Spring 2015 Math 152/201–202
Exam 3H: Solutions
c
Thu, 30/Apr
2015
Art Belmonte
5. Find a power series representation for f (x) =
and determine the interval of convergence.
2/3
2
x n
2
n
= 1−(x/3)
= ∑∞
• Now 3−x
= 23 ∑∞
n=0 3n+1 x
n=0 3
x
for 3 < 1 or |x| < 3 by the Geometric Series
Theorem (GST).
• For x = −3, ∑ (−1)n 23 diverges by the Test
for Divergence.
1. Approximate the sum of the series correct to four
∞
(−1)n−1
decimal places. Cite relevant test(s).
∑ n 4n
n=1
• This series converges by the Alternating Series
Test (AST) since bn = |an | = n14n ↓ 0.
• For x = 3, ∑ 32 diverges for the same reason.
• Hence the interval of convergence is I = (−3, 3).
• For a partial sum of the series to be accurate to 4
decimal places, we need the remainder to satisfy
|Rn | ≤ ε = 5 × 10−5 . By the Alternating Series
Estimation Theorem (ASET),
1
−5 for n ≥ 5.
|Rn | ≤ bn+1 = (n+1)4
n+1 ≤ 5 × 10
• Therefore s5 =
(−1)n−1
∑5n=1 n 4n
t
dt as a power
1 + t3
series. What is its radius of convergence?
Z
6. Evaluate the indefinite integral
3 n = ∞ (−1)n t 3n
= ∑∞
∑n=0
n=0 −t
3
provided −t < 1 or |t| < 1 by the GST.
n 3n+1
∞
t
Thus 1+t
for |t| < 1.
3 = ∑n=0 (−1) t
• Now
≈ 0.2231.
(−1)n
is absolutely
n=2 n ln n
convergent, conditionally convergent, or divergent.
Cite relevant test(s).
∞
2. Determine whether the series
∑
1
1−(−t
3)
R
7. Find the Taylor series for f (x) = sin x centered at
a = π. Compute its radius of convergence.
• The
n sequence
o∞ of derivatives of f at π is
(n)
sin (π)
= {0, −1, 0, 1, 0, −1, 0, 1, . . . }.
• However,
∑ |an | diverges by the Integral Test:
R∞ 1
2 x ln x dx = limb→∞ (ln (ln b) − ln (ln 2)) = ∞.
n=0
n
(−1)
• Hence ∑∞
n=2 n ln n is conditionally convergent.
3. Find the radius of convergence of the series
n
∞ (−1) 3n+2
t
for
• Hence 1+t
3 dt = C + ∑n=0 3n+2 t
|t| < 1; i.e., the radius of convergence is R = 1.
• The series converges by the Alternating Series
1
Test (AST) since bn = |an | = n ln
n ↓ 0.
∞
2
3−x
• Accordingly, the desired Taylor series is
∑∞
n=0
n2 x n
∑ 2n n! .
f (n) (π)
n!
k+1
(−1)
2k+1
.
(x − π)n = ∑∞
k=0 (2k+1)! (x − π)
• The
Test gives
Ratio
ak+1 |x−π|2k+3 (2k+1)!
|x−π|2
ak = (2k+3)! · |x−π|2k+1 = (2k+2)(2k+3) → 0 < 1
as k → ∞ for all x ∈ R. Therefore, the radius of
convergence is R = ∞.
n=1
Cite relevant test(s).
• Apply the Ratio Test. As n → ∞, we have aan+1
n
x3 − 3x + 3 tan−1 x
.
x→0
x5
2
1 + n1 |x|
(n + 1)2 |x|n+1 2n n!
= n+1
·
→0<1
=
2 (n + 1)! n2 |x|n
2 (n + 1)
8. Use series to evaluate the limit lim
• The Maclaurin series for tan−1 x is
3
5
7
x − x3 + x5 − x7 + · · · . Hence as x → 0,
for all x ∈ R; radius of convergence: R = ∞.
x3 −3x+3 tan−1 x
x5
4. Find the radius and interval of convergence of the
∞
(x − 3)n
series ∑
. Cite relevant test(s).
n
n=1
=
3 5 3 7
5 x − 7 x +···
x5
= 35 − 37 x2 + · · · → 53 .
9. Use the Alternating Series Estimation Theorem to
estimate the range of values of x for which the
approximation is accurate to within the stated error.
p
√
• Root Test gives n |an | = |x−3|
n n → |x − 3| < 1 as
n → ∞ provided 2 < x < 4.
n
tan−1 x ≈ x −
• At x = 2, ∑ (−1)
n converges by the AST since
1
n ↓ 0.
x3 x5
+
3
5
(|error| < 0.05)
7
√
• Solve − x7 ≤ 0.05 to get |x| ≤ 7 0.35 ≈ 0.86.
Hence −0.86 ≤ x ≤ 0.86, approximately.
• At x = 4, ∑ n1 diverges (p-series with p = 1 ≤ 1).
• Thus R = 1 and I = [2, 4).
1
10. A car is moving with speed 20 m/s and acceleration
2 m/s2 at a given instant. Using a second-degree
Taylor polynomial, estimate how far the car moves in
the next second. Would it be reasonable to use this
polynomial to estimate the distance traveled in the
next minute? Explain.
13. A wagon is pulled a distance of 100 m along a
horizontal path by a constant force of 70 N. The
handle of the wagon is held at an angle of 35◦ above
the horizontal. Find the work done by the force.
• The work is done is
W = F·D = kFk kDk cos θ = 70·100·cos 35◦ ≈ 5734 J.
• Let f (t) be the car’s position in general. For a
frame of reference, let the given instant be t = 0
and f (0) = 0.
14. Find the scalar and vector projections of b = [2, 4, −1]
onto a = [3, −3, 1].
00
• Now f (t) ≈ T2 (t) = f (0) + f 0 (0)t + f 2(0) t 2
where f 0 (0) = 20 and f 00 (0) = 2. Therefore,
T2 (t) = 20t + t 2 .
• The scalar projection is
√
−7
a·b
7 19
= √ =−
compa b =
≈ −1.61.
kak
19
19
• Since T2 (1) = 21, we estimate the car moves
21 meters in the first second.
• The vector projection is
a·b
a
−7 [3, −3, 1]
√
proja b =
=√
kak kak
19
19
21 21
7
or − 19 , 19 , − 19
≈ [−1.11, 1.11, −0.37].
• It would be unreasonable to use T2 to estimate
the distance traveled in the next minute. Think
about it: after 60 seconds, the car’s speed would
be 140 m/s or 268 mph!
11. Find an equation of the sphere with center (−3, 2, 5)
and radius 4. What is the intersection of this sphere
with the yz-plane?
15. Find the volume of the parallelepiped (sheared box)
with adjacent edges PQ, PR, and PS.
R (5, 1, −1) S (0, 4, 2)
−→ →
−
→ −
• The desired volume is PQ · PR × PS .
Sequentially compute edge vectors, cross
product, dot product, and absolute value.
P (3, 0, 1)
• The yz-plane is x = 0. Plugging into the sphere
with equation (x + 3)2 + (y − 2)2 + (z − 5)2 = 42
yields (y − 2)2 + (z − 5)2√
= 7, a circle in the
yz-plane with radius r = 7 center (0, 2, 5).
12. Find the distance between the spheres x2 + y2 + z2 = 4
and x2 + y2 + z2 = 4x + 4y + 4z − 11.
Q (−1, 2, 5)
−→
PQ = [−4, 2, 4]
−
→
PR = [2, 1, −2]
−
→
PS = [−3, 4, 1]
→
−
→ −
PR × PS = [9, 4, 11]
−→ −
→
→ −
PQ · PR × PS
= 16
dist?
The volume is 16 cm3 .
16. Determine all vectors v = [v1 , v2 , v3 ] such that
[1, 2, 1] × v = [3, 1, −5] .
• The first sphere has center (0, 0, 0) and radius 2.
• Completing the squares for the second equation
gives (x − 2)2 + (y − 2)2 + (z − 2)2 = 1, a sphere
with center (2, 2, 2) and radius 1.
(There are infinitely many of them. Express your
answer in terms of the parameter t.)
• The vector equation yields a set of three linear
scalar equations.
• As illustrated, take the distance between these
centers and subtract the sum of the spheres’ radii
to
between the spheres; namely,
√ get the distance
√
12 − 3 = 2 3 − 3 ≈ 0.46 cm.
2v3 − v2 = 3,
v1 − v3 = 1,
v2 − 2v1 = −5
• Solving for v1 , v2 , v3 gives
v = [v1 , v2 , v3 ] = [t + 1, 2t − 3, t] .
2
17. Find an equation of the plane through the origin and
the points (2, −4, 6) and (5, 1, 3).
• A normal vector to the plane is
n = [2, −4, 6] × [5, 1, 3] = [−18, 24, 22].
• An equation of the plane is n · [x, y, z] = n · O:
−18x + 24y + 22z = 0 or 9x − 12y − 11z = 0.
18. Determine whether the lines L1 and L2 are parallel,
skew, or intersecting. If they are intersecting, find the
point of intersection.
5 − 12t, 3 + 9t, 1 − 3t
L1 (t) =
3 + 8s, −6s, 7 + 2s
L2 (s) =
• Direction vectors for the lines are
v1 = [−12, 9, −3] and v2 = [8, −6, 2],
respectively.
• Since v1 = − 23 v2 , these vectors are parallel.
Hence the corresponding lines are parallel.
(Alternatively, v1 × v2 = 0 shows that the
direction vectors are parallel.)
19. Find the slope of the tangent line to the polar curve
r = 2 + sin 3θ at the point corresponding to θ = π/4.
• First express x and y in terms of r and θ .
x = r cos θ = (2 + sin 3θ ) cos θ
y = r sin θ = (2 + sin 3θ ) sin θ
• The slope is
dy
dx
=
dy/dθ
dx/dθ
or
3 sin θ cos 3θ + cos θ (2 + sin 3θ )
.
3 cos θ cos 3θ − sin θ (2 + sin 3θ )
At θ = π2 ,
dy
dx
=
√
4−3 2
2
√
= 2 − 23 2 ≈ −0.12.
20. Find the length of the polar curve, both exactly and
approximately: r = 5θ , 0 ≤ θ ≤ 2π.
q
R
dr 2
• The arc length is L = αβ r2 + dθ
dθ
√
q
π
R 2π θ
(25 −1) 1+(ln 5)2
= 0 5
1 + (ln 5)2 dθ =
ln 5
or approximately 29015.56 cm.
3