Mathematics 104-184 Midterm Exam 2A
Midterm Exam 2 — November 17
Duration: 1 hour
This test has 7 questions on 10 pages, for a total of 50 points.
• Q1-Q5 are short-answer questions[3 pts each part]; put your answer in the boxes provided.
• Q6 and Q7 are long-answer and each worth 10 pts; you should give complete arguments and
explanations for all your calculations; answers without justifications will not be marked.
• Use the two blank pages is you need extra space, but you must leave a clear note that this
has been done.
• 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.)
• Please circle your course and section:
MATH 104
MATH 184
101 102 103 104 105 106 108
109
101 102 103 104 105 106 I don’t know.
• PRINT your name and ID # very clearly. Failure to do so may result in a grade of 0:
First Name:
Last Name:
Student-No:
Signature:
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Mathematics 104-184 Midterm Exam 2A
Short-Answer Questions. Put your answer in the box provided. Full marks will be given
for a correct answer placed in the box, while part marks may be given for work shown. Unless
otherwise stated, calculator ready answers are acceptable.
1. (a) A go-kart’s position (in metres) along a straight race track is given by
p(t) =
et
− 1,
t+1
where the units of time are seconds. How fast is the go-cart travelling at t = 3?
Answer: 3e3 /16 m/s
Solution:
p(t) =
v(t) =
=
v(3) =
et
−1
t+1
et (t + 1) − et
(t + 1)2
tet
(t + 1)2
3e3
1
16
1
(b) The cost of producing q toasters is given by C = 6000 + ln
to estmate the cost of producing the 13
th
q2
q+3
. Use marginal cost
toaster.
Answer: 1/10 or 0.1
Marking scheme: 1pt for
Solution:
dC
;
dq
1pt for using it to approximate the MUC
q2
C = 6000 + ln
q+3
= 6000 + 2 ln q − ln(q + 3)
dC
2
1
= −
dq
q q+3
Therefore, M U C(12) ≈ M C(12) =
Solution:
1
6
−
1
15
=
9
90
q2
C = 6000 + ln
q+3
dC
q+3
2q(q + 3) − q 2
=
·
dq
q2
(q + 3)2
2
1
q + 6q
=
·
q2
q+3
q+6
=
q(q + 3)
Therefore, M U C(12) ≈ M C(12) =
18
12·15
=
3·6
2·6·3·5
=
1
10
Mathematics 104-184 Midterm Exam 2A
2. (a) Below is a graph of
df
.
dx
f 0 (x)
2
4
6
8
x
10
i. State the intervals on which is the graph of y = f (x) concave up?
2
Answer: (2, 4), (7, ∞) ii. Write all the x−values where a local minimum occurs.
Answer: 2 and 7(local mins of f’), or 8(local min of f)
1
3
(b) Find the absolute minimum value of the function f (x) = ex −x in the interval [0, 2].
√
1
Answer: f √
= e−2/3 3
3
Solution: Marking scheme: 1pt for correct derivative; 1pt for testing the critical
value and both endpoints
f (x) = ex
3 −x
f 0 (x) = ex
3 −x
· (3x2 − 1)
1
1
From which we see the critical values occur when 3x2 − 1 = 0 =⇒ x = ± √ .
3
We now evaluate the function at the critical points (in the interval) and the two
1
end points of the interval where function is defined:
f (0) = e0 = 1
√
1
f √
= e−2/3 3 < 1
3
f (2) = e6 > 1
√
Thus, the absolute minimum f √13 = e−2/3 3
Mathematics 104-184 Midterm Exam 2A
3. (a) The function
y=x
3√x
is defined for all x > 0. Determine the x-values for which it has a horizontal tangent
line.
Answer: x = e−3
Solution:
y = x(x
1/3 )
ln y = ln x(x
1/3 )
1
ln y = x1/3 · ln x
y0
1
x1/3
= x−2/3 · ln x +
y
3
x
ln x + 3
y0 = y ·
3x2/3
1
From which we see that y 0 = 0 ⇐⇒ x = e−3 .
2x2
4
+
is defined on the interval (2, ∞), and has a critical
x−2
3
number c = 3. Use the second derivative to determine if this is a local minimum or
local maximum. Hint: is the function concave up or down at this point?
Answer: Local minimum; concave up.
(b) The function g(x) =
Solution:
4
2x2
+
x−2
3
4
4x
1
g 0 (x) = −
+
(x − 2)2
3
8
4
1
g 00 (x) =
+
3
(x − 2)
3
4
g 00 3x) = 8 + > 0
3
g(x) =
The function is concave up which means that it must be a local minimum.
Mathematics 104-184 Midterm Exam 2A
4. (a) For a community theatre, the typical nightly attendance q is related to the ticket price p
dq p
· q,
in this equation: (p + 10)(q + 20) = p2 q. Find the price elasticity of demand, E = dp
when p = 10 and q = 5.
Answer: -15/8
dq
Solution: From the definition of elasticity E = pq · dp
, we see that we need the
0
derivative q (p) = dq/dp. Implicit-differentiating the given equation with respect to
p, we have
dq
dq
1
= 2pq + p2 ·
(q + 20) + (p + 10) ·
dp
dp
Subbing in p = 10 and q = 5 gives us:
dq
dq
= 100 + 100 ·
dp
dp
dq
−75 = 80
dp
−75
dq
=
dp
80
1 −75 10
dq p −15
E=
· =
·
=
dp q
80
5
8
(25) + (20) ·
√
(b) Compute the limit lim
x→−∞
4x6 − 2x4 + 5x2 − 1
.
5x3 − x2 + 3x + 25
Answer: -2/5
3
Solution:
√ Marking scheme: 1pt for dividing top and bottom by x . 1pt for
x = − x2 when x < 0.
q
√
3
2|x
|
1 − 12 x−2 + 45 x−4 − 14 x−6
6
4
2
4x − 2x + 5x − 1
= lim
.
lim
x→−∞
x→−∞ 5x3 − x2 + 3x + 25
x3 (5 − x−1 + 3x−2 + 25x−3 )
Since x → −∞, x will be negative and |x3 |/x3 = −1. Thus the limit is −2/5.
Solution:
√
lim
x→−∞
√
1
4x6 − 2x4 + 5x2 − 1 4x6 − 2x4 + 5x2 − 1
=
lim
x→∞ −5x3 − x2 − 3x + 25
5x3 − x2 + 3x + 25
q
4
2
4x6
− 2x
+ 5x
− x16
1
x6
x6
x6
= lim −5x3 x2 3x 25
x→∞
− x3 − x3 + x3
3
qx
4 − x24 + x54 − x16
= lim
x→∞ −5 − 1 − 32 + 253
x
x
x
√
4−0+0−0
=
−5 − 0 − 0 + 0
= −2/5
Mathematics 104-184 Midterm Exam 2A
5. (a) The price (in dollars) p and the quantity demanded q for a new transistor radio are
related by the equation: qe2p = 3. If the price is decreasing at a rate of 3 dollars per
week when the price is 10 dollars, how fast is the the quantity demanded changing ?
18
Answer: 20
e
Solution: Marking scheme:
Differentiating q = 3e−2p with respect to time
gives us:
dq
dp
−2p
1
= 3e
· −2
.
dt
dt
Using
dp
dt
1 we get
= −3 and p = 10 ,
dq 1
= 3e−2(10) (−2)(−3) = 18e−20 .
dt p=10
(b) Let f (x) = x5 + x. Find the equation of the tangent line of y = f −1 (x) at x = 2. Recall
that f −1 (x) is the inverse function of f (x) and its derivative can be calculated using
df −1
1
formula
= 0 −1
.
dx
f (f (x))
Answer: y = 1 + 16 (x − 2)
1 Next we’ll need
Solution: First we need the point: f (1) = 2 =⇒ f −1 (2) = 1 .
the slope:
y0 =
1
f 0 (f −1 (2))
=
1
f 0 (1)
1
So, we need f 0 (x) = 5x4 + 1 =⇒ f 0 (1) = 6 =⇒ m = y 0 (2) = 61 .
Mathematics 104-184 Midterm Exam 2A
Full-solution problems - 10 pts each: Justify your answers and show all your work
for problems 6 and 7. Place a box around your final answer. Unless otherwise indicated,
simplification of answers is required in these questions.
6. Consider the function f (x) =
x3
, which has derivatives
ex
x2 (3 − x)
f (x) =
ex
0
and
x(x2 − 6x + 6)
f (x) =
.
ex
00
(a) Find the intervals on which f (x) is increasing or decreasing and classify the local extreme values.
Solution: Marking scheme: 1pt for critical numbers; 1pt for correct +/- of f 0 (x);
1pt for classifying the local max.
Since ex > 0 for all x 6= 0,
Therefore, we have:
f (x) is increasing ⇔ f 0 (x) > 0 ⇔ x < 3 or (−∞, 3)
f (x) is decreasing ⇔ f 0 (x) < 0 ⇔ x > 3 or (3, ∞)
Note that f 0 (x) = 0 at x = 0 and x = 3. These points should have a horizontal
tangent line. There’s a local max at x = 3, and a hump at x = 0.
(b) Determine where f (x) is concave up or down.
Solution: Marking scheme: 1pt for trying to get roots with the quadratic equation; 1pt for correct +/- of f 00 (x); 1pt for interpreting concave & down properly.
Since ex > 0 for all x 6= 0, the sign of x(x2 − 6x + 6) will determine the sign of f 00 (x).
We to find the roots of that parabola; it looks like we’ll need to use the quadratic
equation:
√
√
√
√
√
−b ± b2 − 4ac
6 ± 36 − 24
6 ± 12
6±4 3
x=
=
=
=
= 3 ± 3.
2a
2
2
2
√
3), (3 + 3, ∞)
√
√
f (x) is concave down ⇔ f 00 (x) < 0 ⇔ x is in the intervals (−∞, 0), (3− 3, 3+ 3)
√
There will be an inflection point when x = 0, 3 ± 3.
f (x) is concave up ⇔ f 00 (x) > 0 ⇔ x is in the intervals (0, 3 −
√
Mathematics 104-184 Midterm Exam 2A
(c) Sketch the graph of y = f (x).
Solution:
Marking scheme: 1pt for the max at x=3; 1pt for pts of inflection; 1pt for for
the horizontal asyptote (note that for x > 3 the function is decreasing and positive;
1pt for x < 0
Mathematics 104-184 Midterm Exam 2A
7. At a certain instant an aircraft is flying due east at 400 km/hr and passes directly overhead
a car traveling due southeast (so, it’s travelling 45◦ between south and east) at 100 km/hr
on a straight, level road. If the aircraft is traveling at an altitude of 1 km, how fast is the
distance between the aircraft and the car increasing 36 seconds after the airplane passes
directly over the car?
Hint: Draw a very careful diagram and understand the problem, you may need to use 2
triangles and the Cosine Law: For a triangle with sides A, B, and C and the angle c
opposite the side C, one has C 2 = A2 + B 2 − 2AB cos(c).
Solution: Let’s begin with a diagram! We’ll let x(t) be the distance the plane has
travelled and y(t) be the distance that the car has travelled. C will be the initial point
.
at time and s(t) the distance between the plane and the car. We’re looking for ds
dt
2 for diagram and defined notation 1 defined notation
Using the Law of Cosines we can the lenth of the bottom edge of the red triangle to be:
s
q
p
√
1
1
x2 + y 2 − 2xy cos 45◦ = x2 + y 2 − 2xy √
= x2 + y 2 − 2xy.
2
Next, we can use the Pythagorean Theorem to get the distance between the car and the
plane.
√
1
s2 = 1 + x2 + y 2 − 2xy, Differentiating with respect to t gives
2s
√ dy
ds
dx
dy √ dx
2
= 2x + 2y − 2 y − 2x .
dt
dt
dt
dt
dt
1
We have dx/dt = 400 and dy/dt = 100. Also, when
√ t = 1/100√(36 seconds in hours),
2
1 Solving for ds/dt
we have x = 4, y = 1, and s = 1 + 16 + 1 − 4 2 = 18 − 4 2 .
above gives
√
ds
1
= (2(4)(400) + 2(1)(100) − 2 · [400(1) + 4(100)]
dt
2s
√
1
= (3400 − 800 2)
2s
√
1700 − 400 2
1
= p
√ km/h .
18 − 4 2
Mathematics 104-184 Midterm Exam 2A
Solution: Let’s begin with a diagram! We’ll let x(t) be the distance the plane has
travelled and y(t) be the distance that the car has travelled. C will be the initial point
at time and s(t) the distance between the plane and the car. We’re looking for ds
. The
dt
black triangle is on the ground level and the red triangle in in the air.
2 for diagram and defined notation 1 defined notation
We can use the Pythagorean Theorem to get the distance between the car and the plane.
1
s2 = 1 + D 2 , Differentiating with respect to t gives:
2s
ds
dD
ds
D dD
1
= 2D
=⇒
=
·
.
dt
dt
dt
s dt
So we need to figure out what s, D, and
dD
dt
are equal to after 1 min.
Using the Law of Cosines we can the lenth of D via:
D2 = x2 + y 2 − 2xy cos 45◦ =⇒ D2
1
2
2
= x + y − 2xy √
2
√
2
2
1
= x + y − 2xy.
Differentiating with respect to t gives
2D
√ dy
dD
dx
dy √ dx
1
= 2x
+ 2y − 2 y − 2x .
dt
dt
dt
dt
dt
1
We have dx/dt = 400 and dy/dt = 100. Also, when
√ t = 1/100√(36 seconds in hours),
1 Solving for ds/dt
we have x = 4, y = 1, and s2 = 1 + 16 + 1 − 4 2 = 18 − 4 2 .
above gives
√
ds
1
= (2(4)(400) + 2(1)(100) − 2 · [400(1) + 4(100)]
dt
2s
√
1
= (3400 − 800 2)
2s
√
1700 − 400 2
1
= p
√ km/h .
18 − 4 2