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→∞ (√(x2 + 6x) − x).

f(x) = √(x2 + 6x) − xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 5x)/sin(3x), with x in radians.

f(x) = ln(1 + 5x)/sin(3x)x → 0
L =
3.DerivativesMedium
xyx = 3

Let f(x) = ln(2x2 + 4). Find f′(3).

f(x) = ln(2x2 + 4)x = 3
f′(3) =
4.DerivativesHard
xy

Let f(x) = x3 − 9x2 + 15x − 5. Find the x-coordinate x of the local maximum.

f(x) = x3 − 9x2 + 15x − 5
x =
5.IntegralsMedium
xy01

Evaluate I = ∫01 2x·(x² + 1)3 dx.

f(x) = 2x·(x² + 1)3[0, 1]
I =
6.IntegralsHard
xyy = x2​y = 2x

Find the area A of the region enclosed by y = x2 and y = 2x.

y = x2y = 2x
A =
7.Complex numbersMedium
ReIm55−5−5θz

Find the argument θ of z = −6 − j7 in degrees, as the principal value between −180° and 180°.

z = −6 − j7
θ = °
8.Complex numbersHard
ReIm22−2−244−4−4ABCD

Exactly one of the points A–D is a root of z³ = 27∠0°. Click it.

w = 27∠0°n = 3
  1. 1A (3∠60°)
  2. 2B (3∠−150°)
  3. 3C (5.4∠−120°)
  4. 4D (3∠−120°)
9.Differential equationsMedium
tyy(0) = 1y(1.5) = ?

Solve y″ + 5y′ + 6y = 0 with y(0) = 1 and y′(0) = 4. Find y(1.5).

y″ + 5y′ + 6y = 0y(0) = 1y′(0) = 4
y(1.5) =
10.Differential equationsHard
tyK = 10ymax​ = ?

A system obeys y″ + 2y′ + 10y = 100 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + 2y′ + 10y = 100y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
A =2−400−11−4−42

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

A = [2 −4 0; 0 −1 1; −4 −4 2]
det A =
12.Matrices & determinantsHard
A =11−2−5b =16

Solve the system x + y = 1 and −2x − 5y = 6. Find x.

x + y = 1−2x − 5y = 6
x =
13.Vectors: dot & cross productMedium
xyzab

Find the value of k for which the vectors a = (−3, 5, −1) and b = (k, −3, 4) are perpendicular.

a = (−3, 5, −1)b = (k, −3, 4)
k =
14.Vectors: dot & cross productHard
xyzABC

Find the area A of the triangle with corners A(1, 3, −4), B(2, 1, −3) and C(−1, 4, −2). Coordinates are in metres.

A(1, 3, −4)B(2, 1, −3)C(−1, 4, −2)
A = m²
15.ProbabilityMedium
xp(x)100.25200.15300.3400.3μ = 26.5

For the distribution shown, find the variance Var(X) of X.

x: 10, 20, 30, 40p(x): 0.25, 0.15, 0.3, 0.3
Var(X) =
16.ProbabilityHard
n = 11, p = 0.6: P(X = 5) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 5 correct answers?

n = 11p = 0.6k = 5
P =
17.Descriptive statisticsMedium
2025303540

Find the sample standard deviation s of the data: 18, 24, 20, 40, 26.

n = 5
s =
18.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 50 and standard deviation 15. Find the z-score z of a score of 72.5.

μ = 50σ = 15x = 72.5
z =
19.Normal distributionMedium
171.35179.4P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 171.35 and 179.4 cm? Give P as a decimal.

μ = 178σ = 7x1 = 171.35x2 = 179.4
P =
20.Normal distributionHard
μ = 32x = ?5%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 5% of them lie (the 5th percentile).

μ = 32σ = 6P = 5%
x = min

3Solutions

1.Limits

Answer: L = 3

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).

Multiply by the conjugate

f(x) = 6x/(√(x2 + 6x) + x)

Divide by x

f(x) = 6/(√(1 + 6/x) + 1)

Let x → ∞

L = 6/(1 + 1) = 3

Answer

L = 3
2.Limits

Answer: L = 1.667

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).

L'Hôpital's rule

L = limx→0 [5/(1 + 5x)]/[3·cos 3x]

Substitute x = 0

L = 5/3 = 1.667

Answer

L = 1.667
3.Derivatives

Answer: f′(3) = 0.5455

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.

Chain rule for the logarithm

f′(x) = 4x/(2x2 + 4)

Substitute

f′(3) = 12/22 = 0.5455

Answer

f′(3) = 0.5455
4.Derivatives

Answer: x = 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.

Derivative

f′(x) = 3x2 − 18x + 15 = 3·(x2 − 6x + 5)

Solve f′(x) = 0

x = 1 or x = 5

Which one is which

f″(x) = 6x − 18: negative at a maximum, positive at a minimum.

f″(1) = −12, f″(5) = 12
x = 1

Answer

x = 1
5.Integrals

Answer: I = 3.75

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.

Substitution

u = x² + 1, du = 2x dx
I = ∫12 u3 du

Integrate and evaluate

I = [u4/4]12 = (16 − 1)/4 = 3.75

Answer

I = 3.75
6.Integrals

Answer: A = 1.333

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.

Intersection points

x2 = 2x → x = 0, x = 2

Integrate upper minus lower

A = [x2 − 0.3333x3]02 = 1.333

Answer

A = 1.333
7.Complex numbers

Answer: θ = −130.6 °

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.

Quadrant

a = −6, b = −7 → III

Reference angle

α = atan(7/6) = 49.4°

Argument

θ = −(180° − α) = −130.6°

Answer

θ = −130.6°

In polar form z = 9.22∠−130.6°.

8.Complex numbers

Answer: D (3∠−120°)

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.

Modulus and arguments of the roots

|z| = 3
θ = 0°, 120°, −120°

The point

D = 3∠−120°

Answer

D (3∠−120°)
9.Differential equations

Answer: y(1.5) = 0.2819

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 + 2)(r + 3) = 0 → r = −2, −3

Constants from the initial values

A + B = 1, −2A − 3B = 4
A = 7, B = −6

Evaluate

y(1.5) = 7·e−3 + (−6)·e−4.5 = 0.2819

Answer

y(1.5) = 0.2819
10.Differential equations

Answer: ymax = 13.51

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 = −1 ± 3j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10
y′(0) = −1A + 3B = 0 → B = −3.333

Peak

t = π/3 = 1.047
ymax = 10·(1 + e−1·π/3) = 13.51

Answer

ymax = 13.51
11.Matrices & determinants

Answer: det A = 20

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 = 2, M₁₂ = 4, M₁₃ = −4

Cofactor expansion

det A = 2·2 − (−4)·4 + 0·(−4) = 20

Answer

det A = 20
12.Matrices & determinants

Answer: x = 3.667

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 = 1·(−5) − 1·(−2) = −3

Replace the x column

det Aₓ = 1·(−5) − 1·6 = −11
x = −11/(−3) = 3.667

Answer

x = 3.667
y = −2.667
13.Vectors: dot & cross product

Answer: k = −6.333

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.

Dot product equals zero

a·b = (−3)·k + 5·(−3) + (−1)·4 = 0
−19 + (−3)·k = 0

Solve for k

k = −(−19)/(−3) = −6.333

Answer

k = −6.333
14.Vectors: dot & cross product

Answer: A = 3.536 m²

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.

Edge vectors

AB = B − A = (1, −2, 1)
AC = C − A = (−2, 1, 2)

Cross product

(AB × AC)x = AByACz − ABzACy = (−2)·2 − 1·1 = −5
(AB × AC)y = ABzACx − ABxACz = 1·(−2) − 1·2 = −4
(AB × AC)z = ABxACy − AByACx = 1·1 − (−2)·(−2) = −3
|AB × AC| = √((−5)² + (−4)² + (−3)²) = 7.071

Area of the triangle

A = ½·7.071 = 3.536 m²

Answer

A = 3.536 m²
15.Probability

Answer: Var(X) = 132.8

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 = 10·0.25 + 20·0.15 + 30·0.3 + 40·0.3 = 26.5

Mean of the squares

E(X²) = 100·0.25 + 400·0.15 + 900·0.3 + 1600·0.3 = 835

Variance

Var(X) = 835 − 26.5² = 132.8

Answer

Var(X) = 132.8
16.Probability

Answer: P = 0.1471

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 = 5) = 462·0.65·0.46 = 462·0.07776·0.004096 = 0.1471

Answer

P = 0.1471
17.Descriptive statistics

Answer: s = 8.649

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 = 128/5 = 25.6

Squared deviations

Σ(x − x̄)² = (−7.6)² + (−1.6)² + (−5.6)² + 14.4² + 0.4² = 299.2

Standard deviation

s = √(299.2/4) = 8.649

Answer

s = 8.649
18.Descriptive statistics

Answer: z = 1.5

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 = (72.5 − 50)/15 = 1.5

Answer

z = 1.5
19.Normal distribution

Answer: P = 0.4082

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 = (171.35 − 178)/7 = −0.95
z₂ = (179.4 − 178)/7 = 0.2

Area between

P = Φ(0.2) − Φ(−0.95) = 0.5793 − 0.1711 = 0.4082

Answer

P = 0.4082
20.Normal distribution

Answer: x = 22.13 min

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 for the percentile

Φ(z) = 0.05 → z = −1.645

Back to x

x = 32 + −1.645·6 = 22.13 min

Answer

x = 22.13 min

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