Full Name:
Student Number:
Math 103 Midterm Test 1
Feb. 10, 2006
50 min.
1. Ensure that your full name and student number appear on this page.
2. No calculators, books, or notes, or electronic devices of any kind are permitted.
3. Unless otherwise indicated, show all your work. Answers not supported by calculations or
reasoning may not receive credit. Correct answers that do not demonstrate an appropriate
understanding of the course material might not receive credit. Messy work will not be graded.
Read each question carefully to be sure you are answering the question being asked.
4. Exposing your test paper, copying from another student’s paper, or sharing information about
this test constitutes academic dishonesty. Such behaviour may jeopardize your grade on this
test, in this course, and your standing at this university.
5. Five minutes before the end of the test period you will be given a verbal notice. After that
time, you must remain seated until all test papers have been collected.
6. When the test period is over, you will be instructed to put away writing implements. Put
away all pens and pencils at this point. Continuing to write past this instruction will be
considered as cheating.
7. Please remain seated and pass your test paper down the row to the nearest indicated aisle.
Once all the test papers have been collected, you are free to leave.
Problem #
1
Grade
Value
10
2
6
3
6
4
8
Total
30
I have read and understood the instructions and agree to abide by them.
Signed:
Summation Formulae
N
X
C = NC
k=1
N
X
N
X
k = N (N + 1)/2
k=1
N
X
k=1
k 3 = (N (N + 1)/2)2
k 2 = N (N + 1)(2N + 1)/6
k=1
N
X
k=0
rk = (1 − rN +1 )/(1 − r)
1. Calculate each of the following.
15
X
(a)
(b)
2k .
1−216
=
1−2
e−2x dx.
Solution (2 points)
Solution(2 points)
k
k=3 2
2
0
k=3
P15
Z
=
−
P15
k=0 2
1−23
1−2
k
−
P2
k=0 2
R2
k
0
= 216 − 8 = 65528
e−2x dx =
2
1 −2x e
−2
0
= 12 (1 − e−4 )
(c) Find the area under the curve f (x) = x2 between x = −1 and x = 2.
Solution (2 points)
2
1 3 Area =
x dx = x = 8/3 − (−1/3) = 9/3 = 3
3 −1
−1
Z
2
2
(d) The area between the graphs y1 = 2 − x2 and y2 = |x|.
Solution (2 points)
These two curves cross when 2 − x2 = |x|. Because both curves are symmetric about the
y-axis, we need only find the intersection point with positive x value so solve 2 − x2 = x.
This equation has only one positive solution, x = 1 so y1 and y2 cross at x = ±1. Note
that y1 is always higher than y2 between x = −1 and x = 1 (although we do not actually
need to know this) so the area between them is given by
Z
1
Z
(y1 − y2 )dx = 2
−1
0
1
x2 x3
(2 − x − x )dx = 2 2x −
−
2
3
2
1
= 2(2 − 1/2 − 1/3) = 7/3
0
Had we not known which curve was higher and used y2 − y1 , this would have come out
as −7/3 and we would simply drop the minus sign so as to give a positive value for area.
P
π
(e) An approximation for the sum 40
k=1 sin 80 k . Is your approximation higher or lower
than the actual value?
Solution (2 points)
Note that
Z
b
f (x)dx ≈
a
N
X
f (xk )∆x
k=1
b−a
N
where ∆x =
and xk = a + k∆x for the right-corner sum or xk = a + (k − 1)∆x for
P
π
the left-corner sum. We must interpret 40
k=1 sin 80 k as an example of the right hand
side of this general formula. In this particular case, notice that there is a k but no k − 1
2
appearing so we are in the right-corner case. Also, based on the upper limit of the sum,
N = 40. Moreover, ∆x = 1 as there
is no apparent constant term multiplying the sum.
π
Next, we interpret f (x) = sin 80
x which means xk = k and so the lower limit is a = 0
and the upper limit is b = x40 = 40. Note that because we are in a right-corner case, the
lower limit is not a = x1 = 1 because 1 is the right edge of the first rectangle (although
using a = 1 simply omits one rectangle from the sum which should not introduce a
serious error for N large enough - no points were deducted for this error). Finally, we
have an integral approximation:
40
X
k=1
π π Z 40
π 40 80
80
sin
sin
k ≈
x dx = −
cos
x =
80
80
π
80 0
π
0
π
Because the function sin 80
x is increasing between 0 and 40 and the sum is a rightcorner sum, the integral approximation lies entirely below the rectangles. Thus, this
approximation is lower than the actual value of the sum.
Note that there are several variations that give the same or almost the same approximation.
See Problem 3.6 for a similar example.
3
2. A cone-shaped coffee filter with a height of 10 cm and an opening radius of 10 cm is used to
make coffee. While in use, the filter is full of water and the density of coffee grounds is given
1
by ρ(y) = − 10
y where y = 0 corresponds to the top of the filter. y is measured in cm and ρ
is measured in grams/cm3 .
10cm
10cm
(a) Use the method of disks to find the volume of water in the filter.
Solution (4 points)
When sliced into horizontal disks, a disk at height y has a radius given by 10 + y.
Therefore, the volume is
Z
0
2
V =
Z
0
π(10 + y) dy = π
−10
2
(100 + 20y + y )dy = π
−10
1 3 0
100y + 10y + y 3
−10
2
1
1000
π cm3 .
= π 103 − 103 + 103 =
3
3
(b) Calculate the total mass of coffee grounds in the filter.
Solution (2 points)
The total mass of coffee grounds is found by adding up the mass in a thin disk for each
1
value of y between -10 and 0. Each thin disk has ρ(y) · Vdisk (y) cells where ρ(y) = − 10
is the density in each thin disk and Vdisk (y) = π(10 + y)2 dy is the volume of each thin
disk. Therefore, the total mass is
Z 0
Z 0
1
1
2
M=
π −
(10 + y) dy = −π
(10y + 2y 2 + y 3 )dy
10
10
−10
−10
= −π
2 3
1 4 0
2 1
250
3 1
5y + y + y = π10
− +
=
π grams
3
40
2 3 4
3
−10
2
4
3. A sink can hold a maximum of 20 liters before it overflows. Suppose the sink is initially
empty. The inflow rate from the tap is given by T (t) = A(1 + e−3t ) and the outflow through
the drain is D(t) = A(1 − e−4t ). T (t) and D(t) are both measured in liters per minute and t
is measured in minutes.
(a) How many liters of water are in the sink at time t?
Solution (4 points)
The total change in water volume in the sink is found by integrating its net rate of
change from 0 to t:
Z t
Z t
V (t) = V (t) − V (0) =
(T (s) − D(s))ds =
A(e−3s + e−4s )ds
0
=A
0
1
1 −3s
1 −4s t
1
−3t
−4t
e
+
e
+
1−e
=A 3 1−e
−3
−4
4
0
(b) What is the maximum value of A for which the sink does not eventually overflow?
Solution (2 points)
Notice that the volume of water in the sink V (t) starts at zero and steadily rises to an
asymptotic value of 7A/12. The maximum of A such that the sink does not overflowis
determined by 7A/12 = 20 or equivalently,
Amax =
5
240
.
7
4. Below are graphs of the birth rate and death rate of a particular species of wild hare in the
arctic over a period of ten years starting in 1995 (x = 0). By examining the graphs, answer
the following questions.
b(t)
20
10
0
d(t)
2
4
6
8
10
t
(a) Determine the years in which the population size was increasing and in which years the
population size was decreasing.
Solution (2 points)
From t = 0 to t = 3 (1995 through 1997 inclusive), the birth rate was higher than
the death rate so the population increased. From t = 3 to t = 5 (1998 and 1999) the
population decreased. From t = 5 to t = 8 (2000 through 2002 inclusive) the population
increased. From t = 8 to t = 10 (2003 and 2004) the population decreased.
(b) At the start of which year was the population of hares largest and when was it smallest?
Briefly explain your answer.
Solution (2 points)
The population will be largest at the end of a period of increase so either at t = 3 or
t = 8. Between those two values of t, the integral of b(t) − d(t) (i.e. the change in
population) gains more than it loses (the area between 3 and 5 is smaller than the area
between 5 and 8). This means that the population was largest at t = 8 so at the start
of 2003. By similar reasoning, it was smallest at t = 0 so at the start of 1995.
6