United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
WORKSHEET 1 – SOLUTION
Chapter 2 Differentiation
Calculus I for Engineering
MATH 1110 SECTION 53 CRN 31403
11:00 – 1:00 on Monday & Wednesday
Due Date: Wednesday, March 4, 2009
ID No:
Solution
Name:
Solution
Score:
40/40
Instructions:
1. Read the questions carefully before you start working.
2. Show all your work to get the full credit.
3. Make your work to be neat and organized.
4. Do NOT use the calculator.
5. I do NOT receive any late worksheet.
FIRST WORKSHEET – SOLUTION
Spring, 2009
1. (Total 15 points) Find the derivative of each function. (Do NOT simplify your answer.)
(1.1) (5 points) f (x) =
1
+ 5 sec x2 + ln x2
x
Answer. We simplify first: f (x) = x−1 + 5 sec x2 + 2 ln x. Differentiating it, we get
¡
¢¡
¢
2
f 0 (x) = − x−2 + 5 sec x2 tan x2 (2x) + ,
x
(1.2) (5 points) g(t) =
i.e.,
f 0 (x) = −
1
2
2 2
+
10x
sec
x
tan
x
+ .
x
x2
ä
2
e2t
+ 5t
4t
Answer.
(e2t )0 (4t) − e2t (4t)0
t2
+
5
g (t) =
(ln 5) (t2 )0
2
(4t)
2t
2e (4t) − 4e2t
e2t (2t − 1)
t2
t2
=
+
5
5)
(2t)
=
+
2t5
ln 5.
(ln
16t2
4t2
0
ä
(1.3) (5 points) r(θ ) = 3θ −2 − aθ cos2 (4θ ), where a is a constant.
Answer.
0
r (θ ) = −6θ
−3
h
¢0 i
¡ 2
0
2
− a (θ ) cos (4θ ) + θ cos (4θ )
£
¤
= −6θ −3 − a cos2 (4θ ) + θ (2 cos (4θ )) (− sin (4θ )) (4θ )0
£
¤
= −6θ −3 − a cos2 (4θ ) + θ (2 cos (4θ )) (− sin (4θ )) 4
= −6θ −3 − a cos(4θ ) [cos (4θ ) − 8θ sin (4θ )] .
ä
2. (5 points) Find a such that the tangent line to the graph of f (x) = 2 − ax2 at x = −1 has
the slope 6.
Answer. The tangent line to the graph of f (x) at x = −1 has the slope f 0 (−1). We compute
f 0 (x) = −2ax,
f 0 (−1) = 2a.
ä
The condition f 0 (−1) = 6 implies 2a = 6, i.e., a = 3.
Page 1 of 4
Points Earned:
out of 20 points.
FIRST WORKSHEET – SOLUTION
Spring, 2009
3. (Total 10 points) A particle moves along a line so that at time t its position is given by
s(t) = 6t − t2 , where s(t) in meters and t in seconds.
(3.1) (3 points) What is its initial velocity v(0)?
Answer. The velocity v(t) is obtained by
v(t) = s0 (t) = 6 − 2t.
ä
So the initial velocity is v(0) = 6 m/s.
(3.2) (3 points) When does it change the moving direction?
Answer. The particle changes the direction when the velocity is zero. We observe
v(t) = 6 − 2t = 0
at
t = 3.
ä
Thus, it changes the direction at t = 3.
(3.3) (4 points) How fast is it moving when it returns to the initial position?
Answer. The particle is located at the initial position when s(t) = 0. We observe
s(t) = 6t − t2 = t(6 − t) = 0
at
t=0
or
t = 6.
It implies that the particle returns to the initial position at t = 6. Moreover, the
velocity at t = 6 is v(6) = 6 − 2(6) = −6 m/s. Hence, the desired speed is 6 m/s. (The
negative sign means that the moving direction has been changed.)
ä
Page 2 of 4
Points Earned:
out of 10 points.
FIRST WORKSHEET – SOLUTION
Spring, 2009
4. (Total 5 points) Let m A , m B , m C , m D and m E be the slopes of the tangent lines at the
points A, B, C, D, and E on the given graph, respectively.
(4.1) (1 point) Among the slopes, m A to m E , which one is the smallest one?
(4.2) (1 point) Among the slopes, m A to m E , which one is the largest one?
(4.3) (3 points) List the slopes m A , m B , m C , m D and m E in increasing order from the
smallest one.
Answer. Let us compare the values m A , m B , m C , m D and m E .
(i) It is easy to see that m A < 0, m B > 0, m C > 0, m D = 0, and m E < 0. That is,
m A , mE < mD = 0 < mB , mC .
It gives four possible lists:
m A < mE < mD < mB < mC ,
m A < mE < mD < mC < mB ,
mE < m A < mD < mB < mC ,
mE < m A < mD < mC < mB .
(ii) It looks to me that m A < m E , which can be observed by drawing tangent lines at A
and E. The tangent line at A is steeper than the one at E. So, the possible lists are
m A < mE < mD < mB < mC ,
m A < mE < mD < mC < mB .
(iii) The point D is closer to C than to B. It implies m C < m B . Therefore, the desired list
is
m A < mE < mD < mC < mB .
ä
5. (5 points) Find the second–degree polynomial (i.e., f (x) = ax2 + bx + c) satisfying all the
conditions:
f (0) = 0,
f 0 (0) = 5,
f 00 (0) = 1.
Answer. From f (0) = 0, we get 0 = f (0) = a(02 ) + b(0) + c = c, i.e., c = 0. From f 0 (0) = 5, we
get
f 0 (x) = 2ax + b,
5 = f 0 (0) = 2a(0) + b = b,
i.e.,
b = 5.
Page 3 of 4
Points Earned:
out of 10 points.
FIRST WORKSHEET – SOLUTION
Spring, 2009
From f 00 (0) = 1, we get
f 00 (x) = 2a,
1 = f 00 (0) = 2a,
Therefore, we deduce
f (x) =
i.e.,
1
a= .
2
x2
x
+ 5x = (x + 10) .
2
2
Page 4 of 4
Points Earned:
ä
out of ?? points.
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
WORKSHEET 2 – SOLUTION
Chapter 3 Applications of Differentiation
Calculus I for Engineering
MATH 1110 SECTION 54 CRN 31403
11:00 – 1:00 on Monday & Wednesday
Due Date: Wednesday, March 25, 2009
ID No:
Solution
Name:
Solution
Score:
40/40
Instructions:
1. Read the questions carefully before you start working.
2. Show all your work to get the full credit.
3. Make your work to be neat and organized.
4. Do NOT use the calculator.
Calculus I for Engineering
SECOND WORKSHEET – SOLUTION
Spring, 2009
1. (5 points) Find the absolute extrema of the function f (x) =
x2
on the interval (−1, 1).
x2 − 1
Answer.
x2
x2 − 1 + 1
1
f (x) = 2
=
= 1+ 2
,
2
x −1
x −1
x −1
x
f 0 (x) = −
2x
(x2 − 1)2
.
0
Sign of f 0
Graph of f
0
−
f (0) = 0 Decrease
+
Increase
As x approaches to −1 from the right–hand side of −1, the values of f (x) become negatively
smaller and smaller, i.e., the graph of f (x) goes downward near x = −1 from the right–hand
side of −1. Similarly, as x approaches to 1 from the left–hand side of 1, the values of f (x)
become negatively smaller and smaller, i.e., the graph of f (x) goes downward near x = 1
from the left–hand side of 1. It implies f (x) does not have an absolute minimum value on
the interval (−1, 1). However, by the First Derivative Test and the table above, we deduce
f (x) has the absolute maximum value f (0) = 0.
ä
t ˆ› G j–” —–šŒ™ G XU XU \
› › —a V Vž ž ž U ” ˆ› Š–” —–šŒ™ U Š–”
5
y
4
3
2
1
-5
-4
-3
-2
-1
x
1
2
3
4
5
-1
-2
-3
-4
-5
Graph of f (x) =
x2
x2 − 1
x2 + 1
, x 6= 0, is increasing and
2. (5 points) Find the intervals where the graph of f (x) =
x
decreasing.
Answer.
f (x) =
x2 + 1
1
= x+ ,
x
x
f 0 (x) = 1 −
Page 1 of 6
x2 − 1 (x − 1)(x + 1)
1
=
=
.
x2
x2
x2
Points Earned:
out of 10 points.
Calculus I for Engineering
SECOND WORKSHEET – SOLUTION
x
Sign of f 0
Graph of f
Spring, 2009
−1
1
0
−
f (−1) = −2 Decrease
+
Increase
0
+
f (1) = 2 Increase
We observe that f 0 (x) > 0 for x < −1 or x > 1 and f 0 (x) < 0 for −1 < x < 1.
ä
Hence, f (x) increases on (−∞, −1) ∪ (1, ∞) and f (x) decreases on (−1, 1).
t ˆ› G j–” —–šŒ™ G XU XU \
› › —a V Vž ž ž U ” ˆ› Š–” —–šŒ™ U Š–”
5
y
4
3
2
1
-5
-4
-3
-2
x
-1
1
2
3
4
5
-1
-2
-3
-4
-5
Graph of f (x) =
x2 + 1
1
= x+
x
x
3. (5 points) Find the intervals where the graph of f (x) =
ward and concave downward.
20 3 33 2
x − x + 7x + 6 is concave up3
2
Answer.
f 0 (x) = 30x2 − 33x + 7 = (5x − 7)(4x − 1) = 0
f 00 (x) = 40x − 33 = 0
at
x=
at
33
≈ 0.825
40
x
Sign of f 00
Graph of f
1
7
x= , x=
4
5
(Critical Numbers)
(Candidate for Inflection Point)
33/40
−
Concave Downward
0
+
f (33/40) = 6861/1600 ≈ 4.28813 Concave Upward
From the table, we observe f (x) is concave upward on (0.825, ∞) and concave downward
on (−∞, 0.825). Since the concavity changes around the point (0.825, 4.28813), the point
(0.825, 4.28813) is the inflection point of f (x).
ä
Page 2 of 6
Points Earned:
out of 5 points.
Calculus I for Engineering
SECOND WORKSHEET – SOLUTION
Spring, 2009
t ˆ› G j–” —–šŒ™ G XU XU \
› › —a V Vž ž ž U ” ˆ› Š–” —–šŒ™ U Š–”
5
t ˆ› G j–” —–šŒ™ G XU XU \
› › —a V Vž ž ž U ” ˆ› Š–” —–šŒ™ U Š–”
y
4
8
y
3
7
2
6
1
5
4
-5
3
-4
-3
-2
-1
1
2
3
4
5
-1
2
1
-0.4 -0.2
x
Inflection Point
-2
x
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
-3
-1
-2
-4
-5
20 3 33 2
Graph of f (x) =
x − x + 7x + 6
3
2
Graph of f (x) =
x2 + 1
1
= x+
x
x
1
x
4. (5 points) Use the Second Derivative Test to find the local extrema of f (x) = x + .
Answer.
f 0 (x) = 1 −
f 00 (x) =
2
,
x3
1
x2 − 1
=
=0
x2
x2
at
x = ±1
f 00 (−1) = −2 < 0,
(Critical Numbers)
f 00 (1) = 2 > 0.
By the Second Derivative Test, we conclude f (x) has the local maximum value f (−1) = −2
and the local minimum value f (1) = 2.
ä
5. (5 points) Bubba, a farmer, wants to fence–in a three–sided area next to his barn, using
the side of the barn as the fourth side of the rectangle. He also wants to divide this
rectangular area into three identical pens1 (see figure below). Suppose Bubba has 136
feet of available fencing. Find the dimension of each pen which maximize its area. Round
dimensions to nearest tenth.
Answer. Let x and y be the lengths of width and height of one pen. Then each pen has
area x y and total area of three pens is A = 3x y. Moreover, 3x + 4y = 136, that is,
A = 3x y,
1
3x + 4y = 136,
0<x<
136
≈ 45.3333,
3
0< y<
136
= 34.
4
A pen is a small area with a fence round it in which farm animals are kept for a short time.
Page 3 of 6
Points Earned:
out of 10 points.
Calculus I for Engineering
SECOND WORKSHEET – SOLUTION
Spring, 2009
Barn
Pen
Pen
Pen
Putting 3x = 136 − 4y into A = 3x y, we get
A 0 (y) = 8(17− y) = 0
A(y) = y(136−4y) = 4y(34− y),
at
y = 17 ∈ (0, 34)
(Critical No.)
Since A 00 (y) = −8 < 0 for all y, so A 00 (17) = −8 < 0. By the Second Derivative Test, A(y) has
136 − 4(17) 68
the local maximum value A(17) = 4(17)(34−17) = 1156 with x =
=
≈ 22.6667
3
3
and y = 17. Since lim A(y) = 0 and lim A(y) = 0, hence A(17) = 1156 is in fact the absoy→0
y→136/4
ä
lute maximum value.
6. (5 points) Find the volume of the largest box with no top that can be made from a piece of
cardboard 54 inches square by cutting equal squares from the corners and turning up the
sides.
t ˆ› G j–” —–šŒ™ G XU XU \
› › —a V Vž ž ž U ” ˆ› Š–” —–šŒ™ U Š–”
x
x
x
x
54
x
x
x
x
54
Answer. The box has the height x, while the length and width are 54 − 2x so the volume
V of the box is
V (x) = x(54 − 2x)2 = 4x(x − 27)2 ,
0 < 54 − 2x,
0<x<
54
= 27.
2
To find the maximum value of V , we find the critical numbers of V (x):
V 0 (x) = 12(x − 9)(x − 27) = 0
at
x = 9 ∈ (0, 27) (Critical Number)
Since V 00 (x) = 24(x − 18) and V 00 (9) = −216 < 0, by the Second Derivative Test, V (x) has the
local maximum value V (9) = 11664. Because V (0) = 0 = V (27), the local maximum value
is in fact the absolute maximum value, i.e., the volume of the largest box is V (9) = 11664
with the length and width 36 and height 9.
ä
Page 4 of 6
Points Earned:
out of 10 points.
Calculus I for Engineering
SECOND WORKSHEET – SOLUTION
Spring, 2009
7. (5 points) Find the point (a, b) on the curve of y = x2 which is closest to the point (3, 0).
Answer. The distance between (3, 0) and the point (a, b) on the curve of y = f (x) = x2 is
d=
p
d 2 = (a − 3)2 + b2 .
(a − 3)2 + b2 ,
Since (a, b) is on the curve, we have b = f (a) = a2 , i.e., b = a2 . Putting this into the equation
on the distance, we have d 2 = (a − 3)2 + a4 , which we want to minimize with the condition
0 ≤ a ≤ 3 and 0 ≤ b ≤ 9. (Why?) Let g(a) = (a − 3)2 + a4 and find the absolute minimum
value.
g0 (a) = 2(a − 1)(2a2 + 2a + 3) = 0
at
a = 1 ∈ [0, 3] (Critical Number)
Since g00 (a) = 2(6a2 + 1) > 0 for all a, so g00 (1) = 14 > 0. Hence, by the Second Derivative
Test, g(1) = 5 is the local minimum value with a = 1 and b = 1. Since the values of g(a) at
the endpoints of the interval of a are g(0) = 9 and g(3) = 81, hence, g(1) = 5 is
p in fact the
absolute minimum value, i.e., the point (1, 1) gives the shortest distance d = 5 from the
curve to the given point (3, 0).
ä
8. (5 points) Find P 0 (V ) =
stants.
A
dP
for the equation P e V (V − B) = C, where A, B, and C are condV
The following small problems are preliminaries to solve the given problem. Solve the
following problems Step–by–Step.
2
dy
(8.1) (0 points) Find y0 (x) =
for the equation ye x (x − 3) = 4.
dx
Method 1 Typical Explicit Differentiation. The given equation can be rewritten as
2
4e− x
y=
.
x−3
Applying the Quotient Rule, we have
h
dy 4
y (x) =
=
dx
0
2
2
e− x x22 (x − 3) − e− x
(x − 3)2
i
2
4e− x
=
x−3
2
(x − 3) − 1
x2
x−3
µ
¶
1
2
= 2−
y.
x−3
x
ä
Method 2 Implicit Differentiation. We differentiate the equation with respect to x:
i
d
d h 2
ye x (x − 3)
(4) =
dx
dx
µ
¶
2 d
d 2
dy 2
e x (x − 3) + y
e x (x − 3) + ye x
0=
(x − 3)
dx
dx
dx
µ
¶
2
2
dy 2
2
=
e x (x − 3) + ye x − 2 (x − 3) + ye x
dx
x
Page 5 of 6
Points Earned:
(Product Rule)
(Chain Rule)
out of 5 points.
Calculus I for Engineering
SECOND WORKSHEET – SOLUTION
·
¸
2 2
2
2
2
dy 2
e x (x − 3) = ye x 2 (x − 3) − ye x = 2 (x − 3) − 1 e x y,
dx
x
x
Spring, 2009
µ
¶
2
dy
1
= 2−
y.
dx
x−3
x
ä
Method 3 Logarithmic Differentiation. We take the natural logarithmic function on
both sides of the given equation and differentiate the whole equation with respect to
x:
i
h 2
2
2
ln 4 = ln ye x (x − 3) = ln y + ln e x + ln(x − 3) = ln y + + ln(x − 3)
x
µ ¶
d
d
d 2
d
1 dy 2
1
ln 4 =
ln y +
+
ln(x − 3),
0=
− 2+
dx
dx
dx x
dx
y dx x
x−3
µ
¶
1
2
dy
= 2−
y.
dx
x−3
x
(8.2) (0 points) Find P 0 (V ) =
(Chain Rule)
ä
2
dP
for the equation P e V (V − 3) = 4.
dV
Answer. By using one of the methods above, we can get
µ
¶
2
1
dP
=
−
P.
P (V ) =
dV
V2 V −3
0
(8.3) (0 points) Find P 0 (V ) =
constants.
ä
A
dP
for the equation P e V (V − B) = C, where A, B, and C are
dV
Answer. By using one of the methods above, we can get
µ
¶
A
dP
1
P (V ) =
=
−
P.
dV
V2 V −B
0
Page 6 of 6
Points Earned:
ä
out of ?? points.
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
WORKSHEET 3 – SOLUTION
Midterm Examination
Calculus I for Engineering
MATH 1110 SECTION 54 CRN 31403
11:00 – 1:00 on Monday & Wednesday
Due Date: Wednesday, April 8, 2009
ID No:
Solution
Name:
Solution
Score:
20/20
Calculus I for Engineering
THIRD WORKSHEET – SOLUTION
Spring, 2009
Section I. Multiple Choice Problems (Total 6 Points, 1.5 Point Each)
(No partial credits for this section)
1. (11/2 points) A weight hanging by a spring from the ceiling vibrates up and down. Its vertical
position is given by s(x) = 8 sin(7x). Find the maximum speed of the weight and its position
when it reaches maximum speed.
Answer. The derivative s′ (x) of the position function s(x) is the velocity v(x) representing the
speed of the weight, i.e.,
v(x) = s′ (x) = 56 cos(7x).
We observe that the maximum value of the cosine function is 1 which occurs at 7x = 0, 2nπ with
integer n, i.e., x = 0, 2nπ/7. Thus, the maximum speed is 56 and it is obtained when x = 0 for
the first time.
2. (11/2 points) For f (x) = sin x, find f (158) (x) (i.e., the 158th derivative of f (x)).
Answer. We observe for k = 0, 1, 2, . . . ,
f (4k) (x) = f (0) (x) = sin x,
f (4k+1) (x) = f (1) (x) = cos x,
f (4k+2) (x) = f (2) (x) = − sin x,
f (4k+3) (x) = f (3) (x) = − cos x.
Since 158 = 4(39) + 2, we conclude f (158) (x) = f (2) (x) = − sin x.
3. (11/2 points) Find the equation of the tangent line to the curve of y = 2x at x = 3.
Answer. Recall that the equation of the tangent line to the graph of y = f (x) at x = a is given by
y − f (a) = f ′ (a)(x − a).
With f (x) = 2x , a simple computation shows
f ′ (x) = 2x ln 2,
f ′ (3) = 8 ln 2,
f (3) = 8.
Therefore, the equation of the desired tangent line is
y − 8 = 8 ln 2(x − 3),
i.e.,
y = (8 ln 2)x − 24 ln 2 + 8 = 8 (x ln 2 + 1 − 3 ln 2) .
Page 1 of 10
Points Earned:
out of 41/2 points.
Calculus I for Engineering
THIRD WORKSHEET – SOLUTION
Spring, 2009
4. (11/2 points) Find all locations of horizontal and vertical tangent lines to the curve of x2 −xy 2 = 16.
Answer 1. The given equation can be rewritten by
xy = x − 16,
2
2
x2 − 16
16
y =
=x− .
x
x
2
Differentiating the whole equation with respect to x, we get
dy
16
2y
= 1 + 2,
dx
x
(
)
dy
1
16
x2 + 16
=
1+ 2 =
.
dx
2y
x
2x2 y
(1)
Since x2 + 16 is always positive, dy/dx cannot be zero. It implies there is no point on the curve
at which the tangent line is horizontal.
On the other hand, either at x = 0 or at y = 0, dy/dx does not exist. It implies there is a point
on the curve at which the tangent line is vertical. The point is of form (0, c) or (c, 0) with some
constant c.
Case 1. Point on the curve with the x–component 0, i.e., form (0, c): We observe x = 0 does not
satisfy the given equation. So this case cannot happen.
Case 2. Point on the curve with the y–component 0, i.e., form (c, 0): Putting y = 0 into the
equation, we have
x2 = 16,
x = ±4.
Thus, there are two points (±4, 0) on the curve at which the tangent lines are vertical. Figure is
given on the last page.
Answer 2. Differentiating the whole equation with respect to x, we get
d
d ( 2)
d ( 2)
(16) =
x −
xy ,
dx
dx
dx
(
)
dy
0 = 2x − y + 2xy
,
dx
2
dy
2x − y 2
=
.
dx
2xy
(2)
(One can check that the equations (1) and (2) are exactly same. There must be only one derivative.)
When y 2 = 2x, we have dy/dx = 0, i.e., the horizontal tangent lines occur at points satisfying
y 2 = 2x. In order to find the point on the curve, we put y 2 = 2x into the given equation and then
x2 − x(2x) = 16,
x2 = −16,
which cannot happen because x2 is always positive. For this reason, there is no point on the curve
at which the tangent line is horizontal.
Either at x = 0 or at y = 0, the derivative dy/dx does not exist. By the exactly same argument
as given in Answer 1 above, there are two points (±4, 0) on the curve at which the tangent lines
are vertical. Figure is given on the last page.
Page 2 of 10
Points Earned:
out of 11/2 points.
Calculus I for Engineering
THIRD WORKSHEET – SOLUTION
Spring, 2009
Section II. Multiple Step Problems (Total 12 Points, 3 Points Each)
5. (Total 3 Points)
(5.1) (1 point) Find the point P (a, b) on the graph of y = x2 which is closest to the point Q(0, 1).
Answer. First, we restrict the interval of the point P (a, b) as −1 ≤ a ≤ 1 and 0 ≤ b ≤ 1.
(Why?) It is easy to see that the distance D between P and Q is given by
√
D=
(a − 0)2 + (b − 1)2 =
√
a2 + (b − 1)2 ,
which we want to minimize. Since P (a, b) is on the graph of y = x2 , we have the equation
b = a2 . Plugging this into the equation on D, we get
√
D(a) =
a2 + (a2 − 1)2 =
√
a4 − a2 + 1.
Now, we observe that minimizing D is exactly same as minimizing D2 (a). (Why?) So we find
the absolute minimum value of D2 (a) = a4 − a2 + 1 by using the Closed Interval Method.
Step 1. Critical Number of D2 (a): Differentiating D2 (a), we get
)
d ( 2 )
d ( 4
D (a) =
a − a2 + 1 = 4a3 − 2a = 2a(2a2 − 1) = 0
da
da
1
at a = 0, ± √ .
2
√
Since those values are in the domain given by us, we have the critical numbers 0 and ±1/ 2.
At these points, D2 (a) has the values
(
D (0) = 0 − 0 + 1 = 1,
2
4
2
D
2
1
±√
2
)
(
1
= ±√
2
)4
(
1
− ±√
2
)2
+1=
1 1
3
− +1= .
4 2
4
Step 2. Values at Endpoints: We compute the values of D2 (a) at the endpoints of the interval
−1 ≤ a ≤ 1.
D2 (−1) = (−1)4 − (−1)2 + 1 = 1,
D2 (1) = 14 − 12 + 1 = 1.
Step 3. Comparison: Comparing the values at the critical points and endpoints, we conclude
√
the absolute minimum of D2 (a) occurs at a = ±1/ 2 with b = 1/2. That is, the closest
√
√
√
points on the curve are (a, b) = (±1/ 2, 1/2) with the distance D(±1/ 2) = 3/2.
Page 3 of 10
Points Earned:
out of 1 points.
Calculus I for Engineering
THIRD WORKSHEET – SOLUTION
∫
tan x
(5.2) (1 point) If F (x) =
√
Spring, 2009
t + 3 dt, compute F ′ (x).
2x
Answer. We recall the Fundamental Theorem of Calculus, Part II :
(
)
d ∫ g(x)
f (t) dt = f (g(x))g ′ (x) − f (h(x))h′ (x).
dx h(x)
By the theorem, we have
)
(∫ tan x √
d
t + 3 dt
dx 2x
√
√
√
√
= tan x + 3(tan x)′ − 2x + 3(2x)′ = sec2 x tan x + 3 − 2 2x + 3.
F ′ (x) =
∫
(5.3) (1 point) Evaluate the integral
sin x
dx.
cos x − csc2 (π/2)
Answer. First, we simplify the integrand:
( )
csc
2
π
2
1
=
=
2
sin (π/2)
(
1
sin (π/2)
)2
= 1.
Then the integral becomes
∫
∫
∫
sin x
sin x
sin x
dx =
dx = −
dx.
2
cos x − csc (π/2)
cos x − 1
1 − cos x
Since (1 − cos x)′ = sin x, we observe
∫
∫
∫
sin x
sin x
(1 − cos x)′
dx = −
dx = −
dx = − ln |1 − cos x| + C,
cos x − csc2 (π/2)
1 − cos x
1 − cos x
where C is the constant of integration and the last equality holds by the Theorem 6.1 on
page 397 in the textbook, i.e.,
∫
f ′ (x)
dx = ln |f (x)| + C.
f (x)
Page 4 of 10
Points Earned:
out of 2 points.
Calculus I for Engineering
THIRD WORKSHEET – SOLUTION
Spring, 2009
6. (Total 3 Points) Find the derivative dy/dx. (Do NOT simplify!)
(
)
2
(6.1) (11/2 points) y = 3tan(x ) + ln sin−1 (2x) − e5 , where sin−1 x is the inverse (arcsine) of sin x.
Answer.
(
)′
′
(sin−1 (2x))
−0
sin−1 (2x)
(2x)′
2
√
= (x2 )′ sec2 (x2 )3tan(x ) ln 3 +
sin−1 (2x) 1 − (2x)2
y ′ = tan(x2 ) 3tan(x ) ln 3 +
2
2
2
= 2x sec2 (x2 )3tan(x ) ln 3 +
√
sin−1 (2x) 1 − (2x)2
.
(6.2) (11/2 points) xey − 3y sin x = 1. (Use implicit differentiation.)
Answer.
d
d
(1) =
(xey − 3y sin x)
dx
dx )
(
(
)
[(
)
(
)]
d
d y
dy
d
y
0=
x e +x
e −3
sin x + y
sin x
dx
dx
dx
dx
y dy
[
dy
= e + xe
−3
sin x + y cos x
dx
dx
y
]
dy
dx
y
dy
−e + 3y cos x
=
.
dx
xey − 3 sin x
= ey − 3y cos x + (xey − 3 sin x)
(xey − 3 sin x)
dy
= −ey + 3y cos x,
dx
Page 5 of 10
Points Earned:
out of 3 points.
Calculus I for Engineering
THIRD WORKSHEET – SOLUTION
Spring, 2009
7. (Total 3 Points) Consider f (x) = x2 ln x.
(7.1) (11/2 points) Find all critical numbers, if they exist, and the intervals where the graph of f (x)
increases and decreases.
Answer. First, the domain of the given function is (0, ∞).
f ′ (x) = 2x ln x + x2
1
= 2x ln x + x = x (2 ln x + 1) = 0,
x
2 ln x + 1 = 0
at x = e−1/2 .
So f (x) has only one critical number e−1/2 . We make the sign chart as follows:
e−1/2
x
Sign of f ′
−
Graph of f
Decreases
0
+
Critical Point Increases
Hence, the graph of f (x) increases on (e−1/2 , ∞) while it decreases on (0, e−1/2 ). Please confer
the figure on the last page.
(7.2) (11/2 points) Find all inflection points, if they exist, and the intervals where the graph of f (x)
is concave upward and concave downward.
Answer. From the Answer above, we recall f ′ (x) = x (2 ln x + 1).
f ′′ (x) = (x)′ (2 ln x + 1) + x (2 ln x + 1)′ = 2 ln x + 1 + x
2
= 2 ln x + 3 = 0
x
at x = e−3/2 .
So we have a candidate for the inflection point. We make the sign chart as follows:
e−3/2
x
Sign of f ′′
Graph of f
−
0
+
Concave Downward Inflection Point Concave Upward
Since the concavity changes around x = e−3/2 , it really gives the inflection point
(
− 32
e
(
− 32
,f e
))
(
= e
)
− 23
3e−3
,−
.
2
The graph of f (x) is concave upward on (e−3/2 , ∞) while it is concave downward on (0, e−3/2 ).
Please confer the figure on the last page.
Page 6 of 10
Points Earned:
out of 3 points.
Calculus I for Engineering
THIRD WORKSHEET – SOLUTION
Spring, 2009
8. (3 points) Sand is poured into a conical pile. During all the operation, the height of the pile equals
the diameter of the pile. If the sand is poured at a constant rate of 5 m3 /s, then at what rate is
the height of the pile increasing when the height is 2 meters?
1
(The volume V of a cone is V = πr2 h, where r is the radius of the base and h is the height.)
3
Answer. First, we observe that the radius r and the height h have the relation h = 2r and they
are functions of time t. So the volume becomes
(
1
h(t)
1
V (t) = πr2 (t)h(t) = π
3
3
2
)2
h(t) =
1
πh3 (t).
12
Differentiating it, we get
(
)
dV (t)
d 1
1
dh(t)
=
πh3 (t) = πh2 (t)
.
dt
dt 12
4
dt
The given information says dV (t)/dt = 5 and we are asked to find dh(t)/dt when h(t) = 2. Thus,
5=
dh(t)
1 2 dh(t)
π2
=π
,
4
dt
dt
dh(t)
5
= .
dt
π
Section III. Concept Problems (Total 2 Points, 0.5 Point Each)
9. (Total 2 Points) (True or False) Put T for the correct statement and F for the false one.
(9.1) (1/2 point) Suppose the concentration x(t) of a certain chemical after t seconds of an autocatalytic reaction is given by
12
x(t) =
.
3 + e−3t
Then the concentration x(t) never exceeds 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T
Answer. Since t represents the time, we may assume the domain of x(t) as the interval [0, ∞).
We observe x(0) = 12/4 = 3 and as t → ∞, e3t → ∞ and e−3t = 1/e3t → 0. So, we get
as t → ∞,
x(t) =
12
12
→
= 4.
3 + e−3t
3+0
We use the derivatives to get the information on the shape of the curve:
12(3 + e−3t )′
36e−3t
x (t) = −
=
>0
(3 + e−3t )2
(3 + e−3t )2
′
for all t ∈ [0, ∞),
implying that the graph of x(t) always increases. Moreover,
(
36e−3t
12
x (t) =
= 3e−3t
−3t
2
(3 + e )
3 + e−3t
′
)
= 3e−3t x(t)
Page 7 of 10
Points Earned:
out of 31/2 points.
Calculus I for Engineering
THIRD WORKSHEET – SOLUTION
(
x′′ (t) = −9e−3t x(t) + 3e−3t x′ (t) = −9e−3t x(t) + 3e−3t
(
)
Spring, 2009
)2
x(t)
= −9e−3t x(t) + 9e−6t x(t) = −9e−3t 1 − e−3t x(t) < 0
for all t ∈ [0, ∞),
implying that the graph of x(t) is always concave downward. Therefore, the graph of x(t)
never exceeds 4. Please confer the figure on the last page.
(9.2) (1/2 point) The function g(x) = ln(ax) with ax > 0 has the derivative g ′ (x) =
Answer.
g ′ (x) = (ln(ax))′ =
(ax)′
a
1
=
= .
ax
ax
x
1
. ...... T
x
(9.3) (1/2 point) The graph of f (x) = x1/3 has a vertical tangent line at x = 0. It does not have a
local extremum at x = 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T
Answer. The domain of f (x) is (−∞, ∞).
1
1
f ′ (x) = x−2/3 = √
,
3
3
3 x2
which cannot be zero, but does not exist at x = 0. So f (x) has only one critical value x = 0
at which the tangent line is vertical. To check the local extremum at x = 0, we can use either
the First Derivative Test or the Second Derivative Test. Here we take the First Derivative
Test. We observe
1
>0
for all x,
f ′ (x) = √
3
3 x2
implying that the graph always increases. It means the increment of the graph does not
change around the critical point x = 0 and thus the First Derivative Test says that there is
no local extremum at x = 0.
(9.4) (1/2 point) The graph of the function f (x) = x4 has only one inflection point. . . . . . . . . . F
Answer.
f ′ (x) = 4x3 ,
f ′′ (x) = 12x2 = 0
at x = 0.
So we have a candidate x = 0 for the inflection point. However, since f ′′ (x) ≥ 0 for all x,
the graph of f (x) is always concave upward, thus (0, f (0)) = (0, 0) cannot be the inflection
point. That is, there is no inflection point. The graph of f (x) = x4 looks very similar to the
graph of x2 .
Page 8 of 10
Points Earned:
out of 11/2 points.
Calculus I for Engineering
THIRD WORKSHEET – SOLUTION
Spring, 2009
Graph of x 2 -xy2 =16
10
5
-4
-2
2
4
6
8
10
-5
- 10
Problem #4 Graph of x2 − xy 2 = 16
Graph of y=x 2 Log@xD
0.8
0.6
0.4
0.2
- 0.2
1
1
ã3‘2
ã
1
Inflection Point
Critical Point
- 0.4
Problem #7 Graph of y = x2 ln x
Page 9 of 10
Points Earned:
out of ?? points.
Calculus I for Engineering
THIRD WORKSHEET – SOLUTION
Spring, 2009
12
Graph of xHtL=
3 + e-3 t
5.0
4.5
4.0
3.5
3.0
2.5
0
1
2
3
4
Problem #9.1 Graph of x(t) =
5
12
3 + e−3t
Graph of fHxL=x 13
1.5
1.0
0.5
-4
-2
2
4
- 0.5
- 1.0
- 1.5
Problem #9.3 Graph of f (x) = x1/3
Page 10 of 10
Points Earned:
out of ?? points.
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