# Final fall 2012 solutions

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15 Feb 2019
School
Department
Course
Professor
Northwestern University NetID:
ID #:
Math 230 Final Exam Solutions!
Fall Quarter 2012
Wednesday, December 12, 2012
Complete the front page!!!! Mark your section!! [3 points!]
Do not write your name on this exam!
Write your NetID and your Student ID Number in the spaces provided at
Instructions: Show ALL of your work on the pages provided! Explain your reasoning using
complete sentences and correct grammar, spelling, and punctuation. All numerical answers MUST
be exact; e.g., you should write πinstead of 3.14...,2 instead of 1.414..., and 1
Make sure that your ﬁnal answer is clearly indicated. This test has seventeen pages. Check to make
sure you have pages 1 - 17. Calculators, course notes, textbooks, etc. are NOT allowed. You have 2
hours! Good luck!
XSec. # Time Instructor
21 8:00 Garton
31 9:00 Bohmann
33 9:00 Tosatti
37 9:00 Yang
41 10:00 TiM
47 10:00 Tacy
49 10:00 Yang
57 11:00 Burslem
59 11:00 Garton
61 12:00 Burns
67 12:00 Tacy
71 1:00 Drummond-Cole
81 2:00 Burns
87 2:00 Drummond-Cole
Question Possible Points Score
Front Page 3
1 12
2 15
3 9
4 9
5 18
6 14
7 18
8 16
9 18
10 16
11 18
12 18
13 16
Total 200
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Math 230 Final Exam Solutions! Fall Quarter 2012 Page 2 of 17
Question 1. Suppose that u,vare two vectors in R3such that kuk= 3, kvk= 2, and the angle
between them is θ=π
3.
(a) (3 points) Calculate u·vand ku×vk.
u·v=kukkvkcos(θ) = 3(2)(1
2) = 3
ku×vk=kukkvksin(θ) = 33
(b) (5 points) Calculate (uv)·(u+ 2v).
(uv)·(u+ 2v) = u·(u+ 2v)v·(u+ 2v)
=u·u+ 2u·vv·u2v·v
=kuk2+u·v2kvk2
= 9 + 3 8
= 4
(c) (4 points) Calculate (u+ 3v)·(3u×v).
Answer: Recall that u×vis orthogonal to both uand v, so
u·(u×v) = v·(u×v) = 0
Then,
(u+ 3v)·(3u×v) = 3u·(u×v) + 9v·(u×v)
= 0 + 0
= 0
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Math 230 Final Exam Solutions! Fall Quarter 2012 Page 3 of 17
Question 2. For each of the following, calculate the limit or explain why the limit does not exist:
(a) (5 points) lim
(x,y)(0,0)
y3+ 5x2+ 5y2
x2+y2
Method 1: Polar Coordinates Method 2: Sandwich Theorem
lim
r0
r3sin3(θ) + 5r2
r2= lim
r0rsin3(θ) + 5 = 5
(since 5 rrsin3(θ)5 + r)
=lim
(x,y)(0,0)
y3+ 5x2+ 5y2
x2+y2= 5
5− |y| ≤ y3+ 5x2+ 5y2
x2+y25 + |y|
lim
(x,y)(0,0) 5− |y|= lim
(x,y)(0,0) 5 + |y|= 5
=lim
(x,y)(0,0)
y3+ 5x2+ 5y2
x2+y2= 5
(b) (5 points) lim
(x,y,z)(0,0,0)
yz
x2+y2+z2.
Method 1: Test Directions Method 2: Spherical Coordinates
Suppose y=z= 0 : lim
x0
0
x2= 0
Suppose x= 0, y =z: lim
y0
y2
2y2=1
26= 0
=lim
(x,y,z)(0,0,0)
yz
x2+y2+z2DNE
lim
ρ0
ρ2sin(θ) sin(ϕ) cos(ϕ)
ρ2= lim
ρ0sin(θ) sin(ϕ) cos(ϕ) :
Let θ=π
2,and ϕ= 0 : sin(θ) sin(ϕ) cos(ϕ) = 0
Let θ=π
2,and ϕ=π
4: sin(θ) sin(ϕ) cos(ϕ) = 1
2
Therefore, the limit depends on ϕ
=DNE
(c) (5 points) lim
(x,y)(0,0)
x4+y4
px4+y4+ 1 1
lim
(x,y)(0,0)
x4+y4
px4+y4+ 1 1= lim
(x,y)(0,0)
x4+y4
px4+y4+ 1 1·px4+y4+ 1 + 1
px4+y4+ 1 + 1
= lim
(x,y)(0,0)
(x4+y4)(px4+y4+ 1 + 1)
(x4+y4+ 1) 1
= lim
(x,y)(0,0) px4+y4+ 1 + 1
= 2
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