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 − 2x) − x).

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

Find the limit L = limx→0 (e3x − 1 − 3x)/x².

f(x) = (e3x − 1 − 3x)/x²x → 0
L =
3.DerivativesMedium
xyx = 0.5

Let f(x) = x·sin(3x), with x in radians. Find f′(0.5).

f(x) = x·sin(3x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
xy

Let f(x) = x3 − 13.5x2 + 54x + 1. Find the x-coordinate x of the local maximum.

f(x) = x3 − 13.5x2 + 54x + 1
x =
5.IntegralsMedium
xy1y = 3x

The region under y = 3x from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 3x0 ≤ x ≤ 1
V =
6.IntegralsHard
xyy = 2x2​y = 2x

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

y = 2x2y = 2x
A =
7.Complex numbersMedium
ReIm22−2−244−4−4θz

Find the argument θ of z = −5 − j4 in degrees, as the principal value between −180° and 180°.

z = −5 − j4
θ = °
8.Complex numbersHard
ReIm11−1−122−2−2ABCD

Exactly one of the points A–D is a root of z⁴ = 16∠60°. Click it.

w = 16∠60°n = 4
  1. 1A (1∠105°)
  2. 2B (2∠60°)
  3. 3C (2∠105°)
  4. 4D (2∠150°)
9.Differential equationsMedium
t (min)T (°C)18Ta​33 °C28 °C

A cup of coffee at 82 °C is left in a room at 21 °C. After 18 min its temperature is 33 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 28 °C?

T0 = 82 °CTa = 21 °CT(18) = 33 °CT = 28 °C
t = min
10.Differential equationsHard
t (min)m (kg)t = ?26 kg

A tank holds 500 L of water in which 10 kg of salt is dissolved. Brine with 0.1 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 26 kg of salt?

V = 500 Lm0 = 10 kgc = 0.1 kg/Lr = 5 L/min
t = min
11.Matrices & determinantsMedium
A =−12−52B =3−634

Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).

A = [−1 2; −5 2]B = [3 −6; 3 4]
c12 =
12.Matrices & determinantsHard
A =−3−1−73

Find the larger eigenvalue λ of the matrix A shown.

A = [−3 −1; −7 3]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the dot product a·b of a = (7, −5, 1) and b = (−1, 4, 7).

a = (7, −5, 1)b = (−1, 4, 7)
a·b =
14.Vectors: dot & cross productHard
xyzab

For a = (−6, −6, 0) and b = (0, −2, −1), find the z-component (a × b)z of the cross product a × b.

a = (−6, −6, 0)b = (0, −2, −1)
(a × b)z =
15.ProbabilityMedium
red: 4blue: 9

A bag holds 4 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

4 / 9
P =
16.ProbabilityHard
n = 8, p = 0.2: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 8p = 0.2
P =
17.Descriptive statisticsMedium
202530354045

Find the sample standard deviation s of the data: 37, 44, 47, 21, 17.

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 27.5.

μ = 50σ = 15x = 27.5
z =
19.Normal distributionMedium
85.75107.5P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 85.75 and 107.5 points? Give P as a decimal.

μ = 100σ = 15x1 = 85.75x2 = 107.5
P =
20.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =

3Solutions

1.Limits

Answer: L = −1

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) = −2x/(√(x2 − 2x) + x)

Divide by x

f(x) = −2/(√(1 + (−2)/x) + 1)

Let x → ∞

L = −2/(1 + 1) = −1

Answer

L = −1
2.Limits

Answer: L = 4.5

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

(3·e3x − 3)/(2x) → 0/0

Second application

9·e3x/2

Substitute x = 0

L = 9·1/2 = 4.5

Answer

L = 4.5
3.Derivatives

Answer: f′(0.5) = 1.104

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(3x) + 3x·cos(3x)

Substitute

f′(0.5) = 0.9975 + 1.5·0.07074 = 1.104

Answer

f′(0.5) = 1.104
4.Derivatives

Answer: x = 3

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 − 27x + 54 = 3·(x2 − 9x + 18)

Solve f′(x) = 0

x = 3 or x = 6

Which one is which

f″(x) = 6x − 27: negative at a maximum, positive at a minimum.

f″(3) = −9, f″(6) = 9
x = 3

Answer

x = 3
5.Integrals

Answer: V = 9.425

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∫₀1 9x² dx

Integrate

V = π·9·1³/3 = 9.425

Answer

V = 9.425

Check: a cone of radius 3 and height 1 has (1/3)πr²h = 9.425.

6.Integrals

Answer: A = 0.3333

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

2x2 = 2x → x = 0, x = 1

Integrate upper minus lower

A = [x2 − 0.6667x3]01 = 0.3333

Answer

A = 0.3333
7.Complex numbers

Answer: θ = −141.3 °

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 = −5, b = −4 → III

Reference angle

α = atan(4/5) = 38.66°

Argument

θ = −(180° − α) = −141.3°

Answer

θ = −141.3°

In polar form z = 6.403∠−141.3°.

8.Complex numbers

Answer: C (2∠105°)

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 = 15°, 105°, −165°, −75°

The point

C = 2∠105°

Answer

C (2∠105°)
9.Differential equations

Answer: t = 23.97 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(61/12)/18 = 0.09033 min⁻¹

Time to reach the target

t = ln(61/7)/0.09033 = 23.97 min

Answer

t = 23.97 min
10.Differential equations

Answer: t = 51.08 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.

Result

m(t) = 50 + (−40)·e−t/100

Solve for t

e−t/100 = (50 − 26)/(50 − 10) = 0.6
t = 100·ln(1.667) = 51.08 min

Answer

t = 51.08 min
11.Matrices & determinants

Answer: c12 = 14

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

c12 = (−1)·(−6) + 2·4 = 14

Answer

c12 = 14
12.Matrices & determinants

Answer: λ = 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.

Characteristic equation

λ² − (0)λ + (−16) = 0

Solve

λ = (0 ± √(64))/2 → λ = −4, 4
λ = 4

Answer

λ = 4
13.Vectors: dot & cross product

Answer: a·b = −20

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 = 7·(−1) + (−5)·4 + 1·7 = (−7) + (−20) + 7 = −20

Answer

a·b = −20

Negative: the angle between the vectors is more than 90°.

14.Vectors: dot & cross product

Answer: (a × b)z = 12

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 = (−6)·(−1) − 0·(−2) = 6
(a × b)y = azbx − axbz = 0·0 − (−6)·(−1) = −6
(a × b)z = axby − aybx = (−6)·(−2) − (−6)·0 = 12

Requested component

(a × b)z = 12

Answer

(a × b)z = 12

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

15.Probability

Answer: P = 0.07692

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 = 4/13 · 3/12 = 12/156 = 0.07692

Answer

P = 0.07692
16.Probability

Answer: P = 0.5033

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.

Zero and one failure

P(X = 0) = 0.88 = 0.1678
P(X = 1) = 8·0.2·0.87 = 0.3355

Add them

P = 0.1678 + 0.3355 = 0.5033

Answer

P = 0.5033
17.Descriptive statistics

Answer: s = 13.54

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 = 166/5 = 33.2

Squared deviations

Σ(x − x̄)² = 3.8² + 10.8² + 13.8² + (−12.2)² + (−16.2)² = 732.8

Standard deviation

s = √(732.8/4) = 13.54

Answer

s = 13.54
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 = (27.5 − 50)/15 = −1.5

Answer

z = −1.5
19.Normal distribution

Answer: P = 0.5204

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 = (85.75 − 100)/15 = −0.95
z₂ = (107.5 − 100)/15 = 0.5

Area between

P = Φ(0.5) − Φ(−0.95) = 0.6915 − 0.1711 = 0.5204

Answer

P = 0.5204
20.Normal distribution

Answer: P = 0.9868

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 = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.04)/0.01803 = −2.22

Probability

P = 1 − Φ(−2.22) = 0.9868

Answer

P = 0.9868

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