Download FE Math Review

FE Mathematics Review
Dr. Scott Molitor
Associate Professor
Undergraduate Program Director
Department of Bioengineering
Topics covered

Analytic geometry
–
–
–

Differential equations
Solution and applications
Laplace transforms
Difference equations and Z
transforms
– Numerical methods

Complex numbers
Matrix arithmetic and
determinants
– Vector arithmetic and
applications
– Progressions and series
– Numerical methods for finding
solutions of nonlinear equations
Integrals and applications
Numerical methods
–
–
–
Differential calculus
–
–
Derivatives and applications
Limits and L’Hopital’s rule
Algebra
–
–
Equations of lines and curves
Distance, area and volume
Trigonometric identities
Integral calculus
–
–



Probability and statistics
–
–
–
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–
–
–
Rules of probability
Combinations and permutations
Statistical measures (mean,
S.D., etc.)
Probability density and
distribution functions
Confidence intervals
Hypothesis testing
Linear regression
Tips for taking exam

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)
Equations of lines

What is the general form of
the equation for a line whose
x-intercept is 4 and yintercept is -6?
– (A) 2x – 3y – 18 = 0
– (B) 2x + 3y + 18 = 0
– (C) 3x – 2y – 12 = 0
– (D) 3x + 2y + 12 = 0
Equations of lines

What is the general form of
the equation for a line whose
x-intercept is 4 and yintercept 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  y1  6  0 3
m 2


x 2  x1
04
2
b  6
3
y  x 6
2
2  y  3  x  12
0  3  x  2  y  12
Equations of lines

What is the general form of
the equation for a line whose
x-intercept is 4 and yintercept 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)
(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
Trigonometry

For some angle q, csc q = 8/5. What is cos 2q?
– (A) 7/32
– (B) 1/4
– (C) 3/8
– (D) 5/8
Trigonometry

For some angle q, csc q = 8/5. What is cos 2q?
– (A) 7/32
– (B) 1/4
– (C) 3/8
– (D) 5/8

Use trigonometric identities
on handbook page 5

Answer is (A)

Confirm with calculator
– First find q = csc-1(-8/5)
– Then find cos 2q
1
sin q
cos 2q  1  2  sin 2 q
csc q 
1
csc 2 q
52
25
cos 2q  1  2  2  1  2 
8
64
25 7
cos 2q  1 

32 32
cos 2q  1  2 
Polar coordinates

What is rectangular form of
the polar equation r2 = 1 –
tan2 q?
– (A) –x2 + x4y2 + y2 = 0
– (B) x2 + x2y2 - y2 - y4 = 0
– (C) –x4 + y2 = 0
– (D) x4 – x2 + x2y2 + y2 = 0
Polar coordinates

What is rectangular form of
the polar equation r2 = 1 –
tan2 q?
– (A) –x2 + x4y2 + y2 = 0
– (B) x2 + x2y2 - y2 - y4 = 0
– (C) –x4 + y2 = 0
– (D) x4 – x2 + x2y2 + y2 = 0

Polar coordinate identities on
handbook page 5

Answer is (D)
r  x 2  y2
y
q  tan 1 ( )
x
r 2  1  tan 2 q
y
( x  y )  1  tan (tan ( ))
x
y2
2
2
x  y  1 2
x
x 4  x 2 y2  x 2  y2
2
2 2
2
x 4  x 2  x 2 y2  y2  0
1
Matrix identities

For three matrices A, B and C,
which of the following
statements is not necessarily
true?
– (A) A + (B + C) = (A + B) + C
– (B) A(B + C) = AB + AC
– (C) (B + C)A = AB + AC
– (D) A + (B + C) = C + (A + B)
Matrix identities

For three matrices A, B and C,
which of the following
statements is not necessarily
true?

Should know (A) and (D) are
true from linear algebra

Answer (B) appears as
matrix identity in handbook
page 7

Therefore can eliminate (C)
as being true
– (A) A + (B + C) = (A + B) + C
– (B) A(B + C) = AB + AC
– (C) (B + C)A = AB + AC
– (D) A + (B + C) = C + (A + B)

Matrix identities on handbook
page 7

Answer is (C)
Vector calculations

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 calculations

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 page 6

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
Vector calculations

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 page 6

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
Aside: why is the answer zero? A dot
product is only zero when two vectors A
and (B x C) are perpendicular. But this is
the case! A and C are parallel (A = 2C),
and (B x C) is perpendicular to C, hence
perpendicular to A!
Geometric progression

The 2nd and 6th terms of a
geometric progression are
3/10 and 243/160. What is
the first term of the
sequence?
– (A) 1/10
– (B) 1/5
– (C) 3/5
– (D) 3/2
Geometric progression

The 2nd and 6th terms of a
geometric progression are
3/10 and 243/160. What is
the first term of the
sequence?
– (A) 1/10
– (B) 1/5
– (C) 3/5
– (D) 3/2

Geometric progression on
handbook page 7

Answer is (B)
l n  ar n 1
3
243
, l6 
10
160
l 6 ar 5
243 / 160 81

 r4 

l 2 ar
3 / 10
16
l2 
81 3

16 2
3 3
l2  a  
2 10
1
l1  a 
5
r4
Geometric progression

The 2nd and 6th terms of a
geometric progression are
3/10 and 243/160. What is
the first term of the
sequence?
– (A) 1/10
– (B) 1/5
– (C) 3/5
– (D) 3/2

Geometric progression on
handbook page 7

Answer is (B)
l n  ar n 1
3
243
, l6 
10
160
l 6 ar 5
243 / 160 81

 r4 

l 2 ar
3 / 10
16
l2 
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.
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
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 page 13

Answer is (D)
x n 1  x n 
f (x n )
f ( x n )
f ( x )  ( x  2) 2  1
f ( x )  2  ( x  2)
x1  9.33
(9.33  2) 2  1
x 2  9.33 
2  (9.33  2)
52.73
x 2  9.33 
 5.73
14.66
(5.73  2) 2  1
x 3  5.73 
2  (5.73  2)
12.91
x 3  5.73 
 4 .0
7.46
Limits

What is the limit of (1 – e3x) /
4x as x  0?
– (A) -∞
– (B) -3/4
– (C) 0
– (D) 1/4
Limits

What is the limit of (1 – e3x) /
4x as x  0?
– (A) -∞
– (B) -3/4
– (C) 0
– (D) 1/4

L’Hopital’s rule on handbook
page 8

Answer is (B)
1  e3x 1  e30 1  1 0
lim


 ?
x 0
4x
40
0
0
f (x) 0
f ' (x)
if lim
 , try lim
x 0 g ( x )
x 0 g ' ( x )
0
1  e3x
 3e3 x
lim
 lim
x 0
x 0
4x
4
 3e3 x  3 1
3
lim


x 0
4
4
4
Limits

What is the limit of (1 – e3x) /
4x as x  0?
– (A) -∞
– (B) -3/4
– (C) 0
– (D) 1/4

L’Hopital’s rule on handbook
page 8

Answer is (B)
1  e3x 1  e30 1  1 0
lim


 ?
x 0
4x
40
0
0
f (x) 0
f ' (x)
if lim
 , try lim
x 0 g ( x )
x 0 g ' ( x )
0
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.
You can also use your calculator to
confirm the answer, substitute a small
value of x = 0.01 or 0.001.
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?
– (A) 0.034 m3 / min
– (B) 0.52 m3 / min
– (C) 0.63 m3 / min
– (D) 0.84 m3 / min
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?
–
–
–
–
(A) 0.034 m3 / min
(B) 0.52 m3 / min
(C) 0.63 m3 / min
(D) 0.84 m3 / min

Derivatives on handbook
page 9; volume of sphere on
handbook page 10

Answer is (C)
4 3
r
3
dV dr


dr dt
dr
 4r 2 
dt
V(r ) 
dV
dt
dV
dt
dV
m
 4  0.5m 2  0.2
dt
min
dV
m3
 0.63
dt
min
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?
–
–
–
–
(A) 0.034 m3 / min
(B) 0.52 m3 / min
(C) 0.63 m3 / min
(D) 0.84 m3 / min

Derivatives on handbook
page 9; volume of sphere on
handbook page 10

Answer is (C)
4 3
r
3
dV dr


dr dt
dr
 4r 2 
dt
V(r ) 
dV
dt
dV
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.
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
Evaluating integrals

Evaluate the indefinite
integral of f(x) = cos2x sin x
– (A) -2/3 sin3x + C
v   cos x
– (B) -1/3 cos3x + C
– (C) 1/3 sin3x + C
– (D) 1/2
sin2x
cos2x
u  cos 2 x
du  2  cos x  sin x  dx
dv  sin x  dx
 u  dv  u  v   v  du
 cos x  sin x  dx   cos x   2  cos
3   cos x  sin x  dx   cos x
2
+C

Apply integration by parts on
handbook page 9

Answer is (B)
3
2
3
1
2
cos
x

sin
x

dx


cos 3 x

3
2
x  sin x  dx
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
( sin 3 x  C)  2  sin 2 x  cos x
dx 3
d
1
(B)
( cos 3 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
( D)
( sin 2 x  cos 2 x  C)  sin x  cos 3 x  sin 3 x  cos x
dx 2
(A)
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
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]?
2
/2
– (A) 1
– (B) 2
– (C) 3
– (D) 4


Need absolute value
because sin x is negative
over interval [, 2]
Answer is (C)
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
Differential equations

What is the general solution
to the differential equation
y’’ – 8y’ + 16y = 0?
– (A) y = C1e4x
– (B) y = (C1 + C2x)e4x
– (C) y = C1e-4x + C1e4x
– (D) y = C1e2x + C2e4x
Differential equations

What is the general solution
to the differential equation
y’’ – 8y’ + 16y = 0?
– (A) y = C1
e4x
– (B) y = (C1 + C2x)e4x
y  8 y  16 y  0
y  2  4 y  16 y  0
a  4, b  16
r 2  2  4r  16r  0
– (C) y = C1e-4x + C1e4x
r  4  4 2  16  4
– (D) y = C1e2x + C2e4x
y  (C1  C 2 x )  e 4 x

Solving 2nd order differential
eqns on handbook page 12

Answer is (B)
Differential equations

What is the general solution
to the differential equation
y’’ – 8y’ + 16y = 0?
– (A) y = C1
e4x
– (B) y = (C1 + C2x)e4x


y  8 y  16 y  0
y  2  4 y  16 y  0
a  4, b  16
r 2  2  4r  16r  0
– (C) y = C1e-4x + C1e4x
r  4  4 2  16  4
– (D) y = C1e2x + C2e4x
y  (C1  C 2 x )  e 4 x
Solving 2nd order differential
eqns on handbook page 12
In this case, working backwards could
give an incorrect answer because answer
Answer is (B)
(A) would also work. The answer in (B)
is the sum of two terms that would satisfy
the differential equation, one of these
terms is the same as answer (A).
Laplace transforms

Find the Laplace transform
of the equation f”(t) + f(t) =
sin bt where f(0) and f’(0) = 0
(A) F(s) = b / [(1 + s2)(s2 + b2)]
– (B) F(s) = b / [(1 + s2)(s2 - b2)]
– (C) F(s) = b / [(1 - s2)(s2 + b2)]
– (D) F(s) = s / [(1 - s2)(s2 + b2)]
–
Laplace transforms

Find the Laplace transform
of the equation f”(t) + f(t) =
sin bt where f(0) and f’(0) = 0
(A) F(s) = b / [(1 + s2)(s2 + b2)]
– (B) F(s) = b / [(1 + s2)(s2 - b2)]
– (C) F(s) = b / [(1 - s2)(s2 + b2)]
– (D) F(s) = s / [(1 - s2)(s2 + b2)]
–

Laplace transforms on
handbook page 174 (EECS
section)

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 bt  e 0 t sin bt 
b
s 2  b2
b
(s 2  1)  F(s)  2
s  b2
1
b
F(s)  2
 2
s  1 s  b2
s 2 F(s)  F(s) 
b
s 2  b2
Probability of an outcome

A marksman can hit a bull’seye 3 out of 4 shots. What is
the probability he will hit a
bull’s-eye with at least 1 of
his next 3 shots?
– (A) 3/4
– (B) 15/16
– (C) 31/32
– (D) 63/64
Probability of an outcome

A marksman can hit a bull’seye 3 out of 4 shots. What is
the probability he will hit a
bull’s-eye with at least 1 of
his next 3 shots?
–
–
–
–

(A) 3/4
(B) 15/16
(C) 31/32
(D) 63/64
Answer is (D)

Let H = hit, M = miss,
Prob(H) = ¾, Prob(M) = ¼

Use combinations for next
three shots

Find Prob(HMM + MHM +
MMH + HHM + ...)

Easier method: Prob(at least
one hit) = 1 – Prob(no hits)

1 – Prob(no hits) = 1 –
Prob(MMM)

Prob(MMM) = Prob(M)3 =
(1/4)3 = 1/64

Answer is 1 – 1/64 = 63/64
Normal distribution

Exam scores are distributed
normally with a mean of 73
and a standard deviation of
11. What is the probability of
finding a score between 65
and 80?
– (A) 0.4196
– (B) 0.4837
– (C) 0.5161
– (D) 0.6455
Normal distribution

Exam scores are distributed
normally with a mean of 73 and
a standard deviation of 11.
What is the probability of finding
a score between 65 and 80?
–
(A) 0.4196
– (B) 0.4837
– (C) 0.5161
– (D) 0.6455


Standard normal tables on
handbook page 20

Let X = a random score, find
Prob(65 < X < 80)
–
X is normally distributed with
mean 72 and S.D. 11

(65 – 72) / 11 = -0.73 ≈ -0.7

(80 – 72) / 11 = 0.64 ≈ 0.6

Prob(65 < X < 80) ≈ Prob(-0.7 <
Z < 0.6)

Convert Prob(-0.7 < Z < 0.6)
Prob(Z < 0.6) – Prob(Z < -0.7)
– Prob(Z < 0.6) – Prob(Z > 0.7)
– F(0.6) – R(0.7) from table
–
Answer is (B)

Prob(65 < X < 80) ≈ 0.7257 –
0.2420 = 0.4837
Confidence intervals

What is the 95% confidence
interval for the mean exam
score if the mean is 73 and
the standard deviation is 11
from 25 scores?
– (A) 73 ± 4.54
– (B) 73 ± 0.91
– (C) 73 ± 4.31
– (D) 73 ± 0.86
Confidence intervals

What is the 95% confidence
interval for the mean exam
score if the mean is 73 and
the standard deviation is 11
from 25 scores?



a = 1 – 0.95 = 0.05
– a/2 = 0.025
– n = 25 – 1 = 24 degrees of
freedom
– t0.025, 24 = 2.064 on page 21
– (B) 73 ± 0.91
– (C) 73 ± 4.31
– (D) 73 ± 0.86
Confidence intervals on
handbook page 19

ta values handbook page 21

Answer is (A)
s
Formula is X  t a / 2 
n
Look up ta/2, n
–
– (A) 73 ± 4.54

Use formula for population
standard deviation unknown

Calculate confidence interval
73 ± (2.064) (11) / √25

Answer is 73 ± 4.54
Hypothesis testing

You sample two lots of light bulbs
for mean lifetime. The first lot mean
= 792 hours, S.D. = 35 hours, n =
25. The second lot mean = 776
hours, S.D. = 24 hours, n = 20.
Determine with 95% confidence
whether light bulbs from the first lot
last longer than those from the
second lot. Provide a statistic
value.
–
–
–
–
(A) First lot lasts longer, t0 = -1.96
(B) No difference, z0 = 1.81
(C) No difference, t0 = 1.74
(D) First lot lasts longer, t0 = 1.96
Hypothesis testing

You sample two lots of light bulbs
for mean lifetime. The first lot mean
= 792 hours, S.D. = 35 hours, n =
25. The second lot mean = 776
hours, S.D. = 24 hours, n = 20.
Determine with 95% confidence
whether light bulbs from the first lot
last longer than those from the
second lot. Provide a statistic
value.
–
–
–
–
(A) First lot lasts longer, z0 = -1.96
(B) No difference, z0 = 1.81
(C) No difference, t0 = 1.74
(D) First lot lasts longer, t0 = 1.96

Hypothesis testing in IE section of
handbook page 198

ta values handbook page 21

Answer is (C)

Test H0: m1 = m2 vs. H1: m1 > m2
–

H0: m1 - m2 = 0 vs. H1: m1 - m2 > 0
Use formula for population standard
deviation or variance unknown
(25  1)  352  (20  1)  24 2
Sp 
 30.63
25  20  2
792  776
t0 
 1.74
30.63 1 25  1 20

Look up ta, n
–
–
–
a = 1 – 0.95 = 0.05
n = 25 + 20 – 2 = 43 d.o.f.
t0.05, 43 = 1.96 from page 21 (n > 29)

Accept null hypothesis since
statistic t0 < t0.05, 43

Bulbs from first lot do not last longer