Download Robotics 101: Starting a robotics project and the use of Lego

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FE
Mathematics
Review
Dr. Omar Meza
Assistant Professor
Department of Mechanical Engineering
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Topics covered

Analytic geometry
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Algebra
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Equations of lines and curves
Distance, area and volume
Trigonometric identities
Complex numbers
Matrix
arithmetic
and
determinants
Vector
arithmetic
and
applications
Progressions and series
Numerical
methods
for
finding solutions of nonlinear
equations
Differential calculus


Derivatives and applications
Limits and L’Hopital’s rule
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Integral calculus
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Integrals and applications
Numerical methods
Differential equations
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
Solution and applications
Laplace transforms
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Tips for taking exam
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Use the reference handbook
 Know what it contains
 Know what types of problems you can use it for
 Know how to use it to solve problems
 Refer to it frequently
Work backwards when possible
 FE exam is multiple choice with single correct answer
 Plug answers into problem when it is convenient to do so
 Try to work backwards to confirm your solution as often as
possible
Progress from easiest to hardest problem
 Same number of points per problem
Calculator tips
 Check the NCEES website to confirm your model is allowed
 Avoid using it to save time!
 Many answers do not require a calculator (fractions vs.
decimals)
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Equations of lines
Handbook page:
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Equations of lines
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Equations of lines
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Equations of lines

What is the general form of the equation for a line whose
x-intercept is 4 and y-intercept is -6?

(A) 2x – 3y – 18 = 0

(B) 2x + 3y + 18 = 0

(C) 3x – 2y – 12 = 0

(D) 3x + 2y + 12 = 0
-0-1-2-3-4-5-6-
1 2 3 4 5
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Equations of lines


What is the general form of
the equation for a line
whose x-intercept is 4 and
y-intercept is -6?

(A) 2x – 3y – 18 = 0

(B) 2x + 3y + 18 = 0

(C) 3x – 2y – 12 = 0

(D) 3x + 2y + 12 = 0
Try using standard form

Handbook pg 3: y = mx + b

Given (x1, y1) = (4, 0)

Given (x2, y2) = (0, -6)
Answer is (C)
y = m.x + b
y 2 - y1
-6 - 0 3
m=
=
=
x 2 - x1
0-4
2
b = -6
3
y = .x - 6
2
2.y = 3.x - 12
0 = 3.x - 2.y - 12
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Equations of lines


What is the general form
of the equation for a line
whose x-intercept is 4 and
y-intercept is -6?

(A) 2x – 3y – 18 = 0

(B) 2x + 3y + 18 = 0

(C) 3x – 2y – 12 = 0

(D) 3x + 2y + 12 = 0
Work backwards

Substitute (x1, y1) = (4, 0)

Substitute (x2, y2) = (0, -6)

See what works
Answer is (C)
Alternative Solution
(A) 2  4  3  0  18  10  0
(B) 2  4  3  0  18  26  0
(C) 3  4  2  0  12  0
(D) 3  4  2  0  12  24  0
(C) 3  0  2  (6)  12  0
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Equations of lines
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Equations of lines
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Equations of lines
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Quadratic Equation
Handbook page:
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Quadratic Equation
What are the roots of f(x) = 2x 2 + 5x - 3
A) 1, 2; B) 3, 2; C) 0.5,-3; D) -0.5, -3
Answer is (C)
Handbook page:
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Quadratic Equation
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Equations of curves
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Equations of curves
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Equations of curves
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Equations of curves
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Equations of curves
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Equations of curves
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Equations of curves
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Equations of curves
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Equations of curves
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Equations of curves
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Equations of curves
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Equations of curves
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Equations of curves
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Equations of curves
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Equations of curves
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Equations of curves
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Logarithms
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Logarithms
ln( 7.3891)xy
xy ln( 7.3891)
xy(2.000006)
≈2xy
( A ) 2 / xy
( B) 0.5xy
(C) 0.8686xy
( D) 2xy
Answer is (D)
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Logarithms
ln( 50)
ln 8 50 =
= 1.88
ln( 8)
Answer is (D)
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Trigonometry
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Trigonometry
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Trigonometry
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Trigonometry
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Trigonometry

For some angle , csc  = -8/5.
What is cos 2?

Use trigonometric identities
on handbook.

Confirm with calculator

First find  = csc-1(-8/5)

Then find cos 2
(A) 7/32
(B) 1/4
(C) 3/8
(D) 5/8
Answer is (A)
1
sin 
cos 2  1  2  sin 2 
csc  
1
csc 2 
52
25
cos 2  1  2  2  1  2 
8
64
25 7
cos 2  1 

32 32
cos 2  1  2 
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Trigonometry
cos 2 θ
( 2 )(sin 2 θ)+
sin θ
Answer is (C)
cos 2 θ + cos 2 θ
2 cos 2 θ
1
1
( 2 )
cos θ
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Trigonometry
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Complex Numbers
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Complex Numbers
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Complex Numbers
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Complex Numbers
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Polar coordinates
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Polar coordinates

What is rectangular form of
the polar equation r2 = 1 –
tan2 ?

(A) –x2 + x4y2 + y2 = 0
x2
+
x2y2
-
y2
-
y4

(B)
=0

(C) –x4 + y2 = 0

(D) x4 – x2 + x2y2 + y2 = 0
r = x2 + y2
y
y
θ = tan ( ), tan θ =
x
x
r 2 = 1 - tan 2 θ
1
y
( x + y ) = 1 - tan (tan ( ))
x
y2
2
2
x + y =1 - 2
x
x 4 + x2y2 = x2 - y2
2

Polar coordinate identities
on handbook
Answer is (D)
2
2
2
x 4 - x2 + x2y2 + y2 = 0
1
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Polar coordinates
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Matrices
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Matrices
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Matrices
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Matrices
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Matrices
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Matrices
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Matrices
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Matrices
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Matrices
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Matrices
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Vector
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Vector
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Vector
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Vector
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Vector calculations
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
For three vectors
A = 6i + 8j + 10k
B = i + 2j + 3k
C = 3i + 4j + 5k, what is the
product A·(B x C)?

(A) 0

(B) 64

(C) 80

(D) 216
Vector products on
handbook
Answer is (A)
i j k
B C  1 2 3
3 4 5
B  C  i(2  5  3  4)  j(1  5  3  3)  k (1 4  2  3)
B  C  2i  4 j  2k
A  (B  C)  (6i  8 j  10k )  (2i  4 j  2k )
A  (B  C)  6  (2)  8  4  10  (2)  0
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Vector calculations
Answer is (D)
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Vector calculations
(-16- 8)i – (-8+16)j + (2+8)k
-24i -8j + 10k
Answer is (A)
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Geometric Progression

The 2nd and 6th terms of a l  ar n 1
n
geometric progression are
3/10 and 243/160. What is l  3 , l  243
2
6
10
160
the first term of the
sequence?
l
ar 5
243 / 160
6


(A) 1/10

(B) 1/5

(C) 3/5

(D) 3/2
Geometric progression on
handbook
Answer is (B)
l2

r 
4
ar
3 / 10
81

16
81 3

16 2
3 3
l2  a  
2 10
1
l1  a 
5
r4
Confirm answer by calculating l2
and l6 with a = 1/5 and r = 3/2.
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Roots of nonlinear equations


Newton’s method is being
used to find the roots of the
equation f(x) = (x – 2)2 – 1.
Find the 3rd approximation
if the 1st approximation of
the root is 9.33

(A) 1.0

(B) 2.0

(C) 3.0

(D) 4.0
Newton’s method on
handbook
x n 1  x n 
f ( x )  ( x  2) 2  1
f ( x )  2  ( x  2)
x 1  9.33
x2
x2
x3
x3
Answer is (D)
f (x n )
f ( x n )
(9.33  2) 2  1
 9.33 
2  (9.33  2)
52.73
 9.33 
 5.73
14.66
(5.73  2) 2  1
 5.73 
2  (5.73  2)
12.91
 5.73 
 4 .0
7.46
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Application of derivatives
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Application of derivatives
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Application of derivatives
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Application of derivatives
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Application of derivatives
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Application of derivatives
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Application of derivatives
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Application of derivatives
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Application of derivatives
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Application of derivatives
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Limits


1  e3x 1  e30 1  1 0
What is the limit of (1 – e3x) /
lim


 ?
x 0
4x as x  0?
4x
40
0
0
 (A) -∞
f (x) 0
f ' (x)
if lim
 , try lim
x 0 g ( x )
x 0 g ' ( x )
 (B) -3/4
0

(C) 0

(D) 1/4
L’Hopital’s rule on
handbook
1  e3x
 3e3 x
lim
 lim
x 0
x 0
4x
4
 3e3 x  3 1
3
lim


x 0
4
4
4
You should apply L’Hopital’s rule
iteratively until you find limit of f(x)
/ g(x) that does not equal 0 / 0.
Answer is (B)
You can also use your calculator to
confirm the answer, substitute a
small value of x = 0.01 or 0.001.
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Application of derivatives

The radius of a snowball
rolling down a hill is
increasing at a rate of 20
cm / min. How fast is its
volume increasing when
the diameter is 1 m?
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



(A) 0.034 m3 / min
(B) 0.52 m3 / min
(C) 0.63 m3 / min
(D) 0.84 m3 / min
Derivatives on handbook;
volume of sphere on
handbook page 10
Answer is (C)
4 3
V(r )  r
3
dV dV dr


dt
dr dt
dV
dr
 4r 2 
dt
dt
dV
m
 4  0.5m 2  0.2
dt
min
dV
m3
 0.63
dt
min
Convert cm to m, convert diameter
to radius, and confirm final units
are correct.
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Evaluating integrals
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Evaluating integrals
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Evaluating integrals
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Evaluating integrals
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Evaluating integrals
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Evaluating integrals
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Evaluating integrals
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Evaluating integrals
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Evaluating integrals
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Evaluating integrals
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Evaluating integrals


Evaluate the indefinite
integral of f(x) = cos2x sin x

(A) -2/3 sin3x + C

(B) -1/3 cos3x + C

(C) 1/3 sin3x + C

(D) 1/2 sin2x cos2x + C
Apply integration by parts
on handbook
u  cos 2 x
du  2  cos x  sin x  dx
dv  sin x  dx
v   cos x
 u  dv  u  v   v  du
 cos x  sin x  dx   cos x   2  cos
3   cos x  sin x  dx   cos x
2
3
2
Answer is (B)
3
1 3
2
cos
x

sin
x

dx


cos x

3
2
x  sin x  dx
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Evaluating integrals


Evaluate the indefinite
integral of f(x) = cos2x sin x

(A) -2/3 sin3x + C

(B) -1/3 cos3x + C

(C) 1/3 sin3x + C

(D) 1/2 sin2x cos2x + C
Alternative method is to
differentiate answers
Answer is (B)
d 2 3
( sin x  C)  2  sin 2 x  cos x
dx 3
d 1 3
(B)
( cos x  C)  cos 2 x  sin x
dx 3
d 1 3
( C)
( sin x  C)  sin 2 x  cos x
dx 3
d 1 2
( D)
( sin x  cos 2 x  C)  sin x  cos3 x  sin 3 x  cos x
dx 2
(A)
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Applications of integrals


What is the area of the
curve bounded by the curve
f(x) = sin x and the x-axis
on the interval [/2, 2]?

(A) 1

(B) 2

(C) 3

(D) 4
Need
absolute
value
because sin x is negative
over interval [, 2]
Answer is (C)
2
/2
area  
2
/ 2
sin x  dx

2
/ 2

area   sin x  dx    sin x  dx
area   cos x  / 2  cos x 2
area  (1)  0  1  (1)  3
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Differential Equations
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Differential Equations
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Differential Equations
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Differential Equations
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Differential equations


What is the general solution
to the differential equation
y’’ – 8y’ + 16y = 0?
y  8 y  16 y  0
y  2  4 y  16 y  0

(A) y = C1e4x

(B) y = (C1 + C2x)e4x

(C) y = C1e-4x + C1e4x
r 2  2  4r  16r  0

(D) y = C1e2x + C2e4x
r  4  4 2  16  4
Solving
differential
handbook
2nd
eqns
Answer is (B)
a  4, b  16
order
on
y  (C1  C 2 x )  e 4 x
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Laplace transforms


Find the Laplace transform of
the equation f”(t) + f(t) = sin t
where f(0) and f’(0) = 0

(A) F(s) =  / [(1 + s2)(s2 + 2)]

(B) F(s) =  / [(1 + s2)(s2 - 2)]

(C) F(s) =  / [(1 - s2)(s2 + 2)]

(D) F(s) = s / [(1 - s2)(s2 + 2)]
Laplace transforms on
handbook
Answer is (A)
f ( t )  s 2 F(s)  s  f (0)  s 2  f (0)
f ( t )  s 2 F(s)
f ( t )  F(s)
sin t  e 0 t sin t 

s 2  2

(s 2  1)  F(s)  2
s  2
1

F(s)  2
 2
s  1 s  2
s 2 F(s)  F(s) 

s 2  2
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Preguntas? Comentarios?
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Muchas Gracias !