Engineering Mathematics: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Limits

Direct substitution
try x = a first; 0/0 means simplify
Factor and cancel
xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + a·xⁿ⁻² + … + aⁿ⁻¹)
Conjugate
(√x − √a)(√x + √a) = x − a
Limits at infinity
divide by the highest power of x; equal degrees give the ratio of leading coefficients
Special limit
limx→∞ (1 + a/x)bx = eab
L'Hôpital's rule
lim f/g = lim f′/g′ (0/0 or ∞/∞)

Derivatives

Power rule
(xⁿ)′ = n·xⁿ⁻¹
Product rule
(uv)′ = u′v + uv′
Quotient rule
(u/v)′ = (u′v − uv′)/v²
Chain rule
f(g(x))′ = f′(g(x))·g′(x)
Common derivatives
(eˣ)′ = eˣ, (ln x)′ = 1/x, (sin x)′ = cos x, (cos x)′ = −sin x

Method

  1. Identify the structure: sum, product, quotient or composition.
  2. Apply the power rule to each power of x.
  3. Use the product, quotient or chain rule where needed.
  4. Simplify, and evaluate at the point if asked.

Integrals

Power rule
∫xⁿ dx = xⁿ⁺¹/(n + 1) + C
Definite integral
∫ab f dx = F(b) − F(a)
Common integrals
∫eˣ dx = eˣ, ∫1/x dx = ln|x|, ∫sin x dx = −cos x
Area between curves
∫ab (f − g) dx

Method

  1. Rewrite the integrand as a sum of simple terms where possible.
  2. Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for ex, 1/x, sin and cos.
  3. Use substitution when you see a function and its derivative together.
  4. For a definite integral, evaluate F(b) − F(a).

Complex numbers

Rectangular form
z = a + jb, j² = −1, z̄ = a − jb
Modulus and argument
|z| = √(a² + b²), tan θ = b/a (check the quadrant)
Polar and exponential form
z = r∠θ = r(cos θ + j sin θ) = r·ejθ
Division
z₁/z₂ = z₁z̄₂/|z₂|²
Product and quotient
r₁∠θ₁ · r₂∠θ₂ = r₁r₂∠(θ₁ + θ₂), r₁∠θ₁ / r₂∠θ₂ = (r₁/r₂)∠(θ₁ − θ₂)
De Moivre
(r∠θ)ⁿ = rⁿ∠nθ
n-th roots
zk = R1/n∠((φ + 360°·k)/n), k = 0 … n − 1
Quadratic with p² < 4q
z = −p/2 ± j√(q − p²/4)

Method

  1. Decide which form suits the operation: rectangular a + jb for adding and subtracting, polar r∠θ for multiplying, dividing, powers and roots.
  2. Multiply out with j² = −1; divide by multiplying the top and bottom by the conjugate of the denominator.
  3. Convert with r = √(a² + b²) and θ from a sketch of the point (check the quadrant), and back with a = r cos θ, b = r sin θ.
  4. For powers use De Moivre, (r∠θ)ⁿ = rⁿ∠nθ; for zⁿ = R∠φ list all n roots R1/n∠((φ + 360°k)/n).

Differential equations

Exponential growth/decay
dy/dt = ky → y = y0ekt
Half-life
k = ln 2 / t½
Newton's law of cooling
T(t) = Ta + (T0 − Ta)e−kt
Integrating factor
y′ + p(x)y = q(x): μ = e∫p dx, (μy)′ = μq
Mixing tank
dm/dt = r·c − (r/V)·m

Matrices & determinants

2×2 determinant
det = ad − bc
2×2 inverse
A⁻¹ = (1/det)·[d −b; −c a]
Product
(AB)ij = Σk AikBkj
Eigenvalues
det(A − λI) = 0
Cramer’s rule
xi = det(Ai)/det(A)

Vectors: dot & cross product

Magnitude
|a| = √(ax² + ay² + az²)
Unit vector
û = a/|a|
Components from an angle
Fx = F·cos θ, Fy = F·sin θ
Dot product
a·b = axbx + ayby + azbz = |a||b|cos θ
Scalar projection of a onto b
a·b/|b|
Work
W = F·d
Cross product
a × b = (aybz − azby, azbx − axbz, axby − aybx)
Areas
parallelogram |a × b|, triangle ½|a × b|
Moment about O
MO = r × F, r = A − O
Volume of a parallelepiped
V = |a·(b × c)|

Method

  1. Write every vector in components; for a force given by a magnitude and an angle, use F·cos θ and F·sin θ.
  2. For lengths and directions use the magnitude √(x² + y² + z²) and the unit vector a/|a|.
  3. Use the dot product for angles, projections, perpendicularity and work: a·b = |a||b|cos θ.
  4. Use the cross product for perpendicular vectors, areas and moments: write the three components carefully, keeping the order a × b.

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

2Practice problems

1.LimitsMedium
xyy = L

Find the limit L = limx→∞ (1 − 1/x)x. Give a decimal value.

f(x) = (1 − 1/x)xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (e−x − 1 + x)/x².

f(x) = (e−x − 1 + x)/x²x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = x·sin(2x), with x in radians. Find f′(1).

f(x) = x·sin(2x)x = 1 rad
f′(1) =
4.DerivativesHard
xyx = 2

The tangent line to f(x) = −x2 − 3x − 5 at x = 2 crosses the y-axis at y = b. Find b.

f(x) = −x2 − 3x − 5x = 2
b =
5.IntegralsMedium
xy2y = 0.5x

The region under y = 0.5x from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 0.5x0 ≤ x ≤ 2
V =
6.IntegralsHard
xy−33

Find the area A of the region between the parabola y = 9 − x² and the x-axis.

y = 9 − x²
A =
7.Complex numbersMedium
ReIm11−1−122−2−233−3−344−4−455−5−566−6−677−7−788−8−8zABCD

The diagram shows z = 2 − j3. With w = 0.5∠−90°, click the point that represents z/w.

z = 2 − j3w = 0.5∠−90°
  1. 1A (4 − j6)
  2. 2B (−6 − j4)
  3. 3C (3 + j2)
  4. 4D (6 + j4)
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

One root of z³ − 13z² + 65z − 125 = 0 is z₁ = 4 + j3. The coefficients are real. Find the real root r.

z₁ = 4 + j3
r =
9.Differential equationsMedium
tyy(0) = 3y(1) = ?

Solve y″ + 4y′ + 3y = 0 with y(0) = 3 and y′(0) = -4. Find y(1).

y″ + 4y′ + 3y = 0y(0) = 3y′(0) = -4
y(1) =
10.Differential equationsHard
tyK = 10ymax​ = ?

A system obeys y″ + 4y′ + 29y = 290 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + 4y′ + 29y = 290y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
A =−243−54−2−304

Find the determinant det A of the 3×3 matrix A shown.

A = [−2 4 3; −5 4 −2; −3 0 4]
det A =
12.Matrices & determinantsHard
A =336−3b =−10−11

Solve the system 3x + 3y = −10 and 6x − 3y = −11. Find x.

3x + 3y = −106x − 3y = −11
x =
13.Vectors: dot & cross productMedium
xyzab

Find the dot product a·b of a = (4, −7, −4) and b = (−7, −4, −7).

a = (4, −7, −4)b = (−7, −4, −7)
a·b =
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (−2, 4, −4), b = (2, −2, 5) and c = (4, −4, −5).

a = (−2, 4, −4)b = (2, −2, 5)c = (4, −4, −5)
V =
15.ProbabilityMedium
xp(x)00.210.2520.2530.140.2μ = 1.85

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3, 4p(x): 0.2, 0.25, 0.25, 0.1, 0.2
σ =
16.ProbabilityHard
n = 5, p = 0.7: P(X = 3) = ?k012345

A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 3 correct answers?

n = 5p = 0.7k = 3
P =
17.Descriptive statisticsMedium
15202530354045

Find the sample standard deviation s of the data: 49, 44, 27, 12, 48, 12.

n = 6
s =
18.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 12. Find the z-score z of a score of 164.

μ = 170σ = 12x = 164
z =
19.Normal distributionMedium
174.5186.4P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 174.5 and 186.4 cm? Give P as a decimal.

μ = 178σ = 7x1 = 174.5x2 = 186.4
P =
20.Normal distributionHard
557085100115130145pointsbottom 95%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 95% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points95%
x = points

3Solutions

1.Limits

Answer: L = 0.3679

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

Take the logarithm

ln L = limx→∞ x·ln(1 − 1/x)

Evaluate

ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.

ln L = 1·(−1) = −1

Undo the logarithm

L = e−1 = 0.3679

Answer

L = 0.3679
2.Limits

Answer: L = 0.5

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

First application

(−1·e−x + 1)/(2x) → 0/0

Second application

1·e−x/2

Substitute x = 0

L = 1·1/2 = 0.5

Answer

L = 0.5
3.Derivatives

Answer: f′(1) = 0.077

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Product and chain rule

f′(x) = sin(2x) + 2x·cos(2x)

Substitute

f′(1) = 0.9093 + 2·(−0.4161) = 0.077

Answer

f′(1) = 0.077
4.Derivatives

Answer: b = −1

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Point and slope

f(2) = −15
f′(x) = −2x − 3 → f′(2) = −7

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −15 − (−7)·2 = −1

Answer

b = −1
5.Integrals

Answer: V = 2.094

Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.

Disc method

V = π∫₀2 0.25x² dx

Integrate

V = π·0.25·2³/3 = 2.094

Answer

V = 2.094

Check: a cone of radius 1 and height 2 has (1/3)πr²h = 2.094.

6.Integrals

Answer: A = 36

Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.

Zeros

9 − x² = 0 → x = ±3

Integrate

A = [9x − x³/3]−33 = 18 − (−18) = 36

Answer

A = 36
7.Complex numbers

Answer: D (6 + j4)

Polar form: z = r∠θ = r(cos θ + j sin θ) = r·ejθ, with r = |z| and θ = arg z. Back to rectangular form: a = r cos θ, b = r sin θ. The argument needs the quadrant: θ = atan(b/a) only when a > 0; for a < 0 add or subtract 180°. In polar form multiplication is easy: multiply the moduli and add the arguments; for division divide the moduli and subtract the arguments.

Effect of w

|z/w| = |z|/0.5, arg(z/w) = arg z − (−90)°

The point

z/w = 6 + j4 → D

Answer

D (6 + j4)
8.Complex numbers

Answer: r = 5

De Moivre and roots: (r∠θ)ⁿ = rⁿ∠nθ. The equation zⁿ = R∠φ has exactly n roots, all with modulus R1/n, at the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1: they lie evenly spaced on a circle, 360°/n apart. A quadratic z² + pz + q = 0 with p² < 4q has the complex roots z = −p/2 ± j√(q − p²/4); a polynomial with real coefficients always has its complex roots in conjugate pairs a ± jb.

Conjugate pair

z₂ = 4 − j3
(z − z₁)(z − z₂) = z² − 2·4·z + (4² + 3²) = z² − 8z + 25

Factor out the quadratic

z³ − 13z² + 65z − 125 = (z² − 8z + 25)(z − 5)

Real root

r = 5

Answer

r = 5

Check with the sum of the roots: z₁ + z₂ + r = 8 + 5 = 13 = −(−13).

9.Differential equations

Answer: y(1) = 0.9446

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.

Characteristic roots

(r + 1)(r + 3) = 0 → r = −1, −3

Constants from the initial values

A + B = 3, −1A − 3B = -4
A = 2.5, B = 0.5

Evaluate

y(1) = 2.5·e−1 + 0.5·e−3 = 0.9446

Answer

y(1) = 0.9446
10.Differential equations

Answer: ymax = 12.85

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.

Steady value and roots

yss = 10, r = −2 ± 5j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10
y′(0) = −2A + 5B = 0 → B = −4

Peak

t = π/5 = 0.6283
ymax = 10·(1 + e−2·π/5) = 12.85

Answer

ymax = 12.85
11.Matrices & determinants

Answer: det A = 108

Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.

2×2 minors of the first row

M₁₁ = 16, M₁₂ = −26, M₁₃ = 12

Cofactor expansion

det A = (−2)·16 − 4·(−26) + 3·12 = 108

Answer

det A = 108
12.Matrices & determinants

Answer: x = −2.333

Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.

Determinant of the coefficient matrix

det A = 3·(−3) − 3·6 = −27

Replace the x column

det Aₓ = (−10)·(−3) − 3·(−11) = 63
x = 63/(−27) = −2.333

Answer

x = −2.333
y = −1
13.Vectors: dot & cross product

Answer: a·b = 28

Dot product: a·b = axbx + ayby + azbz = |a|·|b|·cos θ, a scalar. It gives the angle between two vectors, cos θ = a·b/(|a|·|b|), and it is zero when they are perpendicular. The scalar projection of a onto b is a·b/|b|, and a constant force F moving through a displacement d does work W = F·d.

Multiply matching components and add

a·b = 4·(−7) + (−7)·(−4) + (−4)·(−7) = (−28) + 28 + 28 = 28

Answer

a·b = 28

Positive: the angle between the vectors is less than 90°.

14.Vectors: dot & cross product

Answer: V = 60

Cross product: a × b = (aybz − azby, azbx − axbz, axby − aybx) is perpendicular to both vectors, and |a × b| = |a|·|b|·sin θ is the area of the parallelogram they span (the triangle has half of it). Order matters: b × a = −(a × b). The moment of a force F applied at point A about point O is MO = r × F, with r = A − O.

Cross product b × c

(b × c)x = bycz − bzcy = (−2)·(−5) − 5·(−4) = 30
(b × c)y = bzcx − bxcz = 5·4 − 2·(−5) = 30
(b × c)z = bxcy − bycx = 2·(−4) − (−2)·4 = 0

Dot product with a

a·(b × c) = (−2)·30 + 4·30 + (−4)·0 = 60
V = |60| = 60

Answer

V = 60
15.Probability

Answer: σ = 1.388

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Mean

μ = 0·0.2 + 1·0.25 + 2·0.25 + 3·0.1 + 4·0.2 = 1.85

Mean of the squares

E(X²) = 0·0.2 + 1·0.25 + 4·0.25 + 9·0.1 + 16·0.2 = 5.35

Variance

Var(X) = 5.35 − 1.85² = 1.927
σ = √1.927 = 1.388

Answer

σ = 1.388
16.Probability

Answer: P = 0.3087

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

Binomial formula

P(X = 3) = 10·0.73·0.32 = 10·0.343·0.09 = 0.3087

Answer

P = 0.3087
17.Descriptive statistics

Answer: s = 17.4

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Mean

x̄ = 192/6 = 32

Squared deviations

Σ(x − x̄)² = 17² + 12² + (−5)² + (−20)² + 16² + (−20)² = 1514

Standard deviation

s = √(1514/5) = 17.4

Answer

s = 17.4
18.Descriptive statistics

Answer: z = −0.5

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Standardise

z = (164 − 170)/12 = −0.5

Answer

z = −0.5
19.Normal distribution

Answer: P = 0.5764

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z-scores

z₁ = (174.5 − 178)/7 = −0.5
z₂ = (186.4 − 178)/7 = 1.2

Area between

P = Φ(1.2) − Φ(−0.5) = 0.8849 − 0.3085 = 0.5764

Answer

P = 0.5764
20.Normal distribution

Answer: x = 124.6 points

From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.

Area to the left

Φ(z) = 0.95

z from the table

z ≈ 1.64

Back to the original units

x = 100 + 1.64·15 = 124.6 points

Answer

x = 124.6 points

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