MATH 151 COMMON EXAMINATION I Part 1 Name (print):

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MATH 151, FALL SEMESTER 2011
COMMON EXAMINATION I - VERSION B - SOLUTIONS
Name (print):
Instructor’s name:
Signature:
Section No:
Part 1 – Multiple Choice (12 questions, 4 points each, No Calculators)
Write your name, section number, and version letter (B) of the exam on the ScanTron form.
Mark your responses on the ScanTron form and on the exam itself
1.
Let v
a.
b.
c.
d.
e.
4, 2 and w
5
4
3
2
1
2j. Compute 1 v
2
w .
Correct Choice
SOLUTION: 1 v
2
2.
6i
w
1 4, 2
2
6, 2
2, 1
6, 2
1v
2
4, 3
w
42
32
5
A ball whose weight is 50 Newtons hangs from
two wires, one at angle 23° from horizontal,
and the other at angle 37° from horizontal.
Let T 1 be the tension in the first wire, and
T 2 be the tension in the second wire.
Which set of equations can be used to solve
for T 1 and T 2 ?
a. |T 1 | cos 23°
b.
|T 1 | cos 23°
c. |T 1 | cos 23°
|T 2 | cos 37°
|T 1 | cos 37°
|T 2 | cos 37°
0
50
|T 1 | sin 23°
and |T 2 | sin 23°
|T 2 | sin 37°
0
|T 1 | sin 23°
|T 2 | sin 37°
0
50 and
e.
|T 1 | cos 23°
|T 2 | cos 37°
0
and |T 1 | sin 23°
T2
|T 1 | cos 23°, |T 1 | sin 23°
|T 1 | cos 23° |T 2 | cos 37°
|T 2 | sin 37°
1, 3
and w
SOLUTION: |v|
2 3
cos
v w
50
Correct Choice
W
0, 50
|T 2 | cos 37°, |T 2 | sin 37°
0
|T 1 | sin 23° |T 2 | sin 37° 50 0
Find the angle between the vectors v
a. 0°
b. 30°
Correct Choice
c. 45°
d. 60°
e. 90°
2
50
|T 2 | sin 37°
|T 2 | cos 37°
|w|
50
|T 1 | sin 23°
|T 1 | cos 23°
2
|T 2 | sin 37°
and
d.
SOLUTION: T 1
T1 T2 W 0
3.
and
0
3 ,1 .
v w
|v||w|
2 3
4
3
2
30°
1
4.
Find the scalar projection (component) and vector projection of v
a.
scalar projection
b.
scalar projection
c.
scalar projection
d.
scalar projection
e.
scalar projection
SOLUTION:
v w
comp w v
|w|
5.
20
x2
x2
x2
x2
x2
y
y
y
y
y
vector projection
vector projection
16
5
v ww
|w| 2
proj w v
20
52
36 4i
64 i
25
3j
1
t and y
48 j
25
t2
t.
2
2
1
x
y
1
x
2
1
x2
x
x
Find a vector equation for the line which contains the point 3, 4 and is parallel to 1, 2 .
a.
b.
c.
d.
e.
r
r
r
r
r
t
t
t
t
t
4
3
2t, 3 t
t, 4 2t
3 t, 4 2t
Correct Choice
1 3t, 2 4t
1 3t, 2 4t
SOLUTION: r 0
Let f x
a.
x2
x
3, 4
v
b.
and
lim f x
tv
3, 4
t 1, 2
3
t, 4
2t
Correct Choice
lim f x
x 2
and
lim f x
lim f x
x 2
and
lim f x
x 2
e.
r0
lim f x
and
lim f x
x 2
d.
rt
x 2
x 2
c.
1, 2
5x 6 . Which of the following is true?
2 2
x 2
lim f x
x 2
None of these.
SOLUTION:
lim f x
x 2
2
vector projection
36
3j.
Correct Choice
x
x
3x
3x
3x
SOLUTION: t
7.
vector projection
5
4i
64 i
48 j
169
169
80 i 192 j
169
169
64 i 48 j
25
25
64 i 48 j
Correct Choice
25
25
64 i 48 j
25
25
vector projection
Find the Cartesian equation for the graph of the parametric curve x
a.
b.
c.
d.
e.
6.
16
13
16
13
16
5
16
5
16
5
12j onto w
5i
lim x
x
x 2
3
2
1
0
lim f x
x 2
lim x
x
x 2
3
2
1
0
1
1
t
3
8. Compute lim
t 3
t 3
a. 1/6
b. 0
c.
1/9
Correct Choice
d.
1/6
e. Does not exist
1/t 1/3
3 t
1
SOLUTION: lim
lim 1
lim 1
3t
9
t 3
3
t 3
t 3 t
t 3 3t
9. Which interval contains the unique real solution of the equation 2x 3 x 1 0?
a.
2, 1
b.
1, 0
Correct Choice
c.
0, 1
d.
1, 2
e.
2, 3
SOLUTION: Let f x
2x 3 x 1.
f 1
2 1 1
2
f0
1
Since 2 0 1, by the I.V.T. there is a c
1, 0 where f c
2c 3 c 1 0
2x 2 3
10. Which of the following is a horizontal asymptote of f x
?
x 3 x 3
a. y 2
Correct Choice
1
b. y
2
1
c. y
3
d. y
3
e. None of the above
2
1/x 2
2 3/x 2
2x 2 3
SOLUTION: lim
lim 2x2 3
lim
2
2
x 3 x 3
x
x
x 9
1/x
x
1 9/x 2
2
4
11. Evaluate lim 2x
4
x 2 x x
a.
b.
c.
d.
2
0
1
3
Correct Choice
e.
2
lim 2x 4
4
x 2 x x
2
2x
12. Evaluate lim 3x
x 2
x
SOLUTION:
a.
b.
c.
d.
e.
2 4 4
22 4
1
3
3
3
2
1
Correct Choice
2
SOLUTION: lim 3x 2x
x 2
x
2
lim 3x 2x
x 2
x
1/x
1/x
x
lim 3x 2
1 2x
3
Part 2 – Work Out Problems (5 questions. Points indicated. No Calculators)
Solve each problem in the space provided. Show all your work neatly and concisely, and
indicate your final answer clearly. You will be graded, not merely on the final answer, but also on
the quality and correctness of the work leading up to it.
13. (10 points) An object is moving in the xy-plane and its position vector after t seconds is
rt
t
2, t 2
3t .
Find the position vector of the object at time t
a.
SOLUTION: r 6
2, 6 2
6
3 6
6.
4, 18
Does the object pass through the point 1, 0 ? If yes, when? If no, why not?
b.
SOLUTION: t 2, t 2 3t
1, 0
t 2
t 3 satisfies both equations.
YES
t2
1
3t
0
Does the object pass through the point 2, 8 ? If yes, when? If no, why not?
c.
SOLUTION: t 2, t 2 3t
2, 8
t 2 2
t 2 3t 8
t 4 the only solution of the first equation but does not satisfy the second because
t 2 3t 4 2 3 4 4 8.
NO
5
14.
(14 points) Sketch the graph of the function
x if
x
1
3
3
if
x
1
2
|x|
if
1
1
gx
4
x2
3 if
x
x
2
1
2
-3
-2
-1
1
2
3
1
2
3
-1
Then determine each of the following
-2
lim g x
x
lim g x
1
x
g 1
1
lim g x
lim g x
x 2
x 2
g2
List all value(s) of x where g x is NOT differentiable:
5
SOLUTION:
lim g x
x
1
lim g x
x
4
2
2
lim g x
x
2
g 1
1
g2
3
3
1
lim g x
x
1
2
1
2
g x is NOT differentiable at :
1
1,
0,
2
-3
-2
-1
-1
-2
4
15. (9 points) Compute each of the following or prove the limit does not exist.
a.
lim
x 3
3|
3x
|x
x2
SOLUTION: When x 3, we have x
|x 3|
lim 2
lim x 3
lim 1x
3
x 3 x
3x
x 3 x x
x 3
b.
lim
x 3
3
lim
x 3
0 and |x
1
3
3|
3. So
x
3|
3x
|x
x2
3 . So
x
3|
3x
|x
x2
|x 3|
|x
1
lim 2
2
3
x 3x
x 3 x
3|
does not exist.
3x
SOLUTION: Since lim
x 3
the 2-sided limit lim
x 3
|x
x2
x2
x
6
x
16. (9 points) Consider f x
3
p
a.
3|
1
3
SOLUTION: When x 3, we have x 3
x 3
|x 3|
lim 2
lim
lim x1
3
x 3 x
3x
x 3 x x
x 3
c.
0 and |x
if x
3
if x
3
3|
3x
1,
3
Find lim f x or explain why it does not exist.
x 3
SOLUTION: lim f x
x 3
b.
2
lim x
x 3
x
6
x
lim
3
x
3 x 2
x 3
x 3
Find the value(s) of p that make f x continuous at x
exists.
SOLUTION:
For f x to be continuous at x
lim x
2
5
x 3
3 or explain why no such p
3, we must have f 3
lim f x or p
5.
x 3
17. (10 points) Consider the function f x
a.
Find f x , the derivative of f x , using the limit definition of the derivative.
SOLUTION:
fx
f x
lim
h
0
lim
h
b.
x.
0
h
h fx
h
x h x
x h
x
lim
h
x
0
lim
h
0
h
h
x
x
h
1
h
f 4
1
2 4
x
x
f x at x
4.
0
x
h
x
x
h
x
1
x
Find the slope of the tangent line to the curve y
SOLUTION: slope
h
h
lim
2 x
1
4
5
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