MATH 2203 Exam 1 Solutions: Geometry and Vector Calculus - Prof. Sean F. Ellermeyer, Exams of Advanced Calculus

The solutions to exam 1 for math 2203, a college-level mathematics course focusing on geometry and vector calculus. It includes instructions for students, solutions to various problems, and explanations of mathematical concepts. Topics covered include inequalities, vector dot products, finding angles between vectors, orthogonal vectors, and parametric equations.

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2010/2011

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MATH 2203 –Exam 1 (Version 1) Solutions
February 11, 2008
S. F. Ellermeyer Name
Instructions. Your work on this exam will be graded according to two criteria: mathe-
matical correctness and clarity of presentation. In other words, you must know what
you are doing (mathematically) and you must also express yourself clearly. In particular,
write answers to questions using correct notation and using complete sentences where
appropriate. Also, you must supply su¢ cient detail in your solutions (relevant calculations,
written explanations of why you are doing these calculations, etc.). It is not su¢ cient to just
write down an “answer” with no explanation of how you arrived at that answer. As a rule
of thumb, the harder that I have to work to interpret what you are trying to say, the less
credit you will get. You may use your calculator but you may not use any books or notes.
1. Describe in words (writing a complete sentence or sentences) the region of <3repre-
sented by the inequality 4< x2+y2+z2<9.
Answer: This region consists of all points between (but not on) the spheres of radius
2 and 3 centered at the origin.
2. Let aand bbe the vectors a=h2;1iand b=h5;7i.
(a) Compute jaj,jbj, and ab.
(b) Find the angle between aand b(to the nearest tenth of a degree).
Solution:
jaj=q(2)2+ (1)2=p5
jbj=p52+ 72=p74
ab= (2) (5) + (1) (7) = 17.
Also, if is the angle between aand bthen
cos () = ab
jajjbj=17
p5p74
so
= arccos 17
p5p74152:1.
3. Find a unit vector that is orthogonal to both i+jand i+k.
Solution:
(i+j)(i+k) = ik+ji+jk
=jk+i
=ijk
is orthogonal to both i+jand i+k. Since the vector we found had magnitude p3, a
unit vector that is orthogonal to both i+jand i+kis
u=p3
3(ijk).
1
pf3

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MATH 2203 ñExam 1 (Version 1) Solutions February 11, 2008

S. F. Ellermeyer Name

Instructions. Your work on this exam will be graded according to two criteria: mathe-

matical correctness and clarity of presentation. In other words, you must know what

you are doing (mathematically) and you must also express yourself clearly. In particular,

write answers to questions using correct notation and using complete sentences where

appropriate. Also, you must supply su¢ cient detail in your solutions (relevant calculations,

written explanations of why you are doing these calculations, etc.). It is not su¢ cient to just

write down an ìanswerî with no explanation of how you arrived at that answer. As a rule

of thumb, the harder that I have to work to interpret what you are trying to say, the less

credit you will get. You may use your calculator but you may not use any books or notes.

  1. Describe in words (writing a complete sentence or sentences) the region of <^3 repre- sented by the inequality 4 < x 2 + y 2 + z 2 < 9.

Answer: This region consists of all points between (but not on) the spheres of radius 2 and 3 centered at the origin.

  1. Let a and b be the vectors a = h 2 ; 1 i and b = h 5 ; 7 i.

(a) Compute jaj, jbj, and a  b.

(b) Find the angle between a and b (to the nearest tenth of a degree).

Solution:

jaj =

q (2) 2

  • (1) 2 =

p 5

jbj =

p 52 + 7^2 =

p 74

a  b= (2) (5) + (1) (7) = 17.

Also, if  is the angle between a and b then

cos () =

a  b

jaj jbj

p 5

p 74 so

 = arccos

p 5

p 74

 .

  1. Find a unit vector that is orthogonal to both i + j and i + k.

Solution:

(i + j)  (i + k) = i  k + j  i + j  k

= j k + i

= i j k

is orthogonal to both i + j and i + k. Since the vector we found had magnitude

p 3 , a unit vector that is orthogonal to both i + j and i + k is

u =

p 3

3

(i j k).

  1. Find parametric equations for the line, L 1 , that passes through the point (0; 1 ; 2) and

is parallel to the plane x + y + z = 2 and perpendicular to the line, L 2 , with parametric equations

x = 1 + t

y = 1 t

z = 2t

1 < t < 1

Solution: A normal vector to the given plane is n = h 1 ; 1 ; 1 i. This vector is per- pendicular to L 1. A direction vector for L 2 is v 2 = h 1 ; 1 ; 2 i. This vector is also perpendicular to L 1. If v 1 = ha; b; ci is a direction vector for L 1 , then both n  v 1 = 0 and v 2  v 1 = 0 must be true. Thus

a + b + c = 0

and a b + 2c = 0.

Among the many nonñtrivial solutions of the above system of equations is ha; b; ci = h 3 ; 1 ; 2 i. We thus see that parametric equations for L 1 are

x = 3 t

y = 1 + t

z = 2 + 2t

1 < t < 1

  1. (a) Sketch the parametric curve r (t) = cos (t) i sin (t) j, 0  t  2  with an arrow

on the sketch to denote the direction of increasing t.

(b) Find the tangent vector to this curve at the point on the curve corresponding to t = 3= 4.

Solution: The curve is the unit circle, starting at the point (1; 0) and traced out one complete time in the clockwise direction. The tangent vector at any point is

r 0 (t) = sin (t) i cos (t) j

so the tangent vector to this curve at the point on the curve corresponding to t = 3= 4 is

r 0 (t) = sin

i cos

j =

p 2

2

i +

p 2

2

j.

  1. Find both cylindrical and spherical coordinates for the point whose rectangular coor-

dinates are ( 1 ; 1 ; 1). (Round angles to the nearest tenth of a degree.) Show all details of your work.