# King Fahd University of Petroleum & Minerals Department of ... · PDF file King Fahd University of Petroleum & Minerals Department of Mathematics and Statistics MATH 302 FINAL...

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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 ๐ โ ๐ 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, ๐ are eigenvalues of ๐ด.

(b) The corresponding eigenvectors are

๐พ1 = ( 1 0

) ๐๐๐ ๐พ2 = ( 1 1

).

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

๐ = ( 1 1 0 1

)

and the diagonal matrix D is

๐ท = ๐โ1๐๐ = ( ๐ 0 0 ๐

).

(d) We have

๐ด๐ = ๐๐ท๐๐โ1 = ( 1 1 0 1

) ( ๐ 0 0 ๐

) ๐

( 1 โ1 0 1

)

= ( 1 1 0 1

) (๐ ๐ 0

0 ๐๐ ) (

1 โ1 0 1

) = (๐ ๐ ๐๐ โ ๐๐

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 โซ ๐ โ ๐๐ ๐

of the vector field

๐ = (3๐ฅ + ๐ง77) ๐๐ + (๐ฆ 2 โ ๐ ๐๐ (๐ฅ2๐ง)) ๐๐ + (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,

โซ ๐ญ โ ๐๐บ ๐

= โญ div ๐น ๐๐ ๐ท

= โซ โซ โซ (3 + 2๐ฆ + ๐ฅ)๐๐ฅ๐๐ฆ๐๐ง 1

0

3

0

2

0

= โซ โซ (3 + 2๐ฆ + 1

2 ) ๐๐ฆ๐๐ง

3

0

2

0

= โซ (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

โซ ๐ โ ๐๐ฅ ๐ฟ

of the vector field ๐ = 2๐ฆ2๐๐ + ๐ฅ 2๐ฆ ๐๐ around the positively oriented curve L

where L is the boundary of the region D.

Solution:

๐๐บ = ๐๐๐๐ ๐๐

๐๐๐๐ ๐ญ = (๐๐๐ โ ๐๐) ๐๐

By the Stokesโs theorem,

โซ ๐ญ โ ๐๐ = โฌ ๐๐๐๐ ๐ญ โ ๐๐บ ๐น๐ณ

= โซ โซ (๐๐๐ โ ๐๐)๐๐๐๐ ๐โ๐

๐

๐

๐

= ๐

๐ โซ (๐๐ โ ๐)[๐๐]๐=๐

๐โ๐๐๐ ๐

๐

= โซ (๐ โ ๐)๐๐๐ ๐

๐

= ๐

๐ [(๐ โ ๐)๐]๐=๐

๐

= โ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 | 5

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

Solution:

Using

cosh ๐ง = ๐๐ง + ๐โ๐ง

2 , sinh ๐ง =

๐๐ง โ ๐โ๐ง

2 ,

we obtain

๐๐ง + ๐โ๐ง + 5(๐๐ง โ ๐โ๐ง) = 5,

6๐๐ง โ 4๐โ๐ง โ 5 = 0,

6๐2๐ง โ 5๐๐ง โ 4 = 0,

(3๐๐ง โ 4)(2๐๐ง + 1) = 0,

Which implies

๐๐ง = 4

3 ๐๐ ๐๐ง = โ

1

2 .

Therefore,

๐ง = ๐๐ ( 4

3 ) = ๐๐๐๐ (

4

3 ) + 2๐๐๐,

or

๐ง = ๐๐ (โ 1

2 ) = ๐๐๐๐ (

1

2 ) + (๐ + 2๐๐)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) โซ 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 = โ2.

(b) โฎ 1

๐ง(๐ง+1โ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/2 โค ๐ก โค ๐, and ๐๐ง = 2๐๐๐๐ก๐๐ก.

Thus,

โซ ๐๐(๐ง + 1) ๐๐ง ๐ถ

= โซ (2cos ๐ก + 1) 2๐๐๐๐ก๐๐ก ๐

๐ 2

= 2๐ โซ (2 ( ๐๐๐ก + ๐โ๐๐ก

2 ) + 1) ๐๐๐ก๐๐ก

๐

๐ 2

= 2๐ โซ (๐2๐๐ก + 1 + ๐๐๐ก) ๐๐ก ๐

๐ 2

= [๐2๐๐ก + 2๐ ๐ก + 2๐๐๐ก]๐ 2

๐

= (1 + 2๐ ๐ โ 2) โ (โ1 + ๐ ๐ + 2๐)

= ๐ (๐ โ 2).

(b) Both singularities ๐ง = 0 and ๐ง = โ1 + 3๐ are located outside the contour ๐ถ.

Therefore, โฎ 1

๐ง(๐ง+1โ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๐ข

๐๐ฅ = 0,

๐๐ฃ

๐๐ฅ = 2๐ฅ,

๐๐ข

๐๐ฆ = 0,

๐๐ฃ

๐๐ฆ = 2๐ฆ,

Since the Cauchy-Riemann equations

๐๐ข

๐๐ฅ =

๐๐ฃ

๐๐ฆ ,

๐๐ข

๐๐ฆ = โ

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