math 152 calculus ii midterm 1 examination solution...

©2012 Department of Mathematics – Eastern Mediterranean University

MATH 152 – CALCULUS II

MIDTERM‐1 EXAMINATION Solution Set

Department of Mathematics

Spring Term 2012‐2013 – November 06, 2012

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MATH152–CALCULUSIIMIDTERM‐1EXAMINATION

November 06, 2012

Student’s

Name – Surname: Number:

Signature: Group:

Question No

1 2 3 4 5 6 TOTAL

Weight 20 20 20 20 20 20 120

Received

Instructions

1. There are 6 questions in this examination.

2. No books, notes, calculators, cell phones or other electronic devices are allowed.

3. Students has to shut down their cell phones before the exam starts.

4. Duration is 90 minutes.

5. Results of this examination will be announced on Nov., 13, 2012 on http://brahms.emu.edu.tr/calculus

6. You may check your paper. To do so, visit your instructor’s office.

7. You may find the solutions of the examination questions on http://brahms.emu.edu.tr/calculus

Some Useful Information

0 ⟨ , , ⟩

∙‖ ‖

‖ ‖

, ,

cos| ∙ |‖ ‖‖ ‖

| | , , ̂ ̂

∙ ‖ ‖‖ ‖ cos ‖ ‖⟨cos , sin ⟩ ‖ ‖ cos ‖ ‖ sin

‖ ‖‖ ‖ sin

⋯!

∙

, → | | lim→

cos , sin , , tan


QUESTION 1. (10+10 pts)

a) Find the parametric equations of the line passing through point 3, 5, 4 and is parallel to another line given by

⟨1 3 , 1 2 , 3 ⟩, ∈ . Direction vectors of the two lines are also

parallel ⟨3, 2,1⟩, thus the parametric

equations of line are:

3 3

5 2 ∞ ∞

4

b) Show that the planes

1 and 2 2 are neither orthogonal nor parallel. Then find the parametric equations for the line of intersection of these planes.

Plane 1: ⟨1,1,1⟩ Plane 2: ⟨1, 1,2⟩ ∙ 1 1 2 2 0 planes are not

orthogonal

3 2 0 planes are not parallel

Hence they are intersecting. To find the line

of intersection, a point on the line and a

direction vector is required.

To find a point on the line let 0, then 1 2

Solving for and , we obtain,

3 2⁄ and 1 2⁄ and a point on the

line of intersection is: 3 2⁄ , 1 2⁄ , 0

Direction vector is:

3 2

Hence a set of parametric equations for the

line of intersection is,

32

3

∞ ∞

2


a) Use second order Taylor polynomial to

approximate the value of √6.

√ , 4

2!

√1

2√

1

8 ⁄

214

4164

4

≅

√6 6 ≅ 2146 4

164

6 4

≅ 212

116

≅3916

b) Find the center, radius and interval of

convergence of the power series,

2

Center is at 0 Radius is,

lim→

lim→

2

21

1

So the interval of convergence is 1, 1

But analysis at the two end points is required

At the left end point 1, and the series is

2 1convergent alternating series

At the right end point 1, and the series is

2convergent p‐series

Thus the interval of convergence is

1, 1


QUESTION 3. (5+5+10 pts)

a) Show that the vectors 2 3 and 4 6 are parallel.

2 a scalar multiple

Then the two vectors are parallel. (Also you

may find that their cross product is zero

vector, implying that the vectors are parallel)

b) Show that the vectors 3 4 and

4 3 are orthogonal. ∙ 12 12 0 imlying that the two

vectors are orthogonal

c) Let 2 3 and 2 6 Find

∙| |

∙ 2 6 6 2

| | 4 9 1 14

∙| |

214

2 3

27

37

17

QUESTION 4. (5+5+10 pts)

Let

,

a) Find the domain of ,

Domain is: , : , 0,0

b) Find

lim, → ,

,

lim, → ,

, lim, → ,

1 1

1 112

c) Find

lim, → ,

,

Let for any ∈

lim , → , lim →

lim→

lim→ 1 1

Hence for different values of (i.e. on

different paths) limit has different values. By

the two‐path test limit does not exist.



a) Let ⟨ , 6 , 4⟩. Compute the

derivative of the function .

.

∙

⟨2 , 6, 0⟩| ∙

⟨2 , 6, 0⟩ ∙

⟨2 , 6 , 0⟩

b) Evaluate the integral

2 3 5

2 3 5

2 3 5

25

32

5

25

132

QUESTION 6. (4 pts each)

a) Consider the parametric equations 2 ,

3 4, 2 2. Sketch the curve for the values 2, 1, 0, 1,2. On the graph show the positive orientation of the curve.

b) Eliminate , and obtain an equation of the

curve of part (a) in terms of and .

2 →2

3 432

4, 4 4

c) Graph the polar point 4, 3 2⁄ . Write

one alternative set of polar pair for this point.

d) Express the polar point 2, 2⁄ in

Cartesian coordinates.

2 cos2

0, 2 sin2

2

The point is 0, 2 in Cartesian coordinates

e) Express the point 2, 2 in polar

coordinates.

2 2 2√2,

tan 154

thus the point is 2√2,54

1 2 3 4 5-1-2-3-4-5

1

2

3

45

6

7

89

10

11

-1

-2-3

2

1

0

1

2

4,32

4,2

math 152 calculus ii midterm 1 examination solution...

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