1092 Calculus4 06-12 Final Exam Solution
June 19, 2021
1. (12 pts) Use Green’s theorem to evaluate the line integral
∮C
(x2+y3)dx + (x3+y2)dy
where C traces counter-clockwisely the polar curve r2=sin(4θ), 0 ≤ θ ≤ π 4.
Solution:
The following steps of arguments must be shown clearly!
(2%) State what is the Green’s theorem:
∮C(x2+ y3)dx + (x3+ y2)dy = ∫ ∫D[(x3+ y2)x− (x2+ y3)y]dxdy whereD is the region enclosed by r2= sin(4θ), 0 ≤ θ ≤ π/4.
(2%) Find the relevant partial derivatives and the link to polar coordinates:
(x3+ y2)x− (x2+ y3)y = 3(x2− y2) = 3r2cos 2θ
(3%) Iteration for the double integral in terms of polar coordinates:
∫ ∫D[(x3+ y2)x− (x2+ y3)y]dxdy = ∫0π/4∫
√ sin(4θ)
0 3r3cos(2θ)drdθ
= 34∫0π/4cos(2θ) sin2(4θ)dθ
(5%) Work out the trigonometric integration:
= 38∫0π/4cos(2θ)[1 − cos(8θ)]dθ =38∫0π/4[cos(2θ) −12(cos(6θ) + cos(10θ))]dθ
= 38[sin(2θ)2 −sin(6θ)12 −sin(10θ)20 ]π/40 =38[12+121 −201] = 15
2. Let α and β be two constants and set
F(x, y, z) = (y + αzexz)i + βx j − 2xexzk.
(a) (5 pts) Find curl(F) in terms of α and β.
(b) (4 pts) Find the values of α and β such that F is conservative on R3. (c) (6 pts) For the pair of values that you found in (b), evaluate∫
C
F ⋅ dr where C = {(x, y, z) ∈ R3 ∶ z = 2x2=y3, 0 ≤ x ≤ 2}
oriented in increasing values of x.
Solution:
(a) (Total 5%) curl(F) = 0i + (2 + α)(1 + xz)exzj + (β − 1)k (5 pts) Partial credit below is given if the answer is not completely correct:
Write down the correct definition for curl(F) (2 pt)
Write down correct component of i,j,k (1 pt+1 pt+1 pt ) (b) (Total 4%)
State the correct statement: [F is conservative on R3 iff curlF = 0 on R3] or [if curlF = 0 on R3 then F is conservative on R3] (2 pts)
α = −2, β = 1 (2 pts)
(c) (Total6%) Method 1: Find potential function and use fundamental theorem of line integral:
Write down F(x, y, z) = (y − 2zexz)i + xj − 2xexzk or mention anywhere about the correct substitution of α and β into F (1 pt)
Solving for potential function F = ∇f: f(x, y, z) = xy − 2exz (2 pts, if students include any constant c behind f , it is acceptable)
By Fundamental theorem of line integral,
∫C
F ⋅ dr = f (2, 2, 8) − f (0, 0, 0) = 6 − 2e16
(3 pts; 1 pt for correct initial point (0, 0, 0), 1 pt for correct terminal point (2, 2, 8), 1 pt for correct answer)
Method 2: Direct computation of line integral
Write down F correctly or mention anywhere about the correct substitution of α and β into F (1 pt).
Any correct parametrization AND orientation of C (1 pt).
Correct computation of line integral. (3 pts).
Correct answer = 6 − 2e16 (1 pt).
Remark: For Method 2, if answer is incorrect, the maximum score is 2 pts. Only need to check for F and C.
3. The figure on the right gives a surface S which is part of the graph z = y2−x2 enclosed by the curve C with parameterization
r(t) = ⟨2 sin(t), 2 cos(t), 4 cos(2t)⟩ with 0 ≤ t ≤ 2π.
(a) (7 pts) Find the surface area ∬
S
1 dS of S.
(b) (8 pts) By Stokes’ Theorem, evaluate the circulation
∮C
(yex−y3)dx + (x3+ex)dy + (z2ez)dz.
Solution:
Marking Scheme for Question 3a
1% - (1) Parametrisation of S
1% - (2) Correct specification of the ranges of parameters D
1% - (3) Correct value of ∣∣rx×ry∣∣
2% - (4) Correct definition of surface integrals (at most 1% for those who does
not/cannot specify D or makes mistakes in ∣∣rx×ry∣∣)
2% - (5) Correct evaluation (1% for students who made minor sign errors/obvious typos)
Remark. In other words, if a student makes mistakes in (2) or (3), he/she can earn at most 3% for this part of the question.
Sample Solution to Q3(a) Ver 1.
Parametrise S by r(x, y) = ⟨x, y, y2−x2⟩
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
1%
where x, y ∈ D = {(x, y) ∶ x2+y2 ≤4}
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
1%
. Then
∬S
1 dSdef= ∬
D
√
4x2+4y2+1
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
1%
dA
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
2%
Polar
= ∫
2π
0 ∫
2
0
r
√
4r2+1 dr dθ = π
6(1732 −1)
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
2%
.
Sample Solution to Q3(a) Ver 2.
Parametrise S by r(r, θ) = ⟨r sin θ, r cos θ, r2cos(2θ)⟩
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
1%
where 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
1%
.
Then ∣∣rr×rθ∣∣ =r√
4r2+1
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
1%
and
∬S
1 dSdef= ∫
2π
0 ∫
2
0
r
√
4r2+1 dr dθ
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
2%
= π
6(1732 −1)
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
2%
.
Marking Scheme for Question 3b
1% - (1) Statement of Stokes Theorem
1% - (2) Correct computation of curl(F)
1% - (3) Correct (k-component of) ry×rx
3% - (4) Correct transformation of a flux integral into a double integral (at most 1%
for those who does not/cannot specify D in (a) or makes mistakes in (2) or (3))
2% - (5) Correct evaluation (1% for students who made a sign error)
Remark a. In other words, if a student makes mistakes in (2) or (3) or not specifying D, he/she can earn about 3-4% for this part of the question.
Remark b. Sign error will be deducted from (5) only.
Sample Solution to Q3(b) Ver 1.
By Stokes’ Theorem, ∮
C
F ⋅ dr =∬
S
curl(F) ⋅ dS
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
1%
.
To evaluate the RHS, first we compute that curl(F) = (3x2+3y2)k
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
1%
and therefore
∬S
curl(F) ⋅ dS =∬
D
⟨0, 0, 3x2+3y2⟩ ⋅ ⟨−2x, 2y, −1⟩
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
1%
dA = −3∬
D
(x2+y2)dA
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
3%
Polar
= −3∫
2π
0 ∫
2
0
r3dr dθ
= −24π
²
2%
Sample Solution to Q3(b) Ver 2.
By Stokes’ Theorem, ∮
C
F ⋅ dr =∬
S
curl(F) ⋅ dS
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
1%
.
To evaluate the RHS, first we compute that curl(F) = (3x2+3y2)k
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
1%
and therefore
∬S curl(F) ⋅ dS =∫
2π
0 ∫
2
0
⟨0, 0, 3r2⟩ ⋅ ⟨⋆, ⋆, −r⟩
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
1%
dr dθ = −3∫
2π
0 ∫
2
0 r3dr dθ
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
3%
= −24π
²
2%
4. Consider the field of gravitational force F ∶ R3∖ {(0, 0, 0)} → R3 given by F(x, y, z) = −1
(x2+y2+z2)
3 2
⋅ (xi + yj + zk).
(a) (6 pts) Show that div(F(x, y, z)) = 0 for any (x, y, z) ≠ (0, 0, 0).
(b) (7 pts) Evaluate, directly, the flux of F across the sphere SR ∶x2+y2+z2 =R2 where R > 0 endowed with outward orientation.
(c) (5 pts) Hence, find the flux of F across the ellipsoid E ∶ x2+y2+2z2 =1 oriented outward.
Solution:
For (a), let F = F1i + F2j + F3k. The formula of divergence is
div(F) = (F1)x+ (F2)y+ (F3)z. (1) It is straightforward to obtain
(F1)x(x, y, z) =−1 + 3x2(x2+y2+z2)−1 (x2+y2+z2)
3 2
. (2)
Remark : One can obtain (F2)y and (F3)z in a straightforward manner like (2), or just say that F is symmetric in the symbols x, y, z.
For (b), the flux is given by
∬SR
F(x, y, z) ⋅(x, y, z)
R dS. (3)
Applying spherical coordinates yields
∬SR
−(x, y, z) R3 ⋅
(x, y, z)
R dS = −1 R2 ∬
SR
dS (4)
=
−1
R24πR2 (5)
= −4π. (6)
Remark : One can use other methods than spherical coordinates to evaluate the flux.
For (c), let R > 0 be sufficiently small so that SR is contained in the interior of the ellipsoid E.
We endow SR with the orientation in the item (b). Let ΩRbe the space region between SR and E.
Since (0, 0, 0) is not in ΩR, we can apply the divergence theorem on ΩR and use the item (a) to obtain
(∬E− ∬
SR
)F ⋅ d ⃗S =∭
ΩR
div(F) dV (7)
=0. (8)
Therefore, by the item (b) we see
∬E
F ⋅ d ⃗S =∬
SR
F ⋅ d ⃗S (9)
= −4π. (10)
Grading Suggestion.
For (a), get 2/6 by obtaining (1); get 1/6 by obtaining each (F1)x, (F2)y, and (F3)z as shown similarly in (2); get 1/6 by explaining why div(F(x, y, z)) = 0.
For (b), get 2/7 by obtaining (3); get 2/7 by obtaining (4); get 2/7 by obtaining (5); get 1/7 by obtaining (6).
For (c), get 1/5 by explaining the setting in the red paragraph; get 1/5 by obtaining each of (7)–(10).
5. (a) (8 pts) Find the interval of convergence of the power series
∞
∑
n=0
x2n
(2n + 1)(2n + 2).
(b) Determine whether each of the following series is absolutely convergent, conditionally convergent, or divergent. Please state the test(s) that you use.
(i) (6 pts)
∞
∑
n=1
(−1)n⋅ [ 1
n +tan−1( 1 n)].
(ii) (6 pts)
∞
∑
n=1
(−1)n⋅ (21/n2 −1).
Solution:
(a) (Method 1: Use ratio test) Set an=
x2n
(2n + 1)(2n + 2). Then
n→∞lim∣an+1
an ∣ = lim
n→∞
RR RR RR RR RR RR RR RR RR R
x2n+2 (2n + 3)(2n + 4)
x2n (2n + 1)(2n + 2)
RR RR RR RR RR RR RR RR RR R
= ∣x2∣. (2%)
By ratio test, we know that ∑∞n=0 x2n
(2n + 1)(2n + 3) converges for ∣x∣ < 1 (1%) and diverges for
∣x∣ > 1 (1%).
For x = ±1, we consider the series ∑∞n=0 (±1)2n
(2n + 1)(2n + 3) = ∑∞n=0 1
(2n + 1)(2n + 3). We compute that
n→∞lim
1
(2n + 1)(2n + 3) 1
n2
= 1
4. (1%)
Since ∑∞n=0 1
n2 converges, by limit comparison test, ∑∞n=0 1
(2n + 1)(2n + 3) converges (2%). Therefore, the interval of convergence of the power series ∑∞n=0 x2n
(2n + 1)(2n + 3) is [−1, 1] (1%).
(Method 2: Use root test) Set an=
x2n
(2n + 1)(2n + 2). Then, by
n→∞lim
1
n
√
(2n + 1)(2n + 3)
= lim
n→∞e−ln(2n+1)(2n+3)
n = lim
n→∞e−(2n+1)(2n+3)8n+8 =e0 =1, (1%) we have
n→∞lim
n
√
∣an∣ = lim
n→∞
1
n
√
(2n + 1)(2n + 3)
∣x2∣ = ∣x2∣. (1%)
By root test, we know that ∑∞n=0 x2n
(2n + 1)(2n + 3) converges for ∣x∣ < 1 (1%) and diverges for
∣x∣ > 1 (1%).
For x = ±1, we consider the series ∑∞n=0 (±1)2n
(2n + 1)(2n + 3) = ∑∞n=0
1
(2n + 1)(2n + 3). We compute that
n→∞lim
1
(2n + 1)(2n + 3) 1
n2
= 1
4. (1%)
Since ∑∞n=0 1
n2 converges, by limit comparison test, ∑∞n=0 1
(2n + 1)(2n + 3) converges (2%). Therefore, the interval of convergence of the power series ∑∞n=0 x2n
(2n + 1)(2n + 3) is [−1, 1] (1%).
(b)-(i) Set an= ∣(−1)n(1
n+tan−1 1 n)∣ = 1
n +tan−1 1 n by 1
n, tan−1(1
n) >0 for n ≥ 1. Since ∑∞n=0 1 n diverges and 1
n +tan−1 1 n >
1
n, by comparison test, we have ∑∞n=0 1
n +tan−1 1
n diverges (2%). So
∑∞n=1(−1)n( 1
n +tan−1 1
n) is not absolutely convergent (1%). Set bn = 1
n +tan−1 1
n. Let f (x) = x + tan−1x. Then we have f′(x) = 1 + 1
1 + x2 >0 for 0 ≤ x ≤ 1. (Or Since tan−1x is increasing for x ≥ 0, we have tan−1(1/n) is decreasing.) So bn is decreasing and limn→∞ 1
n +tan−1 1 n = 0 (1%). By alternating series test, we obtain that ∑∞n=1(−1)n(
1
n +tan−1 1
n) is convergent (1%).
Therefore, ∑∞n=1(−1)n(1
n+tan−1 1
n) is conditionally convergent (1%).
(b)-(ii) Set an= ∣(−1)n(21/n2 −1)∣ = 21/n2−1. Then we compute that
n→∞lim
21/n2 −1 1/n2 = lim
x→0+
2x−1 x = lim
x→0+ln 2 ⋅ 2x=ln 2.(2%)
Since ∑∞n=1(1/n2)converges (1%), by limit comparison test (1%), we obtain that ∑∞n=1(21/n2−1) converges (1%). Therefore, ∑∞n=1(−1)n(21/n2 −1) is absolutely convergent (1%).
6. Consider the function f (x) =∫
x 0
e−t3dt.
(a) (6 pts)Write down the Maclaurin series of f (x) and specify its radius of convergence.
(b) (4 pts)What is the value of f(691)(0)?
(c) (5 pts)Evaluate lim
x→0
f (x) − x (e2x2−1) ⋅ sin(3x2)
.
(d) (5 pts)Approximate the value of f (0.5) up to an error of 10−4 by some estimation theorem of series.
Solution:
(a)
Since et= ∑∞n=0t
n
n! (1 pt) for t ∈ R, we have e−t3 =
∞
∑
n=0
(−t3)n n! =
∞
∑
n=0
(−1)n
n! ⋅t3n for t ∈ R (2 pts) and
f (x) =∫
x 0
e−t3dt =∫
x 0
(
∞
∑
n=0
(−1)n
n! ⋅t3n)dt =
∞
∑
n=0
(−1)n n! ⋅ ∫
x 0
t3ndt
=
∞
∑
n=0
(−1)n
(3n + 1) ⋅ n! ⋅ (t3n+1∣
x
0
) =
∞
∑
n=0
(−1)n
(3n + 1) ⋅ n!x3n+1 for x ∈ R. (2 pts) The radius of convergence of f (x) is ∞. (1 pt)
(b)
By Taylor theorem, we need to solve the equation 3n + 1 = 691 ⇒ n = 230.(2 pts)The coefficient of x691 in the Maclaurin series of f (x) is 691⋅230!(−1)230. So we have
f(691)(0) = (−1)230
691 ⋅ 230! ⋅ (691!) = 690!
230!. (2 pts) (c)
(Method I) Since
e2x2 =
∞
∑
n=0
(2x2)n n! =
∞
∑
n=0
2n
n! ⋅x2n=1 + 2x2+2x4+ ⋯, (1 pt) sin(3x2) =
∞
∑
n=0
(−1)n
(2n + 1)! ⋅ (3x2)2n+1=
∞
∑
n=0
(−1)n⋅32n+1
(2n + 1)! ⋅x4n+2=3x2− 9
2x6+ ⋯, (1 pt) we have
limx→0
f (x) − x (e2x2 −1) ⋅ sin(3x2)
=lim
x→0
(x − 14x4+ ⋯) −x
[(1 + 2x2+2x4+ ⋯) −1] ⋅ (3x2−92x6+ ⋯)
(1 pt)
=lim
x→0
−14x4+x5(⋯) 6x4+x5(⋯)
=
−1
24. (2 pts) (Method II)
limx→0
f (x) − x (e2x2 −1) ⋅ sin(3x2)
=lim
x→0
∫
x
0 e−t3dt − x (e2x2−1) ⋅ sin(3x2)
( fumdamental theorem of calculus )
L′H
= lim
x→0
e−x3−1
e2x2 ⋅ (4x) ⋅ sin(3x2) + (e2x2−1) ⋅ cos(3x2) ⋅6x (2 pts)
Let g(x) ∶= e2x2 ⋅ (4x) ⋅ sin(3x2), h(x) ∶= (e2x2−1) ⋅ cos(3x2) ⋅6x. Then we have g′(x) = e2x2⋅ (4x)2⋅sin(3x2) +e2x2⋅4 ⋅ sin(3x2) +e2x2 ⋅ (4x) ⋅ cos(3x2) ⋅ (6x)
h′(x) = e2x2⋅4x ⋅ cos(3x2) ⋅6x − (e2x2 −1) ⋅ sin(3x2) ⋅ (6x)2+ (e2x2 −1) ⋅ cos(3x2) ⋅6 limx→0
g′(x)
x2 =0 + 12 + 24 = 36 limx→0
h′(x)
x2 =24 + 0 + 12 = 36 Hence
limx→0
f (x) − x (e2x2 −1) ⋅ sin(3x2)
=lim
x→0
e−x3 −1
e2x2⋅ (4x) ⋅ sin(3x2) + (e2x2 −1) ⋅ cos(3x2) ⋅6x
L′H
= lim
x→0
−3x2⋅e−x3
g′(x) + h′(x)=lim
x→0
−3e−x3
g′(x) x2 +h
′(x) x2
=
−3 36 + 36 =
−1
24.(3 pts) (d)
f (0.5) = ∑∞n=0 (3n+1)⋅n!(−1)n (0.5)3n+1 (1 pt) is an alternating series. Let RN be the error incurred by estimating with the N -th partial sum. Then
RN ≤
1
(3N + 4) ⋅ (N + 1)!⋅ (0.5)3N +4. (2 pts) Let (3N +4)⋅(N +1)!1 ⋅ (0.5)3N +4≤10−4. Then we get N ≥ 2.(1 pt) Therefore,
f (0.5) ≈ 1 2−
1 64+
1
7 ⋅ 256.(1 pt)
Note. If you only give(guess) a number without any explaination, then you get 0 credit.
