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Math 142
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Math 142 - Practice Final 2 Solutions
1. Below is a unit circle with an unknown angle θ in standard position. Label the point A
with its cartesian and polar coordinates, and add two line segments to the picture that have
lengths sin θ and cos θ.
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2. Sketch a graph of cos x for x in [−π, π], and sketch a graph of cos−1 x for whatever domain
is appropriate. (Note that the axes are labeled differently on the two grids.)
Note the domain and range of cos−1 x.
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3. Calculate u · (v × w), where
u = h1, −1, 2i,
v = h−1, 3, 2i,
w = h−2, −2, −3i.
i
j
k 2 u · (v × w) = u · −1 3
−2 −2 −3
= u · h(3)(−3) − (2)(−2), (2)(−2) − (−1)(−3), (−1)(−2) − (3)(−2)i
= u · h−5, −7, 8i
= h1, −1, 2i · h−5, −7, 8i
= (1)(−5) + (−1)(−7) + (2)(8)
= 18.
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4. While will be worth more in 10 years, $10000 invested at 5% interest compounded continuously, or $11000 invested at 6% interest compounded quarterly (i.e. 4 times per year)?
The balance in the continuously compounded account after 10 years is
10000e0.05·10 = 16487.21.
The balance in the quarterly compounded account after 10 years is
11000(1 +
0.06 4·10
4 )
= 19954.20.
So it’s better to take the $11000 compounded quarterly.
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5. What are the possible angles γ that form triangles in the diagram below?
Call the angle at the bottom right α. Then the law of sines gives
sin 30◦
sin α
=
4
3
4
4 1
sin α = sin 30◦ = ·
3
3 2
2
=
3
2
α = sin−1 ≈ 41.8103◦ .
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This is the principal solution. There is also the symmetric solution α = 180 − 41.8103 =
138.1897. Since γ = 180 − 30 − α, we get two possibilities for γ,
γ = 180 − 30 − 41.8103 = 108.1897◦ ,
γ = 180 − 30 − 138.1897 = 11.103◦ .
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6. Find a cartesian equation (i.e. an equation in x and y) for the curve determined by the polar
equation
r = cos(θ + π4 ).
(Hint: remember the sum formula for cosine.) Can you identify what kind of curve this is
from its cartesian equation?
r = cos(θ + π4 )
r = cos θ cos π4 − sin θ sin π4
√
r=
r2 =
x2 + y 2 =
√
2
2
2 cos θ − 2 sin θ,
√
√
2
2
2 r cos θ − 2 r sin θ
√
√
2
2
2 x − 2 y.
now multipling by r,
This is the desired cartesian equation. To identify this, move everything to one side and
complete the square:
x2 + y 2 =
x2 −
√
+ y2 +
√
2
2 x
√
−
2
2 y =0
2
1
(x − 42 )2 − 18 + (y + sqrt2
4 ) − 8 =0
√
2
1
(x − 42 )2 + (y + sqrt2
4 ) = 4
√
2
1 2
(x − 42 )2 + (y + sqrt2
4 ) = (2) .
√
√
a circle with center ( 42 , − 42 ) and radius 21 . Note that
circle with center( 21 , 0) and radius 12 . If you rotate that
√
2
2 x
√
This is
gives a
shifting θ by π/4 (by adding π/4 to θ) rotates the curve!
2
2 y
the polar equation r = cos θ
circle, you get our circle. So
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7. Consider the sequence
−1.001, (−1.001)2 , (−1.001)3 , . . . .
(a) What is the sum of the first 100 terms of the sequence?
First note that this is a geometric sequence with a = r = −1.001. So the sum of the
first 100 terms is
a
1 − r100
1 − (−1.001)100
= (−1.001)
≈ 0.052584114.
1−r
1 − (−1.001)
(b) Does it make sense to add up all terms of this sequence? Why or why not? (You only
need write a sentence or two.)
No, it doesn’t make sense. Since |r| > 1, if we try to imagine what happens to
a
1 − rn
1−r
as n goes to infinity, we don’t zero in on a particular value. The value keeps jumping
around as n increases.
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8. Calculate
50
7π
7π
cos
+ i sin
.
100
100
Recall that when you express complex numbers in polar form (as we have here), complex
multiplication looks like adding angles. So if we raise a complex number to the 50th power,
the complex number’s angle is added to itself 50 times. That is, the angle is multiplied by
50. We get
cos
7π
100
+ i sin
7π
100
50
7π
7π
= cos
· 50 + i sin
· 50
100
100
7π
7π
= cos
+ i sin
2
2
= 0 + i(−1)
= −i.