MAT 111 Exam 1 Fall 2014 Name:SOLUTIONS

Franklin and Marshall College
Department of Mathematics
MAT 111
Exam 1
Fall 2014
Name:SOLUTIONS
Instructions:
• You have 50 minutes to take this exam.
• You are allowed only to have a writing implement (pen, pencil, eraser) while taking this exam.
• No internet, notes, calculators or any other outside aids are permitted. You may use the
backs of pages for scratchwork.
• You do NOT have
to simplify algebraically complicated answers. However, numerical answers
3
π
such as sin 6 , 4 2 , eln 4 , ln(e7 ), e− ln 5 , or e3 ln 3 should be simplified completely. Leave
fractions as they are (do not convert to decimal).
• Please read each question carefully. Show ALL work clearly in the space provided.
• Full credit will be awarded to solutions that are complete, legible, logically presented and
notationally sound.
• The point value of each question is indicated after its statement.
Grading - For Administrative Use Only
Question:
1
2
3
4
5
6
Total
Points:
20
20
20
10
10
20
100
Score:
MAT 111, Manack
Exam 1
1. In R3 , roughly sketch the equation of the sphere
(x − 3)2 + 4y + z 2 = 5 − y 2
Write down its center and radius.
(x − 3)2 + (y + 2)2 + z 2 = 9
Center (3, −2, 0), radius 3
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Math 111, Manack
Exam 1
2. Determine whether the points (0, 1, −1) (1, 3, 1),(2, 1, 1), (3, 1, 3) are coplanar (a collection of
points are coplanar if they are contained in a single plane).
Use 3 points (0, 1, −1) (1, 3, 1),(2, 1, 1) to calculate the plane.
(1, 3, 1) − (0, 1, −1) =< 1, 2, 2 > .
(2, 1, 1) − (0, 1, −1) =< 2, 0, 2 > .
< 1, 2, 2 > × < 2, 0, 2 >=< 4, 2, −4 >
Equation of Plane
4x + 2(y − 1) − 4(z + 1) = 0
Test 4th point (3, 1, 3):
4 ∗ 3 + 2(1 − 1) − 4(3 + 1) = −4 6= 0
So the 4 points are not coplanar.
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Math 111, Manack
Exam 1
3. Define the vector function r(t) by
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r(t) =< cos(t) sin(2t), sin(t) sin(2t), cos(2t) >
(a) Verify directly that this curve r(t) lies on the unit sphere for all t.
Notice that
q
|r(t)| = cos2 (t) sin2 (2t) + sin2 (t) sin2 (2t) + cos2 (2t)
q
q
√
2
2
2
2
= (cos (t) + sin (t)) sin (2t) + cos (2t) = sin2 (2t) + cos2 (2t) = 1 = 1
So r(t) is a unit vector for all t.
(b) Find the equation of the tangent line to the curve r(t) at the point <
t = π/4.
√
√
2/2, 2/2, 0 > .
r0 (t) =< 2 cos(t) cos(2t) − sin(t) sin(2t), 2 sin(t) cos(2t) + cos(t) sin(2t), −2 sin(2t) >
at t = π/4,
√
√
r0 (π/4) =< − 2/2, 2/2, −2 >
Equ of tan line
√
√
√
√
L(t) = r(π/4) + tr0 (π/4) =< 2/2, 2/2, 0 > +t < − 2/2, 2/2, −2 >
√
√
√
√
=< 2/2 − ( 2/2)t, 2/2 + ( 2/2)t, −2t >
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Math 111, Manack
Exam 1
4. Find every unit vector in R2 that is orthogonal to < 1/2, 5 >.
One solution of
< 1/2, 5 > · < x, y >= 0
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is x = 10, y = −1. So the two possible vectors are
√
1
1
< 10, −1 >= √
< 10, −1 >
101
101
and its negative
−1
√
< 10, −1 >
101
5. In R3 , consider the two vectors ~a =< −2, 3, −1 >, ~b =< −4, 1, 1 >. Find every unit vector in
R3 that is parallel to
2(~a − 3~b) + 5~b
Notice
2(~a − 3~b) + 5~b = 2~a − 6~b + 5~b = 2~a − ~b =< 0, 5, −3 >
So the two parallel unit vectors are
1
√ < 0, 5, −3 >
34
and
−1
√ < 0, 5, −3 >
34
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Math 111, Manack
Exam 1
6. Calculate the arclength traced out by the curve
√
2
2 3/2~
r(t) = t cos(t)~i + t sin(t)~j +
t k
3
√ 3/2 as it travels from the point (0, 0, 0) to the point −π, 0, 2 2π
.
3
(Yes, it works out nicely.)
Calculate the arclength from t = 0 to t = π:
√
r0 (t) =< cos(t) − t sin(t), sin(t) + t cos(t), 2t1/2 >
Z πq
√
A=
(cos(t) − t sin(t))2 + (sin(t) + t cos(t))2 + ( 2t1/2 )2 dt
0
π
Z
=
p
1 + t2 + 2t dt
0
π
Z
p
(1 + t)2 dt
=
0
Z
=
π
1 + t dt
0
= π 2 /2 + π
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