MAT 1318X Midterm 2
Summer 2009
August 7
Instructor: Charles Starling
Question Response Points
Family Name:
First Name:
Student Number:
1
E
1
2
D
1
3
D
1
4
A
1
5
F
1
6
×
5
7
×
5
8
×
5
Total
–
20
PLEASE READ THESE INSTRUCTIONS VERY CAREFULLY.
1. You have 90 minutes to complete this exam.
2. This is a closed book exam, and no notes of any kind are allowed. Do not use your
own scrap paper! Use the last page or the backs of pages for rough work.
3. Only use of approved calculators is permitted.
4. Questions 1 through 5 are multiple choice. They are worth 1 point each and no part
marks will be given. Please record your answers in the space provided above.
5. Questions 6, 7 and 8 require a complete solution, and are worth 5 points each, so spend
your time accordingly. The correct answer requires justifcation written legibly and
logically: you must convince me that you know why your solution is correct.
You must answer these questions in the space provided. Use the backs of pages if
necessary.
6. Where it is possible to check your work, do so.
7. Good luck! Bonne chance!
1. Suppose sin t = 35 and the terminal point of t is in the second quadrant. What are
cos t and tan t?
A. cos t = 54 and tan t = 43
D. cos t = − 34 and tan t = − 34
B. cos t =
3
5
and tan t =
3
4
C. cos t = − 54 and tan t =
E. cos t = − 54 and tan t = − 34
3
4
2
F. None of the above.
solution: We know that sin t + cos t = 1, so
r
r
p
16
4
9
2
cos t = ± 1 − sin t = ± 1 −
=±
=±
25
25
5
2
Since t is in the second quadrant, cos is negative, so cos t = − 45 . Thus
tan t =
3/5
3
sin t
=
=−
cos t
−4/5
4
2. Let t ∈ R. Which of the following statements is true?
(a) t and t − 10π have the same terminal point. T
√
(b) The point P ( −3 5 , 23 ) is on the unit circle. T
(c) The terminal point of
3π
2
is P (0, −1). T
(d) If a, b, x > 0, then logb (x) =
loga (x)
.
loga (b)
T
(e) There is a t for which sin(t) = 10 . F
(f) cos(t) = 1 if t = 2kπ with k ∈ Z. T
A. (a), (b), (c), (d), (e), (f).
D. (a), (b), (c), (d), (f).
B. (b) and (d).
E. (b), (e), (c) and (d).
C. (a) and (c).
solution:
F. None of A-E are correct.
(a) T. Adding or subtracting a multiple of 2π doesn’t change the terminal point.
(b) T.
(c) T.
√ !2 2
− 5
2
5+4
9
+
=
= =1
3
3
9
9
3π
2
is three quarters around the unit circle.
(d) T. This is just the change of base formula.
(e) F. The maximum that sin gets is 1.
(f) T. When t = 2kπ, the terminal point is (1, 0).
3. Solve the following equation
e2x − 2ex − 8 = 0
A. x = 4.
D. x = ln(4).
B. x = e.
E. x = 2.
C. x = ln(2).
solution: This is a quadratic in ex :
F. None of the above.
e2x − 2ex − 8 = 0
(ex − 4)(ex + 2) = 0
Thus either ex − 4 = 0 or ex + 2 = 0. The second one is impossible, so we solve
ex − 4 = 0:
ex − 4 = 0
ex = 4
x = ln 4
4. Let
a)f (x) = ln(|1 − x| − 2)
b)g(x) =
1
1 − ex
Find the domains of f and g.
A. a)Df = (−∞, −1) ∪ (3, ∞)
b)Dg = {x | x 6= 0}.
B. a)Df = (−∞, −1) ∩ (3, ∞)
b)Dg = [3, ∞).
D. a)Df = (−∞, −1) ∪ (3, ∞)
b)Dg = [0, 1).
E. a)Df = (−∞, −1)
b)Dg = {x | x 6= 0}.
C. a)Df = (−∞, −1)
F. None of the above.
b)Dg = [−1, 1) ∩ [3, ∞).
solution: We can’t take ln of a negative, so we need
|1 − x| − 2 > 0
|1 − x| > 2
1 − x > 2 or 1 − x < −2
−1 > x or 3 < x
This means Df = (−∞, −1) ∪ (3, ∞). For g, we just need to exclude points that make
the denominator 0. Thus we solve
1 − ex = 0
ex = 1
x = ln 1 = 0
Hence we just need x 6= 0 and so Dg = {x | x 6= 0}
5. What is an equation for the graph given below?
A. y = 3 sin(2x) + 1
D. y = 3 sin(3x)
B. y = 3 cos(2x) − 1
E. y = 2 sin(2x) + 1
C. y = 2 cos(3x) + 1
F. y = 2 sin(3x) + 1
The amplitude is the distance from equilibrium to maximum. Here equilibrium is 1
and the max is 3, so the amplitude is 2.
The period is
2π
,
3
so the number multiplying the x must be
B=
2π
2π
=
=3
period
2π/3
We are shifted up by 1, so we need to add 1 to the end. We start at equilibrium so
the graph must be sin. Putting this all together we get
y = 2 sin(3x) + 1
6. You invest $ 1000 in a bank at 3% interest. How long does it take your investment to
double if interest is compounded
(a) 3 times a year?
solution: The formula is
r nt
A(t) = P 1 +
n
For us P = 1000, A(t) = 2(1000) = 2000 and r = 0.03:
0.03
2000 = 1000 1 +
n
nt
Compounded three times a year gives us n = 3, so we solve
2000 = 1000 (1.01)3t
2 = (1.01)3t
ln 2 = 3t ln(1.01)
ln 2
t =
≈ 23.2
3 ln(1.01)
(b) Continuously?
solution: Here we use
A(t) = P ert
Which becomes
2000 = 1000e0.03t
2 = e0.03t
ln 2 = 0.03t
ln 2
≈ 23.1
t =
0.03
7. (a) A mass on a spring undergoes simple harmonic motion with displacement
y = 10 cos(4πt)
(measured in meters) at time t (measured in seconds). Find the amplitude, period,
and frequency of the motion.
solution: We can read off the amplitude as 10m.
period =
frequency =
1
2π
= s
4π
2
1
1
=
= 2Hz
period
1/2
(b) Sketch the graphs of the following functions (first find their period and, if
applicable, amplitude):
i. f (t) = 5 sin(2t) + 1
solution: Here the amplitude is 5 and the period is
are shifted up by 1. Hence the graph is:
ii. g(x) = 2 tan(3x − π2 )
solution: The period of tan graphs is
2π
2
= π. The +1 means we
π
k
= π3 . Rewriting our equation,
π g(x) = 2 tan 3 x −
6
The graph of 2 tan 3x would have asymptotes at π6 + n π3 . Thus when we shift
we will have an asymptote at the origin and at every multiple of π3 . Thus the
graph is
8. Suppose we have 20kg of a radioactive substance with a half-life of 5 years.
(a) How much of this substance do we have after 10 years?
solution: A half-life is the time it takes for half of the quantity to disappear.
Thus, after 5 years only 10kg will be left. 5 years after that, half of the 10kg will
disappear, leaving us with 5kg.
We could also do this with the radioactive decay formula
m(t) = m0 e−kt
Half-life h is related to k by the formula
ln 2
ln 2
k=
=
h
5
So
ln 2t
m(t) = 20e− 5
After 10 years,
20
=5
4
(b) You can move this substance safely when the quantity of the radioactive material
is 0.1 kg or less. How long does it take before we can move the object in question?
m(10) = 20e−
ln 2·10
5
−2 )
= 20e−2 ln 2 = 20eln(2
= 20(2−2 ) =
solution: We let m(t) = 0.1 and solve for t
0.1 = 20e−
ln 2t
5
ln 2t
0.005 = e− 5
ln 2t
ln(0.005) = −
5
5 ln 0.005
t = −
≈ 38.2
ln 2