Final Exam
Math 1321 - Accelerated Engineering Calc II
May 2, 2016
Answer each question completely in the area below. Show all work and explain your reasoning. If
the work is at all ambiguous, it is considered incorrect. No phones, calculators, or notes are
allowed. Anyone found violating these rules or caught cheating will be asked to leave
immediately. Point values are in the square to the left of the question. If there are any other
issues, please ask the instructor.
By signing below, you are acknowledging that you have read and agree to the above paragraph, as
well as agree to abide University Honor Code:
Name:
Signature:
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Solutions
Question
Points
1
20
2
15
3
15
4
10
5
10
6
20
7
20
8
20
9
15
10
20
11
20
12
20
Total:
205
Score
Note: There are 12 questions on the exam with 205 points available but the exam will be graded
out of 200.
Math 1321: Final Exam
May 2, 2016
Potentially Useful Information
Taylor’s Theorem
M
|x − a|n+1 ,
(n + 1)!
|Rn | <
|f (n+1) | ≤ M
Projections
proja b =
a·b a
kak kak
Curvature
dT kT0 k
kr0 × r00 k
κ=
ds = kr0 k = kr0 k3
Cylindrical Coordinates
x = r cos θ,
y = r sin θ,
z = z,
r2 = x2 + y2
Spherical Coordinates
x = ρ sin φ cos θ,
y = ρ sin φ sin θ,
z = ρ cos φ,
ρ2 = x 2 + y 2 + z 2
2D Surfaces
n=
I
dS = kru × rv kdA
Green’s Theorem
I
ZZ
(∇ × F) · k dA
F · n ds =
∇ · F dA
ZZ
F · dr =
C
ru × rv
,
kru × rv k
D
C
I
Stokes’ Theorem
ZZ
F · dr =
(∇ × F) · dS
C
ZZ
D
S
Divergence Theorem
ZZZ
F · dS =
∇ · F dV
S
E
2/18
/0 pts
Math 1321: Final Exam
20
May 2, 2016
1. Choose one the following series and prove whether it diverges or converges, as well as indicate
which series you’ve chosen.
In either case, make sure to state the requirements of the test you’re using and why they
are satisfied.
∞
X
1
(1)
.
n(ln n)2
n=2
∞
X
(2)
(−1)n−1
n=1
n2
n
.
+1
Solution: This question was taken directly from Exam I.
(1) The integral test considers, in general, a series
∞
X
aj
j=k
and says that we can relate this to the integral of f (x), where f (j) = aj only if f (x) satisfies
three requirements:
(a) f (x) is continuous.
(b) f (x) is positive valued.
(c) f (x) is decreasing.
Here, we see that our choice of f (x) is f (x) =
listed above.
1
x(ln x)2 .
We must check the three conditions
(a) For x ≥ 2, the whole thing is continuous (as x = 0 is the only problematic point).
(b) Since (ln x)2 is always positive and x ≥ 2, the whole quantity is always positive.
(c) For this quiz, it’s perfectly okay to just say the function is decreasing because the
numerator stays 1 and the denominator grows, however we could show this a little
more precisely by considering
f 0 (x) = −
ln x + 2
,
x 2 (ln x)3
which we see, for x ≥ 1/e 2 is always negative, meaning the function is decreasing.
Thus, f (x) satisfies all of the requirements of the integral test meaning that if
Z ∞
Z ∞
1
f (x) dx =
dx
x(ln x)2
2
2
P
2
converges or diverges, so does ∞
j=2 1/(n ln n). To see whether this integral converges or
diverges, consider, as in the hint, u = ln x, so du = dx/x, meaning we have
Z ∞
Z ∞
1
1 u=∞
1
1
1
f (x) dx =
du = −
= lim −
+
=
,
2
u u=ln 2 t→∞
t
ln 2
ln 2
2
ln 2 u
so this integral indeed converges and therefore, by the integral test, as does the series.
3/18
/20 pts
Math 1321: Final Exam
May 2, 2016
(2) The alternating series test considers series of the form
∞
X
n
(−1) bn
∞
X
or
(−1)n−1 bn ,
n=1
n=1
where the terms bn must satisfy the following conditions:
(a) bn is decreasing, that is, bn ≤ bn+1
(b) bn decays to 0 or limn→∞ bn = 0.
Thus, in our case, bn =
n
n2 +1
and we must check these two properties.
(a) It’s pretty clear to see that n2 grows much faster than n so limn→∞ bn = 0. We could
do this more precisely by taking
1
n
n
=
lim
n→∞ n 2 + 1
1+
lim
1
n
= 0.
(b) Again, the above basically already shows its decreasing, but to see this more rigorously,
we can consider f (x) = x/(x 2 + 1) and
f 0 (x) =
1 − x2
,
(x 2 + 1)2
which means that so long as x > 1 (which it is), f 0 (x) < 0 meaning we have a
decreasing function (and therefore terms of the series.)
Thus, the requirements of the alternating series test are satisfied, which means that this
series converges. A very important nuance here: the alternating series test cannot prove
divergence, only convergence, however the integral test can prove either.
4/18
/0 pts
Math 1321: Final Exam
15
May 2, 2016
2. Compute the radius of convergence of the series
∞
X
(−1)n (x − 1)n .
n=1
Solution: This question was taken directly from a quiz. The solution we got to the previous
problem is
∞
X
(−1)n (x − 1)n = 1 − (x − 1) + (x − 1)2 − (x − 1)3 + · · · .
n=0
We now invoke the ratio test to determine for which x values this series converges. The ratio
test says that if you have a series
∞
X
an+1 := L
an
and
lim n→∞ an n=0
then the series converges for L < 1, (and as a reminder, diverges for L > 1 and if L = 1, we
have no clue). Thus, in this case, an = (−1)n (x − 1)n , so we have
(−1)n+1 (x − 1)n+1 (x − 1)n+1 = lim |x − 1|,
L = lim = lim
n→∞
(−1)n (x − 1)n n→∞ (x − 1)n n→∞
thus our condition for convergence is
|x − 1| < 1 =⇒ R = 1,
where R is the radius of convergence. Note that we don’t know if this series converges for the
end points, that is at x = 0, x = 2 but only the radius was asked for this problem. A separate
test could be used for each of these cases if the interval of convergence was asked for.
5/18
/15 pts
Math 1321: Final Exam
15
May 2, 2016
3. Compute T2 (x), the second order Taylor series of
f (x) = ln(1 + x),
around
x = 0.
Solution: This question was taken directly from the practice exam and was discussed in
detail during the review session. There are two ways to approach this problem. The first, is
to just recall the definition of a Taylor series of f (x) around x = a:
∞
T (x) = f (a) + f 0 (a)(x − a) +
X f (j) (a)
f 00 (a)
(x − a)2 + · · · =
(x − a)j .
2!
j!
j=0
Thus, to compute this, we just need to grind out a few derivatives:
f 0 (x) =
1
,
1+x
f 00 (x) = −
1
,
(1 + x)2
f 000 (x) =
2
.
(1 + x)3
And our evaluations are f (0) = 0, f 0 (0) = 1 and f 00 (0) = −1, leading to
T2 (x) = 0 + 1(x − 0) +
1
−1
(x − 0)2 = x − x 2 .
2!
2
An alternative way is to recall the cute construction of geometric series:
1
= 1 + x + x2 + · · ·
1−x
for
|x| < 1
and therefore, we can replace x by −x to get
1
= 1 − x + x2 + · · ·
1+x
for |x| < 1.
Now, we note the relationship
Z
dx
= ln(1 + x),
1+x
meaning that, to construct the series for ln(1 + x), we can simply integrate (term-by-term) the
series for 1/(1 + x):
Z
x2
T (x) =
1 − x + x 2 + · · · dx = C + x −
+ ··· ,
2
but what is C? Note, ln(1 + 0) = 0, so T (0) = 0 must be true, meaning C = 0 and we get
exactly the same series.
6/18
/15 pts
Math 1321: Final Exam
10
May 2, 2016
4. By showing that Rn → 0, use Taylor’s inequality to prove that the Taylor series of
f (x) = sin x
around
x =0
converges to f (x) as n → ∞.
Hint: we know that
P xn
converges, so what can we say about its terms?
n!
Solution: This problem was taken directly from an in-class example.
Taylor’s inequality says that if we chop our Taylor series off at the Tn term, that is, the nth
degree polynomial, then the worst case of the remaining error is
|Rn | ≤
M
|x|n+1 ,
(n + 1)!
where |f (n+1) (x)| ≤ M.
However, the thing that makes this problem doable, is notice that the n + 1th derivative of our
function is either ± cos x or ± sin x, but regardless, | ± sin x| ≤ 1 or | ± cos x| ≤ 1, meaning that
M = 1 here.
Thus, we’re left with:
|Rn | ≤
1
|x|n+1 .
(n + 1)!
Now, we simply take the limit as n → ∞, to see that
|x|n+1
,
n→∞ (n + 1)!
lim |Rn | ≤ lim
n→∞
but by the hint, since
P
x n /n! converges, its terms must go to zero, meaning that
|x|n+1
=0
n→∞ (n + 1)!
lim
which implies
|Rn | → 0,
and then by the squeeze theorem, Rn → 0 as well.
7/18
/10 pts
Math 1321: Final Exam
10
May 2, 2016
5. Find the value(s) of c for which the vectors
u = hc, 1, ci,
v = 2i − 3j + ck
are orthogonal.
Solution: This question was taken directly from a practice exam. What does it mean for
two vectors to be orthogonal? Their dot product is zero. Here, that means
0 = u · v = c(2) + 1(−3) + c(c) = c 2 + 2c − 3 = (c + 3)(c − 1),
or, in other words, c = 1 and c = −3 both work.
8/18
/10 pts
Math 1321: Final Exam
May 2, 2016
6. Consider the ellipse parameterized by
r(t) = ha cos t, b sin ti.
10
(a) Find an expression for the unit tangent vector, T(t).
10
(b) Find the curvature of the ellipse, κ, at the point (a, 0).
Solution: This question was taken directly from an older practice exam.
(a) We know the unit tangent vector can be computed by the expression
T(t) =
r0 (t)
,
kr0 (t)k
meaning that we just need to compute the derivative of our function and it’s magnitude:
p
p
r0 (t) = h−a sin t, b cos ti,
kr0 (t)k = (−a sin t)2 + (b cos t)2 = a2 sin2 t + b2 cos2 t,
and therefore
T(t) =
h−a sin t, b cos ti
r0 (t)
=p
,
0
kr (t)k
a2 sin2 t + b2 cos2 t
which we can’t really simplify any more.
(b) We use the formula on the sheet for curvature κ, which states
κ(t) =
kr0 × r00 k
,
kr0 k3
meaning that we must compute r00 , which here is equal to
r00 = h−a cos t, −b sin ti.
Computing the cross product and its magnitude:
r0 × r00 = h0, 0, ab sin2 t + ab cos2 ti,
kr0 × r00 k = ab.
From this, we can evaluate curvature directly, since we already know kr0 k:
κ(t) =
kr0 × r00 k
ab
,
= 2 2
0
3
kr k
(a sin t + b2 cos2 t)3/2
and note that (a, 0) corresponds to t = 0, so κ(0) = ab/b3 = a/b2 .
9/18
/20 pts
Math 1321: Final Exam
20
May 2, 2016
7. Minimize the function:
f (x, y ) = −x 2 + y 2
subject to the constraint
g(x, y ) = x 2 + y = 0.
Solution: This question was taken directly from Exam II. This problem is a constrained
minimization, meaning that we should use Lagrange multipliers, which say: the minimum occurs
when
∇f = λ∇g,
for some unknown scalar λ. Thus, we implicitly have three unknowns: x, y , λ and three equations:
fx = λgx ,
fy = λgy ,
g(x, y ) = x 2 + y = 0.
We see that
∇f = h−2x, 2y i,
∇g = h2x, 1i,
so our system becomes:
−2x = λ2x,
2y = λ,
x 2 + y = 0.
The first equation is only true when either x = 0 or λ = −1.
If we try the first possibility, x = 0, plugging this into our third equation, we have y = 0 as well
and from the second equation, λ = 0. Thus, the value of f (0, 0) = 0 is a possible minimizer.
Exploring the other possibility: λ = −1, we see that this yields
this into
√
√ y = −1/2. Plugging
2
x + y = 0, we get two values of x that produce this: x = 1/ 2 and x = −1/ 2. When you
plug either of these into f (x, y ), you get −1/4, which clearly is smaller than 0. Thus, these two
values produce the minimization.
10/18
/20 pts
Math 1321: Final Exam
20
May 2, 2016
8. Find a representation for the line describing the intersection of the following two planes:
3x − 6y − 4z = 15,
and
6x + y − 2z = 5.
Solution: This question was taken directly from a practice exam. From the equations of
the two planes, we can immediately conclude their normal vectors are
n1 = h3, −6, −4i
and
n2 = h6, 1, −2i.
The direction vector v of their intersection is then orthogonal to both of these, meaning we
must use the cross product to compute it
i
j
k
v = n1 × n2 = 3 −6 −4 = h16, −18, 39i.
6 1 −2
To define a line, we simply need a direction (which we have) and a point. Thus, we just need to
find ANY point on the intersection. Consider then, when z = 0, we have
3x − 6y = 15
and 6x + y = 5,
which, when we solve we get x = 15/13 and y = −25/13. Thus, the line is then
15 −25
r(t) = r0 + vt =
,
+ th16, −18, 39i.
13 13
Other representations of the line (say, symmetric) are perfectly acceptable.
11/18
/20 pts
Math 1321: Final Exam
May 2, 2016
9. True or false? Either way, justify your answer.
5
(a) div curl F = 0 for all vector fields F.
5
(b) If L is the line segment connecting (−1, 0) and (1, 0), C + is the upper half of the unit circle
and C − is the lower half, and F satisfies
Z
Z
Z
F · dr =
F · dr =
F · dr = 0,
L
C+
C−
then F must be conservative.
5
(c) grad div F = 0 all vector fields F.
Solution: This question was taken directly from a practice exam.
(a) True. We discussed this in class. One way to see it is if we think symbolically: ∇ · (∇ × F),
we know that × produces an orthogonal vector to both ∇ and F, and therefore the dot
product of ∇ with this will be 0.
(b) False. The integral being equal on three paths does not mean it is equal on all paths, which
is the requirement for F to be conservative.
(c) False. No reason to believe this is true. For instance, consider F = hx 2 , 0, 0i, then
∇ · F = 2x and then ∇(∇ · F) = 2 6= 0.
12/18
/15 pts
Math 1321: Final Exam
May 2, 2016
20 10. Compute the line integral
Z
F · dr,
F(x, y ) = x 2 i − xy j,
where
C
and C is the quarter-circle in the top-right half-plane, oriented counter-clockwise.
Hint: parameterize this curve!
Solution: This question was taken directly from a quiz. As the hint suggests, we have to
parameterize the curve C. We know the parameterization of a FULL circle is:
r(t) = cos(t)i + sin(t)j,
0 ≤ t ≤ 2π,
however, we only have a quarter circle. Thus, we take the modification
r(t) = cos(t)i + sin(t)j,
0 ≤ t ≤ π/2.
We can compute the derivative of this
r0 (t) = − sin(t)i + cos(t)j,
and we know the relationship between the infinitesimals is
dr = r0 dt = [− sin(t)i + cos(t)j]dt,
so our integral becomes
Z
Z
Z π
2
0
F(r) · r dt =
F · dr =
C
0
π
2
hcos2 (t), − cos(t) sin(t)i · h− sin(t), cos(t)i dt,
0
where, the dot product evaluates to
hcos2 (t), − cos(t) sin(t)i · h− sin(t), cos(t)i = −2 cos2 (t) sin(t).
so our integral becomes a u-substitution, where u = cos(t)
Z
Z π
2
2 3 t=π/2
2
F · dr = −2
cos2 (t) sin(t) dt =
cos (t) t=0 = − .
3
3
C
0
Note it makes perfect sense we get a negative answer when we notice the line integral goes
against the “circulation” of the vector field:
1.0
0.8
0.6
0.4
0.2
0.2
0.4
13/18
0.6
0.8
1.0
/20 pts
where
y#z
from
z
Math 1321: Final Exam
May 2, 2016
SOLUT
evalu
20 11. Using Stokes’ Theorem, evaluate
Z
F · dr,
where
F = h−y 2 , x, z 2 i,
C
y+z=2
C
and C is the intersection of the plane y + z = 2 and the
cylinder x 2 + y 2 = 1, oriented counterclockwise.
x
x2+y2=1
y
Altho
ellipt
then C
the di
FIGURE 3
Solution: This question was taken directly from an in-class example. We want to use
Stokes’ theorem, which states
Z
ZZ
F · dr =
curl F · dS.
C
S
Thus, we must find some surface S whose boundary is C. In this particular case, the natural
choice is the the ellipse-shaped plane shown in blue. We can then parameterize this surface in
the following way:
r(x, y ) = hx, y , g(x, y )i, 0 ≤ i k
x 2 + y 2 ≤ 1,
where g(x, y ) = 2 − y .
z surface inteIn other words, x, y provide our natural parameterization. We know, to compute
grals:
≈+¥+z@= 4
ZZ
ZZ
ZZ
curl F dS =
curl F · n dS =
curl F · (rx × ry ) dA,
S
S
S
S
C
so we need only compute the appropriate cross product, which, here is:
i j k 0
rx × ry = 1 0 gx = h−gx , −gy , 1i.
0 1 gy y
≈+¥=1
x
We must also compute the curl:
FIGURE 4
i
j
k curl F = ∇ × F = ∂x ∂y ∂z = h0, 0, 1 + 2y i.
−y 2 x z 2 Thus, our integral becomes
ZZ
ZZ
ZZ
ZZ
curl F·dS =
curl F·(rx ×ry ) dA =
−gx ·0−gy ·0+1(1+2y ) dA =
(1+2y ) dA.
S
D
D
D
The region in x, y space that we must integrate over is just the circle with radius 1. Meaning
14/18
/20 pts
v
E
to com
the pa
above
SOLUT
x2 #
circle
so
Also,
There
Math 1321: Final Exam
May 2, 2016
that our integral, under the transformation y = r sin θ, becomes
ZZ
Z 2π Z 1
(1 + 2y ) dA =
(1 + 2r sin θ) r dr dθ
D
0
Z
0
2π
=
0
Z
2π
=
0
r3
r2
+ 2 sin θ
2
3
r =1
dθ
r =0
1 2
+ sin θ dθ
2 3
1
= (2π) = π.
2
15/18
/0 pts
Math 1321: Final Exam
May 2, 2016
20 12. Using the Divergence Theorem, compute
ZZ
F · dS,
where
F = h3x + 4xy , xz − 2y 2 , cos xy 2 i,
S
and S is the surface of a solid cube bounded by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1, oriented
positively.
Solution: This question was taken directly from a practice exam. The divergence theorem
says that
ZZ
ZZZ
F · dS =
∇ · F dV.
S
E
Here, we have that
∇ · F = 3 + 4y − 4y + 0 = 3,
Thus,
ZZ
ZZZ
ZZZ
F · dS =
ZZZ
∇ · F dV =
S
E
3 dV = 3
E
dV = 3V (E),
E
RRR
since integrating
E 1 dV just is the volume of E. In this case, E is just the box described in
the problem, so V (E) = 13 = 1, so
ZZ
ZZZ
F · dS =
∇ · F dV = 3(13 ) = 3.
S
E
16/18
/20 pts
Math 1321: Final Exam
May 2, 2016
Bonus Questions
13. (Math) Find the value of c such that
∞
X
(1 + c)−n =
n=2
1
.
2
Hint: what kind of series is this? Note that it starts at n = 2.
Solution: This is a geometric series, meaning we immediately know that it converges if
1 1 + c < 1.
In this case, we know what a geometric series sums to. In general,
∞
X
a
=
ar n−1 ,
1−r
when
|r | < 1.
n=1
Doing some algebra here:
∞
X
n=2
−n
(1 + c)
n
∞ X
1
=
1+c
n=2
j+1
∞
X
1
=
(just relabeling j = n − 1)
1+c
j=1


2 X
j−1
∞ 1
1


(so it looks like our usual geom. series)
=
1+c
1+c
j=1
2 1
1
=
(by geometric series formula)
1+c
1 − 1/(1 + c)
1
.
=
c(c + 1)
Note, we want this to equal 1/2, so we have
1
1
=
c(c + 1)
2
so
c(c + 1) = 2 =⇒ c 2 + c − 2 = 0.
This has two solutions, c = −2 and c = 1. Note, however, if we plug in c = −2, we see that
our series becomes
∞
X
(−1)−n ,
n=2
which we know diverges. Hence, the only answer is c = 1.
17/18
/0 pts
Math 1321: Final Exam
May 2, 2016
14. (Maze) By popular demand: solve this maze of Isaac Newton, the founder of calculus:
Solution:
15. (Current Events) Last week, which former Republican House Speaker angered Satanists across
the country by not-so-affectionately describing Ted Cruz as “Lucifer in the Flesh”?
Solution: John Boehner.
http://www.rawstory.com/2016/04/satanists-are-furious-that-boehner-compared-ted-cruz-to-the-dark-lord/
18/18
/0 pts