M1C_Sample_Exam_2_Answers
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Subject
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Date
Apr 3, 2024
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MATH 1C: SAMPLE EXAM 2
c Jeffrey A. Anderson
ANSWER KEY
True/False
(15 points: 3 points each) For the problems below, circle T if the answer is true and circle F is the
answer is false. After you’ve chosen your answer, mark the appropriate space on your Scantron
form. Notice that letter A corresponds to true while letter B corresponds to false.
1.
T
F
If
lim
n
→∞
a
n
= 0, then
∞
∑
n
=1
a
n
converges.
2.
T
F
The series
∞
∑
n
=1
3
ne
-
n
2
diverges
3.
T
F
If
f
has a local minimum at point (
a, b
)
∈
R
2
, then
D
u
f
(
a,b
)
= 0
for any unit vector
u
∈
R
2
.
4.
T
F
If
a
n
>
0 and
∞
∑
n
=1
a
n
converges, then
∞
∑
n
=1
(
-
1)
n
a
n
converges.
5.
T
F
Suppose
f
:
R
3
→
R
. If
∇
f
= 0 at a point
x
∈
R
3
, then
f
has a local extreme value
at point
x
.
Multiple Choice
(45 points: 3 points each) For the problems below, circle the correct response for each question.
After you’ve chosen your answer, mark your answer on your Scantron form.
6. Let
z
= sin(
x
·
y
) and let
x
=
x
(
t
) and
y
=
y
(
t
) be functions of
t
. Suppose
x
(1) = 0
,
y
(1) = 1
,
x
0
(1) = 2
,
y
0
(1) = 3
.
Find
dz
dt
when
t
= 1.
A. 1
B.
2
C. 3
D. 4
E. 5
7. The series
∞
∑
n
=0
r
n
converges if and only if:
A.
-
1
< r <
1
B.
-
1
≤
r
≤
1
C.
-
1
≤
r <
1
D.
-
1
< r
≤
1
E.
r <
1
8. Find the direction of maximum increase of the function
f
(
x, y, z
) =
x e
-
y
+ 3
z
at the point (1
,
0
,
4)
.
A.
-
1
-
1
3
B.
1
-
1
3
C.
-
1
3
3
D.
-
1
-
3
3
E.
1
1
3
9. Which of the following series converge?
1)
∞
X
n
=1
1
n
2)
∞
X
n
=1
(
-
1)
n
·
n
ln(
n
)
3)
∞
X
n
=1
(
-
1)
n
n
A. 1
B. 2
C.
3
D. 2
,
3
E. None
10. Find the limit of the sequence
a
n
= 2 +
-
4
5
n
:
A.
2
B.
6
5
C.
-
4
5
D.
-
2
E.
4
5
Math 1C: Sample Exam 2
c Jeffrey A. Anderson
Page 2 of 10
11. Find the shortest distance from the origin to the surface
z
2
= 2
xy
+ 2
A.
1
√
2
B.
√
2
C.
1
2
D. 2
E. 1
12. The series
∞
∑
n
=1
1
n
α
converges if and only if
A.
1
< α
B.
α <
1
C.
-
1
< α <
1
D.
α
≥
1
E.
-
1
< α
13. Find the minimum value of the function
f
(
x, y
) =
x y
subject to the constraint that
x
2
+
y
2
= 2:
A. 1
B. 2
C.
-
1
D.
3
2
E.
-
3
2
14. Find the values of
x
for which the series
∞
∑
n
=1
(
x
-
1)
n
:
A.
-
2
< x <
0
B. 0
< x
≤
2
C.
0
< x <
2
D. 0
≤
x
≤
2
E.
-
2
≤
x <
0
15. Find the directional derivative of the function
f
(
x, y
) =
y
2
·
ln(
x
)
at the point (1
,
2) in the direction of the vector (3
,
4) = 3
i
+ 4
j
:
A.
5
16
B. 12
C.
5
12
D.
16
5
E.
12
5
Math 1C: Sample Exam 2
c Jeffrey A. Anderson
Page 3 of 10
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16. Find an equation of the tangent plane to the surface
√
x
+
√
y
+
√
z
= 4 at the point (4
,
1
,
1)
.
A. 2
x
+
y
-
z
= 1
B.
x
+ 2
y
+ 2
z
= 8
C.
x
-
2
y
+ 4
z
= 0
D.
x
+
y
+
z
= 6
E. 2
x
+
y
+
z
= 10
17. Which of the following series converge?
1)
∞
X
n
=1
3
2
n
2
3
n
2)
∞
X
n
=1
1
(
n
+ 1)
3
3)
∞
X
n
=1
n
+ 1
√
n
3
+ 2
A. None
B. 1
C.
2
D. 3
E. 2
,
3
18. Determine how many critical points the function
f
(
x, y
) =
x y
-
x
2
y
-
x y
2
has:
A. 1
B. 2
C. 3
D.
4
E. 5
19. How many terms of the alternating series
∞
X
n
=1
(
-
1)
n
+1
n
-
2
must we add in order to be sure that the partial sum
s
n
is within 0
.
0001 of the sum
s
.
A. 10
B. 300
C. 30
D.
100
E. 1000
20. Let
f
(
x, y
) =
x
y
+
y
x
. Find the gradient vector
∇
f
:
A.
2
y
2
x
B.
x
y
C.
1
y
-
y
x
2
1
x
-
x
y
2
D.
-
y/x
2
-
x/y
2
E.
y
x
Math 1C: Sample Exam 2
c Jeffrey A. Anderson
Page 4 of 10
Free Response
21. (10 points) A company test-markets a new canned energy drink made of all natural ingredients in 5
cities of equal size on the West Coast of the US. The selling price (in dollars) and the number of drinks
sold per week is each of the cities is listed as follows
City
Price
Sales/Week
1
0
.
79
6000
2
0
.
89
3980
3
0
.
99
3300
4
1
.
09
2440
5
1
.
19
1990
This data is plotted in the figure next to the table above. Although the data do not exactly lie on a
straight line, we can create a linear model to fit this data.
A. Set up the least squares problem to fit this data to a linear model
S
(
p
) =
c
1
+
c
2
p
where
S
is the sales per week and
p
is the price.
Solution:
Recall that the least squares problem is designed to fit data collected during an
experiment to a particular mathematical model. In this case, we are told that our company
collected five data points
{
(
p
i
, s
i
)
}
5
i
=1
,
where
p
i
= the price per energy drink sold in city
i
for
i
= 1
,
2
, ...,
5
s
i
= the number of cans of the energy drink sold in city
i
for
i
= 1
,
2
, ...,
5
We notice that the model appears to fit a linear model
S
(
p
) =
c
1
+
c
2
p
for unknown parameters
c
1
, c
2
∈
R
.
This model can be used to predict the number of cans sold in city
i
based on the
selling price
p
i
as follows:
S
(
p
i
) =
c
1
+
c
2
·
p
i
.
The difference between the observed data and the model prediction is known as the model error
in the
i
th term, given by:
e
i
= (
S
(
p
i
)
-
s
i
) = (
c
1
+
c
2
·
p
i
-
s
i
)
.
To create the model of best fit for unknown parameters
c
1
, c
2
∈
R
, we want to minimize the
Math 1C: Sample Exam 2
c Jeffrey A. Anderson
Page 5 of 10
sum of the squared error terms:
f
(
c
1
, c
2
) =
5
X
i
=1
e
2
i
=
5
X
i
=1
(
c
1
+
c
2
·
p
i
-
s
i
)
2
= (
c
1
+ 0
.
79
·
c
2
-
6000)
2
+ (
c
1
+ 0
.
89
·
c
2
-
3980)
2
+ (
c
1
+ 0
.
99
·
c
2
-
3300)
2
+ (
c
1
+ 1
.
09
·
c
2
-
2440)
2
+ (
c
1
+ 1
.
19
·
c
2
-
1990)
2
Thus, the least squares problem is to minimize the function
f
(
c
1
, c
2
).
B. Explain how you would use multivariable calculus to find the line of best fit.
Solution:
We apply multivariable calculus to solve this problem by recalling the second deriva-
tive test for the multivariable function
f
(
c
1
, c
2
). In particular, we know
f
has a local minimum
if and only if
A.
∇
f
=
0
B.
∂f
∂c
1
·
∂
2
f
∂c
2
-
∂f
∂c
1
∂c
2
2
<
0 with
∂
2
f
∂c
2
1
>
0.
Thus, to find the local minimum of
f
using multivariable calculus, we need to find the critical
points of this function and apply the second derivative test for multivariable function appro-
priately.
Remark (preview of coming attractions): There are two drawbacks of this method worth men-
tioning:
I. The method of minimizing the square of the modeled error is algebraically intensive. It
requires us to expand the multivariable function
f
(
c
1
, c
2
) into quadratic terms in
c
1
and
c
2
. Further to find the zeros of this polynomial requires non-linear methods.
II. Although multivariable calculus can be used to verify that the critical point where
∇
f
=
0
is a local minimum, there is theoretical result that can conclude that this point will also
be a global minimum. Thus, without further analysis of the function
f
, this method will
not always guarantee a unique absolute minimum error term.
In Math 2B (Linear Algebra), we will revisit this problem using least squares techniques to
improve the methods we discussed in this class.
Math 1C: Sample Exam 2
c Jeffrey A. Anderson
Page 6 of 10
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22. (10 points)
A. Find the value of
∞
∑
n
=2
3
n
+ 5
n
15
n
. Show your work.
Solution:
∞
X
n
=2
3
n
+ 5
n
15
n
=
∞
X
n
=2
3
n
15
n
+
5
n
15
n
=
∞
X
n
=2
1
5
n
+
1
3
n
=
∞
X
n
=1
1
5
n
+1
+
1
3
n
+1
=
∞
X
n
=1
1
5
2
·
1
5
n
-
1
+
1
3
2
·
1
3
n
-
1
=
1
25
·
1
1
-
1
5
+
1
9
·
1
1
-
1
3
=
13
60
B. Show that series
∞
∑
n
=2
1
n
·
(ln(
n
))
p
converges if
p >
1 and diverges if
p
≤
1.
Solution:
In this problem, we are asked to find values of
p
such that our given series converges.
To this end, let
f
(
x
) =
1
x
·
(ln(
x
))
p
.
For
x
≥
2, we confirm that
f
(
x
) is positive, decreasing and continuous using methods from
Math 1A. Thus, by the integral test for convergence, we know that our given series and the
integral
∞
Z
2
f
(
x
)
dx
have identical convergence behavior (the either both converge or they both diverge). With this
in mind, consider the two cases:
Math 1C: Sample Exam 2
c Jeffrey A. Anderson
Page 7 of 10
I. Case
p
= 1:
∞
Z
2
f
(
x
)
dx
=
∞
Z
2
1
x
·
ln(
x
)
dx
=
∞
Z
ln(2)
1
u
du
Let
u
= ln(
x
) =
⇒
du
=
1
x
dx
and
u
(2) = ln(2) while
u
(
∞
) =
∞
= lim
t
→∞
t
Z
ln(2)
1
u
du
= lim
t
→∞
ln(
u
)
t
ln(2)
=
lim
t
→∞
ln(
t
)
-
ln(2) =
∞
.
Thus, for
p
= 1 our integral diverges. By the integral test for series, we conclude that
the given series also diverges for
p
= 1.
II. Case
p
6
= 1:
∞
Z
2
f
(
x
)
dx
=
∞
Z
2
1
x
·
(ln(
x
))
p
dx
=
∞
Z
ln(2)
1
u
p
du
Let
u
= ln(
x
) =
⇒
du
=
1
x
dx
and
u
(2) = ln(2) while
u
(
∞
) =
∞
= lim
t
→∞
t
Z
ln(2)
u
-
p
du
= lim
t
→∞
u
1
-
p
1
-
p
t
ln(2)
=
lim
t
→∞
t
1
-
p
1
-
p
+
c
where
c
=
-
(ln(2))
1
-
p
1
-
p
.
Thus we see that the integral converges if
p >
1 and diverges for all
p
≤
1. This is what
was to be shown.
Math 1C: Sample Exam 2
c Jeffrey A. Anderson
Page 8 of 10
23. (10 points) Let
f
:
R
2
→
R
be a differentiable, multivariable function. Let
u
be a unit vector.
A. Derive the dot product formula for the limit definition of the directional derivative
D
u
f
.
Solution:
Let
f
:
R
2
→
R
be a multivariable function with two input variables.
Suppose
u
=
a
i
+
b
j
= (
a, b
) be a unit vector.
To find the directional derivative of
f
at a point
x
0
= (
x
0
, y
0
)
∈
R
2
in the direction of
u
, let
x
=
x
0
+
t
u
us consider the following limit:
D
u
f
(
x
0
) =
lim
x
→
x
0
f
(
x
)
-
f
(
x
0
)
k
x
-
x
0
k
2
= lim
t
→
0
f
(
x
0
+
at, y
0
+
bt
)
-
f
(
x
0
, y
0
)
t
=
g
0
(0)
where we introduce the auxiliary function
g
(
t
) =
f
(
x
0
+
a t, y
0
+
b t
)
.
For reference, the conversion from the first limit to the second follows from function evaluation
at the vectors
x
and
x
0
along with the calculation:
k
x
-
x
0
k
2
=
k
t
u
k
2
=
|
t
|
.
We can now use the auxiliary function
g
(
t
) combined with the chain rule for multivariable
functions to find an equivalent representation for
g
0
(
t
):
g
0
(
t
) =
∂f
∂x
·
dx
dt
+
∂f
∂y
·
dy
dt
=
f
x
(
x
0
+
a t, y
0
+
b t
)
a
+
f
y
(
x
0
+
a t, y
0
+
b t
)
b
Substituting the value
t
= 0 into this equation leads to
D
u
f
(
x
0
) =
f
x
(
x
0
, y
0
)
a
+
f
y
(
x
0
, y
0
)
b
=
∇
f
·
u
.
B. Prove that the gradient vector is in the direction of maximum increase.
Solution:
From Part A above, we know can use the dot product formula for the directional
derivative of
f
in the direction of
u
to find
D
u
f
=
k∇
f
·
u
k
2
=
k∇
f
k
2
k
u
k
2
cos(
θ
)
=
k∇
f
k
2
cos(
θ
)
≤ k∇
f
k
2
We used the cosine formula for the dot product to get this series of equalities where
θ
is the
angle between the gradient vector and the vector
u
. Thus we see that the directional derivative
is maximized when cos(
θ
) = 1, which happens when
θ
= 90
◦
.
In other words, the gradient
points in the direction of maximal increase.
Math 1C: Sample Exam 2
c Jeffrey A. Anderson
Page 9 of 10
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Challenge Problem
24. (Optional, Extra Credit, Challenge Problem)
Suppose that a sequence
{
a
n
}
∞
n
=1
satisfies
0
< a
n
≤
a
2
n
+
a
2
n
+1
for all
n
≥
1. Prove that the series
∞
∑
n
=1
a
n
diverges.
Solution:
Let us begin this problem by considering the significance of the stated property that
0
< a
n
≤
a
2
n
+
a
2
n
+1
for our sequence
{
a
n
}
∞
n
=1
. Consider
a
1
≤
a
2
+
a
3
a
2
+
a
3
≤
a
4
+
a
5
+
a
6
+
a
7
a
4
+
a
5
+
a
6
+
a
7
≤
a
8
+
a
9
+
a
10
+
a
11
+
a
12
+
a
13
+
a
14
+
a
15
Notice we can find a pattern in the indices for these inequalities:
1
≤
2 + 3
2 + 3
≤
4 + 5 + 6 + 7
4 + 5 + 6 + 7
≤
8 + 9 + 10 + 11 + 12 + 13 + 14 + 15
Let’s use this pattern to generalize to the
k
th inequality. Set
p
k
=
2
k
-
1
X
j
=2
k
-
1
a
j
and notice we can write
p
k
≤
p
k
+1
. Thus, we must have
2
N
-
1
X
n
=1
a
j
=
N
X
k
=1
p
k
≥
Np
1
=
Na
1
.
As
N
→ ∞
we see the partial sums are unbounded since
a
1
>
0 and thus the original series must
diverge as claimed.
Math 1C: Sample Exam 2
c Jeffrey A. Anderson
Page 10 of 10