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→∞ (√(4x2 + 3x) − 2x).

f(x) = √(4x2 + 3x) − 2xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 4x)/sin(3x), with x in radians.

f(x) = ln(1 + 4x)/sin(3x)x → 0
L =
3.DerivativesMedium
xyx = 0.5

Let f(x) = x3·e2x. Find f′(0.5).

f(x) = x3·e2xx = 0.5
f′(0.5) =
4.DerivativesHard
xy

Let f(x) = x3 + 9x2 + 24x − 6. Find the x-coordinate x of the local maximum.

f(x) = x3 + 9x2 + 24x − 6
x =
5.IntegralsMedium
xy1y = x·sin x

Evaluate I = ∫₀1 x·sin x dx (x in radians).

∫₀1 x·sin x dx
I =
6.IntegralsHard
xy−22

Find the area A of the region between the parabola y = 4 − x² and the x-axis.

y = 4 − x²
A =
7.Complex numbersMedium
ReIm11−1−122−2−233−3−3z₁z₂w

Given z₁ = 1.5∠−35° and z₂ = 3∠80°, find the real part of w = z₁/z₂.

z₁ = 1.5∠−35°z₂ = 3∠80°
Re(w) =
8.Complex numbersHard
ReIm11−1−122−2−2θz

Use De Moivre's theorem to find the imaginary part of z⁵ for z = 1 + j√3.

z = 1 + j√3n = 5
Im(z⁵) =
9.Differential equationsMedium
t (min)T (°C)24Ta​16 °C12 °C

A can of soda at 5 °C is taken out of the fridge into a room at 26 °C. After 24 min its temperature is 16 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 12 °C?

T0 = 5 °CTa = 26 °CT(24) = 16 °CT = 12 °C
t = min
10.Differential equationsHard
ty1y(1) = ?

Solve y′ + 3y = 10e−t with y(0) = 4. Find y(1).

y′ + 3y = 10e−ty(0) = 4
y(1) =
11.Matrices & determinantsMedium
A =−1311B =1−6−1−2

Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).

A = [−1 3; 1 1]B = [1 −6; −1 −2]
c22 =
12.Matrices & determinantsHard
A =2−246b =35

Solve the system 2x − 2y = 3 and 4x + 6y = 5. Find x.

2x − 2y = 34x + 6y = 5
x =
13.Vectors: dot & cross productMedium
xyzab

Find the value of k for which the vectors a = (−1, −2, 3) and b = (5, 6, k) are perpendicular.

a = (−1, −2, 3)b = (5, 6, k)
k =
14.Vectors: dot & cross productHard
xyzABC

Find the area A of the triangle with corners A(−4, −3, −2), B(1, 1, −1) and C(2, 0, 3). Coordinates are in metres.

A(−4, −3, −2)B(1, 1, −1)C(2, 0, 3)
A = m²
15.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6P(six) = 1/6 on each roll

A fair die is rolled 6 times. What is the probability P of getting at least one six?

n = 6
P =
16.ProbabilityHard
€p(x)−20.8560.1980.05

A game costs €2 to play. You win €8 with probability 0.1 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€8: p = 0.1€100: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 57, 54, 44, 16, 21.

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

A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is above 74.8?

μ = 70σ = 12x = 74.8
P =
19.Normal distributionMedium
320 g316.6 gT = X₁ + … + X₈?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 8 of them is below 316.6 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 8< 316.6 g
P =
20.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =

3Solutions

1.Limits

Answer: L = 0.75

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) = 3x/(√(4x2 + 3x) + 2x)

Divide by x

f(x) = 3/(√(4 + 3/x) + 2)

Let x → ∞

L = 3/(2 + 2) = 0.75

Answer

L = 0.75
2.Limits

Answer: L = 1.333

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 [4/(1 + 4x)]/[3·cos 3x]

Substitute x = 0

L = 4/3 = 1.333

Answer

L = 1.333
3.Derivatives

Answer: f′(0.5) = 2.718

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 rule

f′(x) = u′v + uv′ = e2x·(3x2 + 2·x3)

Substitute

f′(0.5) = 2.718·(0.75 + 0.25) = 2.718

Answer

f′(0.5) = 2.718
4.Derivatives

Answer: x = −4

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 + 24 = 3·(x2 + 6x + 8)

Solve f′(x) = 0

x = −4 or x = −2

Which one is which

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

f″(−4) = −6, f″(−2) = 6
x = −4

Answer

x = −4
5.Integrals

Answer: I = 0.3012

Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.

Choose u and dv

u = x, du = dx; dv = sin x dx, v = −cos x

Antiderivative

∫x sin x dx = −x cos x + ∫cos x dx = −x cos x + sin x

Limits

(−1·cos 1 + sin 1) − (0 + 0) = 0.3012

Answer

I = 0.3012
6.Integrals

Answer: A = 10.67

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

4 − x² = 0 → x = ±2

Integrate

A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67

Answer

A = 10.67
7.Complex numbers

Answer: Re(w) = −0.2113

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.

Modulus and argument of w

|w| = 1.5 / 3 = 0.5
θ = −35° − 80° = −115°

Rectangular form

Re(w) = 0.5·cos(−115°) = −0.2113
Im(w) = 0.5·sin(−115°) = −0.4532
Re(w) = −0.2113

Answer

Re(w) = −0.2113
8.Complex numbers

Answer: Im(z⁵) = −27.71

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.

Polar form

z = 1 + j√3 = 2∠60°

De Moivre

z⁵ = (2)⁵∠300° = 32∠−60°

Rectangular form

z⁵ = 32·(cos −60° + j sin −60°) = 16 − j27.71
Im(z⁵) = −27.71

Answer

Im(z⁵) = −27.71
9.Differential equations

Answer: t = 13.12 min

First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.

Cooling constant

k = ln(-21/-10)/24 = 0.03091 min⁻¹

Time to reach the target

t = ln(-21/(−14))/0.03091 = 13.12 min

Answer

t = 13.12 min
10.Differential equations

Answer: y(1) = 1.79

First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.

Integrating factor

μ = e3t
(e3ty)′ = 10e2t

General solution

y = 10/2·e−t + C·e−3t = 5·e−t + C·e−3t

Initial condition

4 = 5 + C → C = −1

Evaluate

y(1) = 5·0.3679 + (−1)·0.04979 = 1.79

Answer

y(1) = 1.79
11.Matrices & determinants

Answer: c22 = −8

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.

Row times column

c22 = 1·(−6) + 1·(−2) = −8

Answer

c22 = −8
12.Matrices & determinants

Answer: x = 1.4

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 = 2·6 − (−2)·4 = 20

Replace the x column

det Aₓ = 3·6 − (−2)·5 = 28
x = 28/20 = 1.4

Answer

x = 1.4
y = −0.1
13.Vectors: dot & cross product

Answer: k = 5.667

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 = (−1)·5 + (−2)·6 + 3·k = 0
−17 + 3·k = 0

Solve for k

k = −(−17)/3 = 5.667

Answer

k = 5.667
14.Vectors: dot & cross product

Answer: A = 13.52 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 = (5, 4, 1)
AC = C − A = (6, 3, 5)

Cross product

(AB × AC)x = AByACz − ABzACy = 4·5 − 1·3 = 17
(AB × AC)y = ABzACx − ABxACz = 1·6 − 5·5 = −19
(AB × AC)z = ABxACy − AByACx = 5·3 − 4·6 = −9
|AB × AC| = √(17² + (−19)² + (−9)²) = 27.04

Area of the triangle

A = ½·27.04 = 13.52 m²

Answer

A = 13.52 m²
15.Probability

Answer: P = 0.6651

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.

Complement

P(no six) = (5/6)6 = 0.3349
P = 1 − 0.3349 = 0.6651

Answer

P = 0.6651
16.Probability

Answer: E(G) = 3.8 €

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.

Net outcomes

G: 6 (p = 0.1), 98 (p = 0.05), −2 (p = 0.85)

Expected value

E(G) = 6·0.1 + 98·0.05 − 2·0.85 = 3.8 €

Answer

E(G) = 3.8 €

A positive value means you win money on average.

17.Descriptive statistics

Answer: s = 18.88

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 = 192/5 = 38.4

Squared deviations

Σ(x − x̄)² = 18.6² + 15.6² + 5.6² + (−22.4)² + (−17.4)² = 1425

Standard deviation

s = √(1425/4) = 18.88

Answer

s = 18.88
18.Descriptive statistics

Answer: P = 0.3446

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446

Answer

P = 0.3446
19.Normal distribution

Answer: P = 0.1587

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the total

μT = 320 g, σT = 1.2·√8 = 3.394 g

Standardise

z = (316.6 − 320)/3.394 = −1

Probability

P = Φ(−1) = 0.1587

Answer

P = 0.1587
20.Normal distribution

Answer: P = 0.7881

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the difference

μ = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881

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