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

Topics covered Analytic geometry Integral calculus 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 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 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)

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)

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? (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

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)

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 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)

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) 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)

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) 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)

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? (B) -3/4 (C) 0 (D) 1/4 L’Hopital’s rule on handbook page 8 Answer is (B)

Limits What is the limit of (1 – e3x) / 4x as x  0? (B) -3/4 (C) 0 (D) 1/4 L’Hopital’s rule on handbook page 8 Answer is (B) 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)

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) 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 (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)

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)

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 [p/2, 2p]? (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 [p/2, 2p]? (A) 1 (B) 2 (C) 3 (D) 4 Need absolute value because sin x is negative over interval [p, 2p] Answer is (C)

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 = 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)

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 Solving 2nd order differential eqns on handbook page 12 Answer is (B) 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 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)

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 ta values handbook page 21 Answer is (A) Use formula for population standard deviation unknown Formula is Look up ta/2, n 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 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 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