math 28_sample midterm_answer key.pdf

7/29/2019 Math 28_SAMPLE MIDTERM_answer key.pdf

http://slidepdf.com/reader/full/math-28sample-midtermanswer-keypdf 1/3

REVIEW ITEMS FOR MIDTERM EXAM

PART 1. FILL THE BLANKS WITH CORRECT EXPRESSIONS OR WORDS.

1. If 2 12 −−= ,, A , then 3 = A

2. The unit vector in the same direction as 2 12 −−= ,, A is3

2

3

1

3

2 −−= ,,

Au r

r

3. If 14 3 ,,−=C and ,,, 3 6 8 −= Dr

then 90 7 2 3 ,,−=+ DC r r

4. The direction angles of 3 0 0 −,, are2 2

π ππ π β ββ β

π ππ π α αα α == , and π ππ π γ γγ γ =

5. If 2

1

2

3 ,−= A and ,,2 0 = B then 1=⋅ B Ar

6. In problem no.5, the radian measure of the angle between A and B is

=

2

1

3 cos Arc

π ππ π

7. In problem no.5, the scalar projection of A onto B is2

1

8. In problem no.5, the vector projection of A onto B is2

10 ,

9. If the direction angle of a vector G is4

5 π ππ π and its magnitude is 4, then

2 2 2 2 −−= ,G

10. Consider the points ( )5 4 −,C and ( )2 3 ,− D . If DC is a representation of E ,

then 7 7 −= , E

11. An equation of a plane that is parallel to the xz-plane and which passes

through the point ( )3 2 1 ,, is 2 = y

12. The distance between ( )3 2 1 ,, A and ( )4 3 2 −− ,, B is 59 .

13. The midpoint of the segment whose endpoints are ( )3 2 1 ,, A and

( )4 3 2 −− ,, B is

−−

2

1

2

5

2

1,,

14. The standard equation of the sphere with ( )3 2 1 ,, A and ( )4 3 2 −− ,, B as

endpoints of a diameter is4

59

2

1

2

5

2

12 2 2

=

++

−+

+ z y x

15. The point ( )3 2 1 ,, lies

outside the sphere given by ( ) 5 1

2 2 2 =−++ z y x .

16. The graph of 0 10 2 6 4 2 2 2 =−−+−++ z z y y x x is a sphere.

17. A standard equation of the plane passing through ( )3 2 1 ,, and having

4 3 2 −− ,, as a normal vector is given by ( ) ( ) ( ) 0 3 4 2 3 12 =−−−+−− z y x

18. The distance between the parallel planes given by 0 5 2 2 =++− z y x and

0 6 2 4 4 =++− z y x is3

2

19. The distance from the point ( )3 2 1 ,, to the plane given by 0 5 2 2 =++− z y x

is 2

20. The parametric equations of the line passing through ( )3 2 1 ,, and is parallel

to 6 5 4 ,, are given by t z t yt x 6 3 5 2 4 1 +=+=+= ,,

21. If 3 2 1 ,,= A and ( )4 3 2 −− ,, Br

, then k j i A 7 2 17 +−−=× Br

22. In 3 R , the graph of 14

2 =− y x is called a parabolic cylinder.

23. The trace of 1

92

2 2 2

=−− z y x

on the xz-plane is called a hyperbola

24. The limit of the sequence 1,-1,1,-1,1,-1,… does not exist

25. The limit of the sequence

n

nsinas ∞→n is 0

26. 2

3 1

12

n

n

n −

+

∞→lim is equal to 0



27. The k -th partial sum of the geometric series ∑∞

=

−

1

1

k

k ar is

r

r ak

−

−

1

1

28. The series ∑∞

=

1

2

13

n

n

is absolutely convergent.

29. The sum ...... ++++++n

1

4

1

3

1

2

11 is equal to +∞ .

30. The sequence( )

−

n

n1

is convergent

31. If ( ) 0 < x f ' for all 1≥ x , then ( ){ }n f is decreasing

32. The sum of the infinite series ∑∞

=

+

1

1

3

1

n

n

is6

1

33. The sum of the infinite series

( )∑

∞

= +1 1

2

n nn

is 2

34. The p-series ∑∞

=1

1

n

pn

converges if 1> p

35. The series1

1 1

2n

nn

∞

=

+

∑ is divergent

36. If ∑∞

=1nn

k converges, then k = 0.

37. The series ∑∞

=1

2

2

2 nn

nis absolutely convergent. Using the ratio test, the value

of n

n

n u

u 1+

∞→lim is 0

38. In its interval of convergence, the sum of the power series ( )∑∞

=

−

0

1

n

n x is

expressed by x −2

1

39. The interval of convergence of the power series ( )n

x n

n

n∑∞

=

−−

1

11 is ( 11,−

40. The Maclaurin series expansion of the function ( ) x x f sin= is

( )( )∑

∞

=

+

+

−

0

12

12

1

n

nn

n

x

!

PART 2. PROBLEM SOLVING. WRITE YOUR SOLUTIONS NEATLY, COMPLETELY

AND LOGICALLY.1. Determine the general equation of the plane through the point

( )12 0 − , , P and parallel to the plane 0 8 3 2 =++− zy x .

Solution:

Consider point ( )12 0 0 −= ,, P and a normal vector 3 12 ,,−= N which is a

vector orthogonal to the plane 0 8 3 2 =++− zy x Thus, the needed equation of the plane is given by 0 5 3 2 =++− z y x .

2. Find an equation of a plane containing the point ( )113 −,, and parallel to the

lines1

3

3

2

1

2 −=

−=

− z y x and

2

4

4

3

1

2 −=

−=

− z y x .

Solution:

We need to find a vector perpendicular both to the lines, which is also a

normal vector to the plane. The needed vector is given by

112 2 4 113 1 ,,,,,, −=×= N .

An equation of the plane is given by ( ) ( ) ( ) 0 113 2 =++−−− z y x .



In general, we have 0 4 2 =−+− z y x .

3. Identify and sketch the graph of the following surfaces in 3 R :

a. 36 94 2 2 =− z x

b. 12 16 4

2 2 2

=−+z y x

c. x z y 4 4 2 2 =+

Solution:

a. Hyperbolic cylinder

b. Elliptic hyperboloid of one sheet

c. Elliptic paraboloid

4. Given the series ...+⋅

+⋅

+⋅

=∑∞

=7 5

1

5 3

1

3 1

1

0 n

nu

a. Find nu .

Solution:( )( ) ( ) ( )3 2 2

1

12 2

1

3 2 12

1

+−

+=

++=

nnnnun

(Use method of partial fractions)

b. Let nn uuu S +++= ...2 1 . Find a formula for n S .

Solution:( ) ( )

+−

+++

−+

−=

3 2 2

1

12 2

1

10

1

6

1

6

1

2

1

nn S n ...

( )

+−=

+−=

3 2

11

2

1

3 2 2

1

2

1

nn S n

c. Find nn

S ∞→

lim , if it exists.

Solution:2

1

3 2

11

2

1=

+−=

∞→∞→ n S

nn

nlimlim

d. Is the series ∑∞

=0 n

nu convergent? Why?

Solution: Since nn

S ∞→

lim exists, then the series ∑∞

=0 n

nu is convergent.

5. Use Ratio Test to determine whether the series ∑+∞

= 1

2

3

n

n

nis convergent or

divergent.

Solution: Let2

3

nu

n

n = . Consider( ) ( )2

2 2

2

11

1

3

3 1

3

+=⋅

+=

++

n

nn

nu

u

n

n

n

n

Using ratio test,( )

3

1

3

2

2 1 =

+==

∞→

+

∞→ n

n

u

u L

nn

n

nlimlim . Since 1> L , then the

series is divergent.

6. Consider the power series( ) ( )

2 1

0

1 3

2

n n

n

n

x

∞

+

=

− −

∑. Find its radius of

convergence and determine its interval of convergence.

Solution: Applying the ratio test, let( )

( ) 4

3

3

2

2

3 12

3 2

11

−=

−⋅

−=

+

+

++ x

x

x

u

u

n

n

n

n

n

n

7 14 3 14

3 1 <<−⇔<−⇒<

−== +

∞→ x x

x

u

u L

n

n

nlim

At 7 = x : ( )∑

∞

=

−

0

2

1

n

n

which is a divergent series since ( )2

1n

n−

∞→lim is not

equal to zero (nth term test for divergence)

At 1−= x : ∑∞

=0 2

1

n

which is a divergent series since2

1

∞→nlim is not equal to

zero (nth term test for divergence). Thus, IOC is ( )7 1,−

End of items

math 28_sample midterm_answer key.pdf

Documents