UNIVERSITY OF MANITOBA
DEPARTMENT OF MATHEMATICS
MATH 1510A02 Applied Calculus I
SECOND TERM EXAMINATION
November 30, 2016 5:30 pm
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Question Points
1
15
2
5
3
4
4
5
5
10
6
8
Total:
47
Score
INSTRUCTIONS TO STUDENTS:
Fill in clearly all the information above.
The duration of this exam is 75 minutes.
No calculators, texts, notes, cellphones or other aids are permitted.
Show your work clearly for full marks.
This exam has a title page, 5 pages of questions and 1 blank page at the end for rough work.
Please check that you have all pages.
The value of each question is indicated in the left-hand margin beside the statement of the
question. The total value of all questions is 47.
Answer all questions on the exam paper in the space provided. If you need more room, you
may continue your work on the reverse side of the page, but clearly indicate that your
work is continued there.
UNIVERSITY OF MANITOBA
SECOND TERM EXAMINATION
DATE: November 30, 2016
PAGE: 1 of 6
DEPARTMENT & COURSE NO: MATH 1510
TIME: 75 minutess
EXAMINATION: Applied Calculus I
EXAMINER: A. Prymak
1. Find the derivative of each of the following functions. You do not need to simplify your
answer.
[4]
(a) y =
p
√
3
x
5
Solution:
√
1 √ −2
1
y 0 (x) = (5 x ) 3 ln 5 · 5 x · √ .
3
2 x
1√
Alternatively, y = 5 3
x
, and then
1√
y 0 (x) = 5 3
[6]
(b) y = tan2 x · (sec x)x ,
0<x<
x
ln 5 ·
1
1
· √ .
3 2 x
π
2
Solution:
Both tan x and sec x are positive on (0, π/2), we can use logarithmic differentiation.
ln y = 2 ln(tan x) + x ln(sec x)
2
1
y0
=
sec2 x + ln(sec x) + x
sec x tan x
y
tan x
sec x
2 sec2 x
0
2
x
y = tan x · (sec x)
+ ln(sec x) + x tan x
tan x
[5]
(c) y = y(x) given implicitly by
ln(y − x) = xy 2 − 4
Solution:
Assuming y = y(x), we use implicit differentiation to obtain
y0 − 1
= 1 · y 2 + x · 2yy 0
y−x
y 0 − 1 = y 3 − xy 2 + 2xy 2 y 0 − 2x2 yy 0
y0 =
1 + y 3 − xy 2
.
1 − 2xy 2 + 2x2 y
UNIVERSITY OF MANITOBA
SECOND TERM EXAMINATION
DATE: November 30, 2016
PAGE: 2 of 6
DEPARTMENT & COURSE NO: MATH 1510
TIME: 75 minutess
EXAMINATION: Applied Calculus I
EXAMINER: A. Prymak
cot 4x
sin θ
. You can use that lim
= 1.
x→0 cot 3x
θ→0 θ
[5] 2. Evaluate lim
Solution:
We first note that
cot 4x
cos 4x sin 3x
=
·
cot 3x
sin 4x cos 3x
3 cos 4x
4x
sin 3x
= ·
·
·
.
4 cos 3x sin 4x
3x
θ
sin θ
= lim
= 1. Therefore
θ→0 sin θ
θ→0 θ
We have lim
cot 4x
3 cos 4x
4x
sin 3x
= lim ·
·
·
x→0 cot 3x
x→0 4 cos 3x sin 4x
3x
lim
=
cos 4x
4x
sin 3x
3
· lim
· lim
· lim
4 x→0 cos 3x x→0 sin 4x x→0 3x
=
3 cos 0
·
· (1) · (1)
4 cos 0
3
= .
4
UNIVERSITY OF MANITOBA
SECOND TERM EXAMINATION
DATE: November 30, 2016
PAGE: 3 of 6
DEPARTMENT & COURSE NO: MATH 1510
TIME: 75 minutess
EXAMINATION: Applied Calculus I
EXAMINER: A. Prymak
[4] 3. Let f (x) =
1
. Show that there is no value of c such that
3x3
f (1) − f (−1) = f 0 (c)(1 − (−1)).
Does this contradict the Mean Value Theorem? Explain why.
Solution:
1
−1
2
As f 0 (x) = −x−4 , f (1) = , f (−1) =
, the condition becomes = −c−4 · 2, or
3
3
3
c4 = −3, which has no solutions. This does not contradict the MVT. One cannot
apply the MVT to f on [−1, 1] as f is undefined (and, hence, not continuous) at
x = 0.
π π
x2
[5] 4. Are the curves y = cos x and y + =
perpendicular at the point
,0 ?
4
π
2
Justify your answer.
Solution:
The curves are perpendicular at a point if their tangent lines at this point are perpendicular, which means that the product of the slopes has to be equal to −1. So
π
we need to compute the derivatives at x = . For the first function: y 0 (x) = − sin x,
2
x2
π
2x
0 π
− , so y 0 (x) =
, and
y ( 2 ) = −1. The second curve is expressed as y =
π
4
π
y 0 ( π2 ) = 1. As the product of the slopes found is −1, the curves are perpendicular
at the given point.
UNIVERSITY OF MANITOBA
SECOND TERM EXAMINATION
DATE: November 30, 2016
PAGE: 4 of 6
DEPARTMENT & COURSE NO: MATH 1510
TIME: 75 minutess
EXAMINATION: Applied Calculus I
EXAMINER: A. Prymak
5. Suppose f (x) =
[1]
x3
x2 (x2 − 3)
0
.
Then
f
(x)
=
.
x2 − 1
(x2 − 1)2
(a) Find the domain of f (x).
Solution:
The domain of f is {x : x 6= ±1} (or (−∞, −1) ∪ (−1, 1) ∪ (1, ∞)).
[2]
(b) Find all critical points of f (x).
Solution:
√
√
x2 (x2 − 3)
x2 (x − 3)(x + 3)
f (x) =
=
.
(x2 − 1)2
(x2 − 1)2
0
The
√ critical points are the zeroes of the derivative: x = 0, x =
− 3.
[4]
√
3, and x =
(c) Find all interval(s) where f (x) is decreasing.
Solution:
Due to
of x and x2 − 1, the sign of f 0 (x) equals
to√the sign of
√ even powers
√
√
(x − 3)(x + 3), which
is
positive
when
x
∈
(−∞,
−
3)
∪
( 3, ∞) and
√ √
negative
when x ∈ (− √
3, 3) excluding x = ±1. Therefore, f is increasing on
√
(− 3, −1), (−1, 1), (1, 3).
[3]
(d) Classify the critical points of f (x) as relative maxima, relative minima, or neither.
Solution:
We use the first derivative test for relative extrema.
f 0 does not change the sign when passing through x = 0, so there is no relative
extremum at x = 0.
√
f 0√changes from positive to negative when passing through x = − 3, so x =
− 3 is a relative maximum.
√
√
f 0 changes from negative to positive when passing through x = 3, so x = 3
is a relative minimum.
UNIVERSITY OF MANITOBA
SECOND TERM EXAMINATION
DATE: November 30, 2016
PAGE: 5 of 6
DEPARTMENT & COURSE NO: MATH 1510
TIME: 75 minutess
EXAMINATION: Applied Calculus I
EXAMINER: A. Prymak
k
dx
[8] 6. A particle is moving along the curve y = in such a way that
= x3 , where k > 0 is
x
dt
constant and time is measured in seconds. When y = 2, the y-coordinate of the particle
is decreasing at the rate of 8 units per second. Find the value of k.
Solution:
Both x and y are functions of t and we know that
dx
dy 3
= x and
= −8.
dt
dt y=2
Differentiating y =
k
w.r.t. t and using the chain rule, we have
x
−k dx
dy
= 2 · .
dt
x
dt
When y = 2, we can express x as x =
k
k
= , and substitution in the previous
y
2
equation gives
−8 =
so k 2 = 16 and (as k > 0) k = 4.
Answer: k = 4.
−k 3
k
· x = −kx = −k ,
2
x
2
UNIVERSITY OF MANITOBA
SECOND TERM EXAMINATION
DATE: November 30, 2016
PAGE: 6 of 6
DEPARTMENT & COURSE NO: MATH 1510
TIME: 75 minutess
EXAMINATION: Applied Calculus I
EXAMINER: A. Prymak
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