Engineering Calculus, 22m031, Exam 2, Fall, 2012
1. The largest and smallest values attained by the function
f (x) = 2x3 + 3x2 12x on the interval [0; 2] is:
a. Largest 36,
Smallest
b. Largest 36,
Smallest 0
c. Largest 4,
Smallest
d. Largest 4,
Smallest 0
Engineering Calculus, 22m031, Exam 2, Fall, 2013, Camillo
1. The largest and smallest values attained by the function
f (x) = x4 2x2 on the interval [0; 2] is:
a. Largest 0,
Smallest
b. Largest 8,
Smallest 0
c. Largest 8,
Smallest
1
1
d. Largest 16,
Small
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ENGINEERING MATH I:SINGLE VARIABLE CALCULUS
22M:031/MATH:1550
NOVEMBER 13, 2012
CAMILLO
FORM A
Directions for completing your answer sheet
1. Use a no. 2 or softer PENCIL. Marks should be dark and completely ll
22m031, Final Exam, Fall, 2011. Camillo.
1.
a.
b.
c.
d.
e.
p
x 1
lim
x!1 x2 x
=
1
2
1
2
1
4
1
4
2
1
2. If f (x) = x +x+1
then f (x) =
x
a. x1 x3
2
2
2
b.
c.
d.
e.
1
x2
0
3
+
1
x3
1
x2
1
x3
1
x2
2
x3
1
x2 (1
+ x2 )
2
3. The derivative of xe
a. e 2x(1 x)
b.
22m031, nal exam, fall, 2012, Camillo
lim p x
1. x!0
x+1
a. 1
1
=
b. 2
c.
1
d.
2
e. 0
1
2. If f (x) =
a. px1 + 2x
b.
1
3
2x 2
c.
p1
x
d.
p1
x
e.
p
1+2
p x
x
then the derivative of f (x) =
+2
+2
1
3
2x 2
2
3. The derivative of
a.
sin 2
=
1
2
b. 2
c.
1
2
d.
22m031, Engineering Calculus, Final exam, fall, 2013, Camillo
lim p x+2
1. x!
2 x+3
a. 1
1
=
b. 2
c.
1
d.
2
e. 0
1
2. If f (x) =
a.
1
x2
b.
1
x2
c.
d.
then the derivative of f (x) =
1
3
2x 2
+
1
x2
1
2x2
e. ln x
p
1+2 x
x
1
3
x2
1
3
x2
1
p
2 x
p1
x
2
3. T
Engineering Calculus
Fall 2014, Camillo
Final Exam, Form A
16x4
(1) lim x2 4 =
x2
(a) 8
(b) 8
(c) 16
(d) 16
(e) 4
1
(2) If f (x) =
1+x
2 x
then f 0 (x) =
(a)
1
2 x
1+
1
x
(b)
1
4 x
1+
1
x
(c)
1
4 x
1+
1
x
(d)
1
4 x
1
1
x
(e)
1
4 x
1
x
1
sin
(3) The deriv
122
CHAPTER 9. ROTATIONAL DYNAMICS
F
(a)
F
F
(c)
(b)
Figure 9.5: Forces exerted on a door: (a) Force exerted far from the hinge, perpendicular to the door face.
(b) Force is exerted perpendicular to the door face but close to the hinge! (c) Force is exert
121
9.1. THE IMPORTANT STUFF
R
R
(a)
2
5
(b)
MR2
2
3
MR2
R
b
(c)
1
2
(d)
a
MR2
2
2
1
)
12 M(a +b
Figure 9.3: More moments of inertia.
Icm
I
M
cm
D
Figure 9.4: The Parallel Axis Theorem: Axis through the cm gives Icm , and we want the moment of inertia
I a
120
CHAPTER 9. ROTATIONAL DYNAMICS
R
R1
R2
(b)
(a)
1
2
2
MR
2
M(R1+R22)
L
R
R
(d)
(c)
1
2
MR2
1
4
1
MR2 + 12 ML
L
L
(e)
1
12
ML2
2
(f)
1
3
ML2
Figure 9.2: Some values of the moment of inertia for various shapes and choices of axes.
118
CHAPTER 9. ROTATIONAL DYNAMICS
mi
r
Figure 9.1: A little piece of the rotating object has mass mi and is a distance ri from the axis.
Into this we substitute vi = ri , giving
1
m (r )2
2 i i
KErot =
i
Now we take the common factors of
1
2
and 2 outsid
119
9.1. THE IMPORTANT STUFF
9.1.3
More on the Moment of Inertia
Most of the time we are interested in the moment of inertia of some continuous object like
a rod or a disk, and for these shapes, even though we use the principle given in Eq. 9.2 we
actuall
Chapter 9
Rotational Dynamics
9.1
9.1.1
The Important Stu
Introduction
Having worked with the kinematics of rotation in the last chapter we now move on to study
the dynamics of rotation. In eect, we have to re-do that past chapters on dynamics (force,
ene
115
8.2. WORKED EXAMPLES
54 m/s
47 rev/s
r
Figure 8.6: Rotating string in Example 4.
The string rotates at an angular speed of 47 rev , and its tip has a tangential speed
s
of 54 m . What is the length of the rotating string? [CJ7 8-29]
s
The rotating str
113
8.2. WORKED EXAMPLES
aT
ac
w, a
Figure 8.5: Acceleration of a point on a rotating object has centripetal and tangential components, ac and
aT .
So in general for a point on a rotating object, the acceleration has two components, as
illustrated in Fig.
114
CHAPTER 8. ROTATIONAL KINEMATICS
Use the relations 1 rev = 2 rad as well as 1 min = 60 s to convert the units:
rev
rev
33.3 min = (33.3 min )
8.2.3
2 rad
1 rev
1 min
60 s
= 3.49 rad
s
Rotational Motion with Constant Angular Acceleration
3. A ywheel ha
110
CHAPTER 8. ROTATIONAL KINEMATICS
s
r
q
Figure 8.4: Object turns through an angle ; a point on the object at a distance r from the axis moves
through a distance s = r.
8.1.3
Angular Velocity
We follow the same steps as in Chapter 2 to study the relatio
112
CHAPTER 8. ROTATIONAL KINEMATICS
Here we measure such that = 0 at t = 0.
We can get other relations between , and with some algebra. One can show that
Eqs. 8.7 and 8.8 give:
2
2 = 0 + 2
(8.9)
And one can show
1
= 2 (0 + )t
(8.10)
but you are caution
111
8.1. THE IMPORTANT STUFF
should include it to emphasize that we are not using degrees. When we do a calculation
where the answer needs to come out in joules, then it should be dropped. However it would
be OK (with me) to express an angular velocity in
109
8.1. THE IMPORTANT STUFF
q
Figure 8.3: Orientation of a rotating object is given by the angle .
8.1.2
Angular Displacement
The orientation of a rotating object is given by a single number, an angle , which can be
taken as the angle between some some r
108
CHAPTER 8. ROTATIONAL KINEMATICS
(a)
(b)
Figure 8.1: (a) Football thrown the right way. (b) Football thrown the way I always throw one. It has
several kinds of rotation all at once!
v
(a)
(b)
Figure 8.2: The kinds of (simple) rotations we will conside
105
7.1. THE IMPORTANT STUFF
CM
CM
CM
(a)
(b)
Figure 7.3: (a) Diver jumps o diving board and does some of that crazy diver-type stu; action looks
very complicated! (b) If we plot the center of mass of the diver, its motion is fairly simple (a parabolic pa
Chapter 8
Rotational Kinematics
8.1
8.1.1
The Important Stu
Rigid Bodies; Rotating Objects
So far we have had much to say about the motion of objects and the way it is determined by
the forces acting on those objects. But in all of our examples we were tr
106
CHAPTER 7. MOMENTUM
(x3, y3 ). . . then the x coordinate of its center of mass is given by
xcm =
m1 x1 + m2 x2 +
=
m1 + m2 +
i
mi xi
M
with M =
mi
(7.17)
mi
(7.18)
i
Likewise the y coordinate of the center of mass is
ycm =
m 1 y1 + m 2 y2 +
=
m1 +
103
7.1. THE IMPORTANT STUFF
v1i
v2i = 0
m1
m2
v2f
v1f
m1
m2
(b)
(a)
Figure 7.2: Elastic collision in one dimension. (a) Mass m1 has initial velocity v1i and mass m2 is at rest.
(b) Final velocities of the two masses are v1f and v2f . Formulae for these v
104
CHAPTER 7. MOMENTUM
If we consider the case where m2 is enormous compared to m1, then in the solution for
v1f we can replace (m1 m2) by m2 and (m1 + m2) by m2 , giving
v1f =
m2
v1i = v1i
m2
so that mass m1 just reverses its motion. In this case, m2 wi
101
7.1. THE IMPORTANT STUFF
vAf
vBf
A
vBi
vAi
B
B
A
B
A
(a)
(b)
(c)
Figure 7.1: A collision between masses A and B. (a) Masses move freely with velocities vAi and vBi . (b)
For a brief time the masses exert a force on one another. The force changes their
102
CHAPTER 7. MOMENTUM
With this denition, then the left side of 7.10 is the initial value of P and the right side is
the nal value of P. So we have:
Pi = Pf
(7.12)
that is, the value of the total momentum stays the same, or as we say in physics, is cons
100
CHAPTER 7. MOMENTUM
then combining the two equations gives
Fxt = px
Or, dividing by t,
px
(7.2)
t
so that the force on a particle is equal to the rate of change of its momentum.
Eq. 7.2 gives us a denition for the average force acting on a particle. I
97
6.2. WORKED EXAMPLES
v
Rcosq
R
q
Figure 6.7: Boy at position given by has speed v and height R cos
This is a classic and somewhat challenging problem but it does not require any more
math than simple trig.
We focus on the point at which the loss of co