FE Mathematics Review Dr. Scott Molitor Associate Professor Undergraduate Program Director...

FE Mathematics Review

Dr. Scott Molitor

Associate Professor

Undergraduate Program Director

Department of Bioengineering

Topics covered Analytic geometry

– Equations of lines and curves– Distance, area and volume– Trigonometric identities

Integral calculus– Integrals and applications– Numerical methods

Differential equations– Solution and applications– Laplace transforms– Difference equations and Z

transforms– Numerical methods

Differential calculus– Derivatives and applications– Limits and L’Hopital’s rule

Algebra– Complex numbers– Matrix arithmetic and

determinants– Vector arithmetic and

applications– Progressions and series– Numerical methods for finding

solutions of nonlinear equations Probability and statistics

– 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 y-intercept 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


– (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)

12y2x30

12x3y2

6x2

3y

6b

2

3

40

06

xx

yym

bxmy

12

12

Equations of lines


– (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)

012)6(203)C(

024120243)D(

0120243)C(

026180342)B(

010180342)A(

Trigonometry

For some angle , csc = -8/5. What is cos 2?

– (A) 7/32– (B) 1/4– (C) 3/8– (D) 5/8

Trigonometry

For some angle , csc = -8/5. What is cos 2?

– (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 = csc-1(-8/5)– Then find cos 2

32

7

32

2512cos

64

2521

8

5212cos

csc

1212cos

sin212cos

sin

1csc

2

2

2

2

Polar coordinates

What is rectangular form of the polar equation r2 = 1 – tan2 ?

– (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 ?

– (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)0yyxxx

yxyxx

x

y1yx

))x

y((tantan1)yx(

tan1r

)x

y(tan

yxr

22224

22224

2

222

12222

22

1

22

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?

– (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)

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

Vector calculations

For three vectorsA = 6i + 8j + 10kB = i + 2j + 3kC = 3i + 4j + 5k, what is the product A·(B x C)?

– (A) 0– (B) 64– (C) 80– (D) 216

Vector calculations


– (A) 0– (B) 64– (C) 80– (D) 216

Vector products on handbook page 6

Answer is (A)

0)2(1048)2(6)CB(A

)k2j4i2()k10j8i6()CB(A

k2j4i2CB

)3241(k)3351(j)4352(iCB

543

321

kji

CB

Vector calculations


– (A) 0– (B) 64– (C) 80– (D) 216

Vector products on handbook page 6

Answer is (A)

0)2(1048)2(6)CB(A

)k2j4i2()k10j8i6()CB(A

k2j4i2CB

)3241(k)3351(j)4352(iCB

543

321

kji

CB

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



– (A) 1/10– (B) 1/5– (C) 3/5– (D) 3/2

Geometric progression on handbook page 7

Answer is (B) 5

1al

10

3

2

3al

2

3

16

81r

16

81

10/3

160/243r

ar

ar

l

l

160

243l,

10

3l

arl

1

2

4

45

2

6

62

1nn



– (A) 1/10– (B) 1/5– (C) 3/5– (D) 3/2

Geometric progression on handbook page 7

Answer is (B) 5

1al

10

3

2

3al

2

3

16

81r

16

81

10/3

160/243r

ar

ar

l

l

160

243l,

10

3l

arl

1

2

4

45

2

6

62

1nn

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)0.4

46.7

91.1273.5x

)273.5(2

1)273.5(73.5x

73.566.14

73.5233.9x

)233.9(2

1)233.9(33.9x

33.9x

)2x(2)x(f

1)2x()x(f

)x(f

)x(fxx

3

2

3

2

2

2

1

2

n

nn1n

Limits

What is the limit of (1 – e3x) / 4x as x 0?

– (A) -∞– (B) -3/4– (C) 0– (D) 1/4

Limits


– (A) -∞– (B) -3/4– (C) 0– (D) 1/4

L’Hopital’s rule on handbook page 8

Answer is (B)4

3

4

13

4

e3lim

4

e3lim

x4

e1lim

)x('g

)x('flimtry,

0

0

)x(g

)x(flimif

?0

0

0

11

04

e1

x4

e1lim

x3

0x

x3

0x

x3

0x

0x0x

03x3

0x

Limits


– (A) -∞– (B) -3/4– (C) 0– (D) 1/4

L’Hopital’s rule on handbook page 8

Answer is (B)4

3

4

13

4

e3lim

4

e3lim

x4

e1lim

)x('g

)x('flimtry,

0

0

)x(g

)x(flimif

?0

0

0

11

04

e1

x4

e1lim

x3

0x

x3

0x

x3

0x

0x0x

03x3

0x

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




Derivatives on handbook page 9; volume of sphere on handbook page 10

Answer is (C)

min

m63.0

dt

dV

min

m2.0m5.04

dt

dVdt

drr4

dt

dVdt

dr

dr

dV

dt

dV

r3

4)r(V

3

2

2

3




Derivatives on handbook page 9; volume of sphere on handbook page 10

Answer is (C)

min

m63.0

dt

dV

min

m2.0m5.04

dt

dVdt

drr4

dt

dVdt

dr

dr

dV

dt

dV

r3

4)r(V

3

2

2

3

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




Apply integration by parts on handbook page 9

Answer is (B)

xcos3

1dxxsinxcos

xcosdxxsinxcos3

dxxsinxcos2xcosdxxsinxcos

duvvudvu

xcosv

dxxsindv

dxxsinxcos2du

xcosu

32

32

232

2




Alternative method is to differentiate answers

Answer is (B)

xcosxsinxcosxsin)Cxcosxsin2

1(

dx

d)D(

xcosxsin)Cxsin3

1(

dx

d)C(

xsinxcos)Cxcos3

1(

dx

d)B(

xcosxsin2)Cxsin3

2(

dx

d)A(

3322

23

23

23

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]?

– (A) 1– (B) 2– (C) 3– (D) 4

Need absolute value because sin x is negative over interval [, 2]

Answer is (C)

/2

2

3)1(10)1(area

xcosxcosarea

dxxsindxxsinarea

dxxsinarea

22/

2

2/

2

2/

Differential equations

What is the general solution to the differential equationy’’ – 8y’ + 16y = 0?

– (A) y = C1e4x

– (B) y = (C1 + C2x)e4x

– (C) y = C1e-4x + C1e4x

– (D) y = C1e2x + C2e4x



– (A) y = C1e4x

– (B) y = (C1 + C2x)e4x

– (C) y = C1e-4x + C1e4x

– (D) y = C1e2x + C2e4x

Solving 2nd order differential eqns on handbook page 12

Answer is (B)

x421

2

2

e)xCC(y

41644r

0r16r42r

16b,4a

0y16y42y

0y16y8y



– (A) y = C1e4x

– (B) y = (C1 + C2x)e4x

– (C) y = C1e-4x + C1e4x

– (D) y = C1e2x + C2e4x

Solving 2nd order differential eqns on handbook page 12

Answer is (B)

x421

2

2

e)xCC(y

41644r

0r16r42r

16b,4a

0y16y42y

0y16y8y

In this case, working backwards could give an incorrect answer because answer (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 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

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 page 174 (EECS section)

Answer is (A)

222

222

222

22t0

2

22

s1s

1)s(F

s)s(F)1s(

s)s(F)s(Fs

stsinetsin

)s(F)t(f

)s(Fs)t(f

)0(fs)0(fs)s(Fs)t(f

Probability of an outcome

A marksman can hit a bull’s-eye 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’s-eye 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

Answer is (B)

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

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) 73 ± 4.54– (B) 73 ± 0.91– (C) 73 ± 4.31– (D) 73 ± 0.86

Confidence intervals on handbook page 19

t values handbook page 21

Answer is (A)

Use formula for population standard deviation unknown

Formula is

Look up t/2,

– = 1 – 0.95 = 0.05– /2 = 0.025– = 25 – 1 = 24 degrees of

freedom

– t0.025, 24 = 2.064 on page 21

Calculate confidence interval 73 ± (2.064) (11) / √25

Answer is 73 ± 4.54

n

stX 2/

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

t values handbook page 21

Answer is (C)

Test H0: 1 = 2 vs. H1: 1 > 2

– H0: 1 - 2 = 0 vs. H1: 1 - 2 > 0

Use formula for population standard deviation or variance unknown

Look up t,

– = 1 – 0.95 = 0.05– = 25 + 20 – 2 = 43 d.o.f.– t0.05, 43 = 1.96 from page 21 ( > 29)

Accept null hypothesis since statistic t0 < t0.05, 43

Bulbs from first lot do not last longer

74.120125163.30

776792t

63.3022025

24)120(35)125(S

0

22

p

FE Mathematics Review Dr. Scott Molitor Associate Professor Undergraduate Program Director...

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