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 + 0.5/x)3x. Give a decimal value.

f(x) = (1 + 0.5/x)3xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 4x)/(3x²), with x in radians.

f(x) = (1 − cos 4x)/(3x²)x → 0
L =
3.DerivativesMedium
xyx = 2

Let f(x) = (2x + 3)5. Find f′(2).

f(x) = (2x + 3)5x = 2
f′(2) =
4.DerivativesHard
xy

Let f(x) = x3 − 15x2 + 63x − 2. Find the x-coordinate x of the local minimum.

f(x) = x3 − 15x2 + 63x − 2
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
ReIm22−2−244−4−4θz

Write z = 5∠−115° in rectangular form a + jb and give the real part a.

r = 5θ = −115°
a =
8.Complex numbersHard
ReIm22−2−244−4−4ABCD

Exactly one of the points A–D is a root of z⁶ = 64∠180°. Click it.

w = 64∠180°n = 6
  1. 1A (2∠180°)
  2. 2B (2∠90°)
  3. 3C (3.6∠90°)
  4. 4D (2∠120°)
9.Differential equationsMedium
tyy(0) = 4y(0.5) = ?

Solve y″ + 7y′ + 10y = 0 with y(0) = 4 and y′(0) = -4. Find y(0.5).

y″ + 7y′ + 10y = 0y(0) = 4y′(0) = -4
y(0.5) =
10.Differential equationsHard
tyK = 1y(1.5) = ?

A system obeys y″ + 4y′ + 20y = 20 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.

y″ + 4y′ + 20y = 20y(0) = 0y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
A =112410−5−54

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

A = [1 1 2; 4 1 0; −5 −5 4]
det A =
12.Matrices & determinantsHard
A =3364b =−10−1

Solve the system 3x + 3y = −10 and 6x + 4y = −1. Find x.

3x + 3y = −106x + 4y = −1
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (6, −5, 1) onto b = (4, 2, −4), that is, the component of a in the direction of b.

a = (6, −5, 1)b = (4, 2, −4)
compb a =
14.Vectors: dot & cross productHard
xyzab

For a = (3, 4, 4) and b = (−5, −6, 0), find the x-component (a × b)x of the cross product a × b.

a = (3, 4, 4)b = (−5, −6, 0)
(a × b)x =
15.ProbabilityMedium
red: 5blue: 6

A bag holds 5 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 6
P =
16.ProbabilityHard
xp(x)00.3510.320.35μ

X has the distribution shown, and Y = 3X + 5. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 3X + 5
σY =
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 26, 56, 11, 13, 34, 52, 58.

n = 7
s =
18.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 10. Find the z-score z of a score of 160.

μ = 170σ = 10x = 160
z =
19.Normal distributionMedium
27.235.3P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 27.2 and 35.3 min? Give P as a decimal.

μ = 32σ = 6x1 = 27.2x2 = 35.3
P =
20.Normal distributionHard
μ = 178x = ?25%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 25% of them lie (the 25th percentile).

μ = 178σ = 7P = 25%
x = cm

3Solutions

1.Limits

Answer: L = 4.482

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→∞ 3·x·ln(1 + 0.5/x)

Evaluate

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

ln L = 3·0.5 = 1.5

Undo the logarithm

L = e1.5 = 4.482

Answer

L = 4.482
2.Limits

Answer: L = 2.667

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).

First application

4·sin 4x/(6x) → 0/0

Second application

16·cos 4x/6

Substitute x = 0

L = 16/6 = 2.667

Answer

L = 2.667
3.Derivatives

Answer: f′(2) = 24010

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

f′(x) = 5·(2x + 3)4·2

Substitute

f′(2) = 5·74·2 = 24010

Answer

f′(2) = 24010
4.Derivatives

Answer: x = 7

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 − 30x + 63 = 3·(x2 − 10x + 21)

Solve f′(x) = 0

x = 3 or x = 7

Which one is which

f″(x) = 6x − 30: negative at a maximum, positive at a minimum.

f″(3) = −12, f″(7) = 12
x = 7

Answer

x = 7
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: a = −2.113

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.

Components

a = 5·cos(−115°) = −2.113
b = 5·sin(−115°) = −4.532

Rectangular form

z = −2.113 − j4.532
a = −2.113

Answer

a = −2.113
8.Complex numbers

Answer: B (2∠90°)

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| = 2
θ = 30°, 90°, 150°, −150°, −90°, −30°

The point

B = 2∠90°

Answer

B (2∠90°)
9.Differential equations

Answer: y(0.5) = 1.853

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 + 5) = 0 → r = −2, −5

Constants from the initial values

A + B = 4, −2A − 5B = -4
A = 5.333, B = −1.333

Evaluate

y(0.5) = 5.333·e−1 + (−1.333)·e−2.5 = 1.853

Answer

y(0.5) = 1.853
10.Differential equations

Answer: y(1.5) = 0.9592

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 = 1, r = −2 ± 4j

Constants from the initial values

y(0) = 1 + A = 0 → A = −1
y′(0) = −2A + 4B = 0 → B = −0.5

Evaluate

y(1.5) = 1·(1 − e−3(cos 6 + 0.5·sin 6)) = 0.9592

Answer

y(1.5) = 0.9592
11.Matrices & determinants

Answer: det A = −42

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 = 4, M₁₂ = 16, M₁₃ = −15

Cofactor expansion

det A = 1·4 − 1·16 + 2·(−15) = −42

Answer

det A = −42
12.Matrices & determinants

Answer: x = 6.167

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 = 3·4 − 3·6 = −6

Replace the x column

det Aₓ = (−10)·4 − 3·(−1) = −37
x = −37/(−6) = 6.167

Answer

x = 6.167
y = −9.5
13.Vectors: dot & cross product

Answer: compb a = 1.667

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

a·b = 6·4 + (−5)·2 + 1·(−4) = 10

Length of b

|b| = √(4² + 2² + (−4)²) = 6

Scalar projection

compb a = a·b/|b| = 10/6 = 1.667

Answer

compb a = 1.667
14.Vectors: dot & cross product

Answer: (a × b)x = 24

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, component by component

(a × b)x = aybz − azby = 4·0 − 4·(−6) = 24
(a × b)y = azbx − axbz = 4·(−5) − 3·0 = −20
(a × b)z = axby − aybx = 3·(−6) − 4·(−5) = 2

Requested component

(a × b)x = 24

Answer

(a × b)x = 24

Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.

15.Probability

Answer: P = 0.1818

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.

First and second draw

P = 5/11 · 4/10 = 20/110 = 0.1818

Answer

P = 0.1818
16.Probability

Answer: σY = 2.51

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.

Spread of X

μ = 1, Var(X) = 0.7, σX = 0.8367

Linear change

σY = |3|·0.8367 = 2.51

Answer

σY = 2.51

The mean does move: E(Y) = 3·1 + 5 = 8.

17.Descriptive statistics

Answer: s = 19.99

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 = 250/7 = 35.71

Squared deviations

Σ(x − x̄)² = (−9.714)² + 20.29² + (−24.71)² + (−22.71)² + (−1.714)² + 16.29² + 22.29² = 2397

Standard deviation

s = √(2397/6) = 19.99

Answer

s = 19.99
18.Descriptive statistics

Answer: z = −1

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 = (160 − 170)/10 = −1

Answer

z = −1
19.Normal distribution

Answer: P = 0.497

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 = (27.2 − 32)/6 = −0.8
z₂ = (35.3 − 32)/6 = 0.55

Area between

P = Φ(0.55) − Φ(−0.8) = 0.7088 − 0.2119 = 0.497

Answer

P = 0.497
20.Normal distribution

Answer: x = 173.3 cm

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) = 0.25 → z = −0.6745

Back to x

x = 178 + −0.6745·7 = 173.28 cm

Answer

x = 173.28 cm

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