Mathematics 100-201
Page 1 of 8
Student-No.:
Midterm 2
Duration: 50 minutes
This test has 5 questions on 8 pages, for a total of 40 points.
• Read all the questions carefully before starting to work.
• Q1 and Q2 are short-answer questions; put your answer in the boxes provided.
• All other questions are long-answer; you should give complete arguments and explanations
for all your calculations; answers without justifications will not be marked.
• Continue on the back of the previous page if you run out of space.
• Attempt to answer all questions for partial credit.
• This is a closed-book examination. None of the following are allowed: documents,
cheat sheets or electronic devices of any kind (including calculators, cell phones, etc.)
Last Name:
First Name:
Student-No:
Section: 201
Signature:
Question:
1
2
3
4
5
Total
Points:
10
6
8
8
8
40
Score:
Student Conduct during Examinations
1. Each examination candidate must be prepared to produce, upon the
request of the invigilator or examiner, his or her UBCcard for identification.
2. Examination candidates are not permitted to ask questions of the
examiners or invigilators, except in cases of supposed errors or ambiguities in examination questions, illegible or missing material, or the
like.
3. No examination candidate shall be permitted to enter the examination
room after the expiration of one-half hour from the scheduled starting
time, or to leave during the first half hour of the examination. Should
the examination run forty-five (45) minutes or less, no examination
candidate shall be permitted to enter the examination room once the
examination has begun.
4. Examination candidates must conduct themselves honestly and in accordance with established rules for a given examination, which will
be articulated by the examiner or invigilator prior to the examination
commencing. Should dishonest behaviour be observed by the examiner(s) or invigilator(s), pleas of accident or forgetfulness shall not be
received.
5. Examination candidates suspected of any of the following, or any other
similar practices, may be immediately dismissed from the examination
by the examiner/invigilator, and may be subject to disciplinary action:
(i) speaking or communicating with other examination candidates,
unless otherwise authorized;
(ii) purposely exposing written papers to the view of other examination candidates or imaging devices;
(iii) purposely viewing the written papers of other examination candidates;
(iv) using or having visible at the place of writing any books, papers
or other memory aid devices other than those authorized by the
examiner(s); and,
(v) using or operating electronic devices including but not limited to telephones, calculators, computers, or similar devices
other than those authorized by the examiner(s)(electronic devices other than those authorized by the examiner(s) must be
completely powered down if present at the place of writing).
6. Examination candidates must not destroy or damage any examination
material, must hand in all examination papers, and must not take any
examination material from the examination room without permission
of the examiner or invigilator.
7. Notwithstanding the above, for any mode of examination that does
not fall into the traditional, paper-based method, examination candidates shall adhere to any special rules for conduct as established and
articulated by the examiner.
8. Examination candidates must follow any additional examination rules
or directions communicated by the examiner(s) or invigilator(s).
Mathematics 100-201
Page 2 of 8
Student-No.:
Short-Answer Questions. Question 1 is a short-answer question. Put your answer in the box
provided. Each part is worth 2 marks, but not all parts are of equal difficulty. Simplify your
answers as much as possible.
1. Complete justification is required for full marks
2 marks
(a) Let y = sin(cos(x2 )). Find
dy
.
dx
dy
Answer: dx
= −2x · sin(x2 ) ·
cos(cos(x2 ))
Solution: Use the chain rule twice
dy
d
= cos(cos(x2 )) ·
cos(x2 )
dx
dx
= cos(cos(x2 )) · (− sin(x2 )) ·
d 2
(x )
dx
= −2x sin(x2 ) · cos(cos(x2 ))
2 marks
(b) Let f (x) = arcsin (log(2x)). Find f 0 (x). Note: arcsin is inverse sine and log is the
natural logarithm.
Answer: f 0 = √ 1
2
x
Solution: Use the chain rule
1
d
f 0 (x) = p
log(2x)
1 − (log(2x))2 dx
1
= p
x 1 − (log(2x))2
1−(log(2x))
Mathematics 100-201
Page 3 of 8
2 marks
(c) Simplify arcsin cos
23π
16
Student-No.:
.
π
Answer: − 16
Solution:
cos
2 marks
(d) If y = x1/x find
dy
.
dx
23π
7π
= − cos
16
16
π 7π
−
= − sin
2
16
π
= − sin
16
π
= sin −
16
Your answer must be expressed in terms of x only.
Answer:
dy
dx
.
Solution: Take log of both sides:
y = x1/x
log x
log y =
x
now differentiate
x/x − 1 · log x
1 dy
=
y dx
x2
1 − log x
=
x2
dy
1 − log x
= x1/x ·
dx
x2
2 marks
(e) Use the quotient rule to find the derivative of cot(x).
Solution:
d
d cos x
cot x =
dx
dx sin x
− sin2 x − cos2 x
1
=
=− 2
2
sin x
sin x
= x1/x ·
1−log x
x2
Mathematics 100-201
4 marks
Page 4 of 8
2. (a) A radioactive sample decays according to the law
left after time t.
Student-No.:
dy
dt
= ky(t), where y(t) is the quantity
1. Suppose that one-quarter of the sample remains after 10 hours. Assuming we
started with 100 grams at time t = 0 express the quantity left as a function of
time.
2. What is the half-life?
Solution:
1. The general solution is given by y(t) = y0 ekt , y0 = y(0) = 100
We are given that y(10) = 1/4y0 . Comparing with y(10) = y0 e10k and cancelling y0 : e10k = 1/4 ⇒ 10k = ln(1/4) = − ln 4 ⇒ k = − ln104 .
ln 4
Thus, y(t) = 100e−t 10 grams.
2. t1/2 is such that y(t1/2 ) = 12 y0 = 50.
ln 4
ln 4
100e−t1/2 10 = 50 ⇒ e−t1/2 10 = 12 ⇒ −t1/2 ln104 = − ln 2 ⇒ t1/2 =
hours.
2 marks
10 ln 2
2 ln 2
=5
(b) A particle moves along the curve y 2 − x2 = 3. Its x-coordinate changes at a constant
rate of 1 unit per second. How fast is the distance from the origin changing when the
particle is at (1, 2)?
Answer:
Solution: The distance z satisfies
z 2 = x2 + y 2
√2
5
units per second
Mathematics 100-201
Page 5 of 8
but y 2 = 3 + x2 , so z 2 = 2x2 + 3. Hence
2z
Thus at (1, 2), z =
√
5 and
dx
dt
dx
dz
= 4x
dt
dt
= 1, so
√ dz
2 5
=4·1·1
dt
dz
2
=√
dt
5
Student-No.:
Mathematics 100-201
Page 6 of 8
Student-No.:
Full-Solution Problems. In questions 3–5, justify your answers and show all your work. If
a box is provided, write your final answer there. Unless otherwise indicated, simplification of
answers is not required in these questions.
8 marks
3. A large circular swimming pool with vertical sides is being filled from a fire hose at rate
of 5 cubic metres per minute. If the pool is 40 metres across, how fast is the water level
increasing when the pool is one third full? You can use that the volume of a circular cylinder
of height h is V = πr2 h.
Solution: You can think of the swimming pool as a cylinder with radius r and water
depth h, in which case the volume of water is
V = πr2 h
dV
dr
dh
= 2πr h + πr2
dt
dt
dt
and so
= 0. In other words r is constant
The diameter is given as 40 so r = 20 and of course dr
dt
so
dV
dh
= πr2
dt
dt
The rate of increase of the volume is dV
= 5. We can now solve for dh
as
dt
dt
5 = π(20)2
from which we deduce
dh
dt
=
1
;
80π
dh
dt
the water level is increasing at a rate of 1/(80π) m/min.
Note that the fact that the pool is 1/3 full is not used.
Mathematics 100-201
8 marks
Page 7 of 8
Student-No.:
4. A right circular cone is turned upside-down (vertex is at the bottom); the height of cone is
20 cm and the diameter of the top is 16 cm. The cone is filled half-way (by depth), and
water is draining out. The depth of the water is dropping at a rate of 2 cm per second. How
fast is the volume of the water changing at this time? You may use the fact that the volume
of a cone with radius r and height h is πr2 h/3.
Solution: Let h be the height of the water at time t and let r be the radius of water at
time t. We need to find dV /dt at a moment in time when h = 10 and dh/dt = −2.
The height of the cone is 20cm and the radius is 8cm. By similar triangles (use a picture)
we have that 20/8 = h/r and so r = 52 h. Now we use this to get a formula for V in
terms of only h. That is
1
1 2
4π 3
V = πr2 h = π( h)2 h =
h
3
3 5
75
Thus
12π 2 dh
4π 2 dh
dV
=
h ·
=
h ·
dt
75
dt
25
dt
Since we were given that h = 10 and dh/dt = −2, we plug in these numbers to evaluate
dV
at our moment in time: dV
= 4π
(10)2 · (−2) = −32π and the units are cm3 /s. Thus
dt
dt
25
the volume of water is decreasing at a rate of 32π cubic cm per second.
Mathematics 100-201
8 marks
Page 8 of 8
Student-No.:
5. You are sitting outside a fine English pub on a 20◦ C summer day. The publican brings you
a pint straight from the 5◦ C cellar. Just as your beer arrives, your phone goes off and it
distracts you for 1 minute. By that time your beer has warmed to 8◦ C. Just at that point
a “friend” interrupts you and it takes another 2 minutes to get them to leave.
What temperature is your beer when you finally get to drink it?
Assume that your drink is warming according to Newton’s law of cooling. Your answers
may be left in “calculator-ready” form. Be sure to define all variables and units used.
Solution:
• Let T be the temperature of the can in C and t be the time in minutes measured
from the arrival of the beer.
• Let y be the difference in temperature between the beer and the surroundings. So
y = 20 − T .
• Assuming Newton’s law of cooling we have
y(t) = y(0)e−kt
• Initially T (0) = 5, so y(0) = 15 and after 1 minute, T (1) = 8, so y(1) = 12.
y(1) = 12 = 15e−k
5
15
=
ek =
12
4
k = log(5/4)
• Thus after a total of 3 minutes
3 /53 )
y(3) = 15e−3k = 15e−3 log(5/4) = 15elog(4
3 × 43
3 × 64
192
=
=
=
52
25
25
and the temperature is then
T (3) = 20 − y(3) = 20 −
192
308
=
C.
25
25