T H 3 0 2 - T e r m 1 7 2 F I N A L E X A M P a g e | 3

    Q2. Let D be the region described by the box 0 < ๐‘ฅ < 1, 0 < ๐‘ฆ < 3, 0 < z < 2 .

    Find the outward flux โˆซ ๐… โˆ™ ๐’…๐’ ๐‘†

    of the vector field

    ๐… = (3๐‘ฅ + ๐‘ง77) ๐’‚๐’™ + (๐‘ฆ 2 โˆ’ ๐‘ ๐‘–๐‘› (๐‘ฅ2๐‘ง)) ๐’‚๐’š + (๐‘ฅ๐‘ง + ๐‘ฆ๐‘’

    ๐‘ฅ5) ๐’‚๐’› ,

    where S is the boundary of D.

    Solution:

    Given the ugly nature of the vector field, it would be hard to compute this integral directly.

    However, the divergence of ๐‘ญ is nice:

    div ๐น = 3 + 2๐‘ฆ + ๐‘ฅ.

    By the divergence theorem,

    โˆซ ๐‘ญ โˆ™ ๐’…๐‘บ ๐‘†

    = โˆญ div ๐น ๐‘‘๐‘‰ ๐ท

    = โˆซ โˆซ โˆซ (3 + 2๐‘ฆ + ๐‘ฅ)๐‘‘๐‘ฅ๐‘‘๐‘ฆ๐‘‘๐‘ง 1

    0

    3

    0

    2

    0

    = โˆซ โˆซ (3 + 2๐‘ฆ + 1

    2 ) ๐‘‘๐‘ฆ๐‘‘๐‘ง

    3

    0

    2

    0

    = โˆซ (9 + 9 + 3

    2 ) ๐‘‘๐‘ง

    2

    0

    = 36 + 3 = 39

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  • M A T H 3 0 2 - T e r m 1 7 2 F I N A L E X A M P a g e | 4

    Q3. Let D be the region shown in the below figure. Use Stokesโ€™s theorem to find the circulation

    โˆซ ๐… โˆ™ ๐’…๐ฅ ๐ฟ

    of the vector field ๐… = 2๐‘ฆ2๐’‚๐’™ + ๐‘ฅ 2๐‘ฆ ๐’‚๐’š around the positively oriented curve L

    where L is the boundary of the region D.

    Solution:

    ๐’…๐‘บ = ๐’…๐’™๐’…๐’š ๐’‚๐’›

    ๐’„๐’–๐’“๐’ ๐‘ญ = (๐Ÿ๐’™๐’š โˆ’ ๐Ÿ’๐’š) ๐’‚๐’›

    By the Stokesโ€™s theorem,

    โˆซ ๐‘ญ โˆ™ ๐’…๐’ = โˆฌ ๐’„๐’–๐’“๐’ ๐‘ญ โˆ™ ๐’…๐‘บ ๐‘น๐‘ณ

    = โˆซ โˆซ (๐Ÿ๐’™๐’š โˆ’ ๐Ÿ’๐’š)๐’…๐’š๐’…๐’™ ๐Ÿโˆ’๐’™

    ๐ŸŽ

    ๐Ÿ

    ๐ŸŽ

    = ๐Ÿ

    ๐Ÿ โˆซ (๐Ÿ๐’™ โˆ’ ๐Ÿ’)[๐’š๐Ÿ]๐’š=๐ŸŽ

    ๐Ÿโˆ’๐’™๐’…๐’™ ๐Ÿ

    ๐ŸŽ

    = โˆซ (๐’™ โˆ’ ๐Ÿ)๐Ÿ‘๐’…๐’™ ๐Ÿ

    ๐ŸŽ

    = ๐Ÿ

    ๐Ÿ’ [(๐’™ โˆ’ ๐Ÿ)๐Ÿ’]๐’™=๐ŸŽ

    ๐Ÿ

    = โˆ’๐Ÿ’.

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  • M A T H 3 0 2 - T e r m 1 7 2 F I N A L E X A M P a g e | 5

    Q4. Solve the equation 2 cosh ๐‘ง + 10 sinh ๐‘ง = 5.

    Solution:

    Using

    cosh ๐‘ง = ๐‘’๐‘ง + ๐‘’โˆ’๐‘ง

    2 , sinh ๐‘ง =

    ๐‘’๐‘ง โˆ’ ๐‘’โˆ’๐‘ง

    2 ,

    we obtain

    ๐‘’๐‘ง + ๐‘’โˆ’๐‘ง + 5(๐‘’๐‘ง โˆ’ ๐‘’โˆ’๐‘ง) = 5,

    6๐‘’๐‘ง โˆ’ 4๐‘’โˆ’๐‘ง โˆ’ 5 = 0,

    6๐‘’2๐‘ง โˆ’ 5๐‘’๐‘ง โˆ’ 4 = 0,

    (3๐‘’๐‘ง โˆ’ 4)(2๐‘’๐‘ง + 1) = 0,

    Which implies

    ๐‘’๐‘ง = 4

    3 ๐‘œ๐‘Ÿ ๐‘’๐‘ง = โˆ’

    1

    2 .

    Therefore,

    ๐‘ง = ๐‘™๐‘› ( 4

    3 ) = ๐‘™๐‘œ๐‘”๐‘’ (

    4

    3 ) + 2๐‘›๐œ‹๐‘–,

    or

    ๐‘ง = ๐‘™๐‘› (โˆ’ 1

    2 ) = ๐‘™๐‘œ๐‘”๐‘’ (

    1

    2 ) + (๐œ‹ + 2๐‘›๐œ‹)๐‘–,

    where ๐‘› = 0, ยฑ1, ยฑ2, ยฑ3, โ€ฆ .

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  • M A T H 3 0 2 - T e r m 1 7 2 F I N A L E X A M P a g e | 6

    Q5. Evaluate the following integrals:

    (a) โˆซ ๐‘…๐‘’(๐‘ง + 1) ๐‘‘๐‘ง, ๐ถ

    where ๐ถ is the circular arc (in the second quadrant) along |๐‘ง| = 2

    from ๐‘ง = 2๐‘– to ๐‘ง = โˆ’2.

    (b) โˆฎ 1

    ๐‘ง(๐‘ง+1โˆ’3๐‘–) ๐‘‘๐‘ง

    ๐ถ , where ๐ถ is the positively oriented circle |๐‘ง + 2| = 1.

    Solution:

    (a) On curve C, we have ๐‘ง = 2cos ๐‘ก + 2๐‘– sin ๐‘ก = 2๐‘’๐‘–๐‘ก , ๐œ‹/2 โ‰ค ๐‘ก โ‰ค ๐œ‹, and ๐‘‘๐‘ง = 2๐‘–๐‘’๐‘–๐‘ก๐‘‘๐‘ก.

    Thus,

    โˆซ ๐‘…๐‘’(๐‘ง + 1) ๐‘‘๐‘ง ๐ถ

    = โˆซ (2cos ๐‘ก + 1) 2๐‘–๐‘’๐‘–๐‘ก๐‘‘๐‘ก ๐œ‹

    ๐œ‹ 2

    = 2๐‘– โˆซ (2 ( ๐‘’๐‘–๐‘ก + ๐‘’โˆ’๐‘–๐‘ก

    2 ) + 1) ๐‘’๐‘–๐‘ก๐‘‘๐‘ก

    ๐œ‹

    ๐œ‹ 2

    = 2๐‘– โˆซ (๐‘’2๐‘–๐‘ก + 1 + ๐‘’๐‘–๐‘ก) ๐‘‘๐‘ก ๐œ‹

    ๐œ‹ 2

    = [๐‘’2๐‘–๐‘ก + 2๐‘– ๐‘ก + 2๐‘’๐‘–๐‘ก]๐œ‹ 2

    ๐œ‹

    = (1 + 2๐‘– ๐œ‹ โˆ’ 2) โˆ’ (โˆ’1 + ๐‘– ๐œ‹ + 2๐‘–)

    = ๐‘– (๐œ‹ โˆ’ 2).

    (b) Both singularities ๐‘ง = 0 and ๐‘ง = โˆ’1 + 3๐‘– are located outside the contour ๐ถ.

    Therefore, โˆฎ 1

    ๐‘ง(๐‘ง+1โˆ’3๐‘–) ๐‘‘๐‘ง

    ๐ถ = 0, by Cauchy-Goursat Theorem.

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  • M A T H 3 0 2 - T e r m 1 7 2 F I N A L E X A M P a g e | 7

    Q6. Let ๐‘“(๐‘ง) = ๐‘–|๐‘ง|2. Show that ๐‘“(๐‘ง) is differentiable only at ๐‘ง0 = 0 and find ๐‘“โ€ฒ(0).

    Solution:

    Let ๐‘ง = ๐‘ฅ + ๐‘– ๐‘ฆ. We have

    ๐‘“(๐‘ง) = ๐‘–|๐‘ง|2 = ๐‘–(๐‘ฅ2 + ๐‘ฆ2).

    Put ๐‘ข = 0 and ๐‘ฃ = ๐‘ฅ2 + ๐‘ฆ2. We obtain

    ๐œ•๐‘ข

    ๐œ•๐‘ฅ = 0,

    ๐œ•๐‘ฃ

    ๐œ•๐‘ฅ = 2๐‘ฅ,

    ๐œ•๐‘ข

    ๐œ•๐‘ฆ = 0,

    ๐œ•๐‘ฃ

    ๐œ•๐‘ฆ = 2๐‘ฆ,

    Since the Cauchy-Riemann equations

    ๐œ•๐‘ข

    ๐œ•๐‘ฅ =

    ๐œ•๐‘ฃ

    ๐œ•๐‘ฆ ,

    ๐œ•๐‘ข

    ๐œ•๐‘ฆ = โˆ’

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