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 + 3/x)2x. Give a decimal value.

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

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

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

Let f(x) = (4x − 1)3. Find f′(0).

f(x) = (4x − 1)3x = 0
f′(0) =
4.DerivativesHard
xy

Let f(x) = x3 − 10.5x2 + 30x + 1. Find the x-coordinate x of the local minimum.

f(x) = x3 − 10.5x2 + 30x + 1
x =
5.IntegralsMedium
xy2y = √(4x)

The region under y = √(4x) from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).

y = √(4x)0 ≤ x ≤ 2
V =
6.IntegralsHard
xyy = 2x2​y = 3x

Find the area A of the region enclosed by y = 2x2 and y = 3x.

y = 2x2y = 3x
A =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6z₁z₂w

Given z₁ = 2.5∠70° and z₂ = 3∠65°, find the real part of w = z₁·z₂.

z₁ = 2.5∠70°z₂ = 3∠65°
Re(w) =
8.Complex numbersHard
ReIm11−1−122−2−233−3−3θz

Find the imaginary part of (2 − j3)⁴.

z = 2 − j3n = 4
Im(z⁴) =
9.Differential equationsMedium
t (min)T (°C)28Ta​159 °C

A forged steel part at 440 °C is left to cool in a workshop at 23 °C. After 28 min its temperature is 159 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 440 °CTa = 23 °CT(28) = 159 °C
k = min⁻¹
10.Differential equationsHard
xy13(1, −2)y(3) = ?

Solve the initial value problem y′ + 2y/x = 5x2 for x > 0, with y(1) = −2. Find y(3).

y′ + 2y/x = 5x2y(1) = −2
y(3) =
11.Matrices & determinantsMedium
A =−34−60B =2215

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

A = [−3 4; −6 0]B = [2 2; 1 5]
c12 =
12.Matrices & determinantsHard
A =21−3−2

Find the larger eigenvalue λ of the matrix A shown.

A = [2 1; −3 −2]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (−6, −5, 2) and b = (2, −1, 5), in degrees.

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

The unit vector n̂ is perpendicular to both a = (2, −5, 2) and b = (−2, 3, 5) and has a positive z-component. Find its y-component n̂y.

a = (2, −5, 2)b = (−2, 3, 5)n̂z > 0
n̂y =
15.ProbabilityMedium
P(at least one six) = ??1?2?3P(six) = 1/6 on each roll

A fair die is rolled 3 times. What is the probability P of getting at least one six?

n = 3
P =
16.ProbabilityHard
n = 10, p = 0.4: P(X ≤ 1) = ?k012345678910

10 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?

n = 10p = 0.4
P =
17.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 16, 29, 46, 55, 53.

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

A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is below 62?

μ = 60σ = 5x = 62
P =
19.Normal distributionMedium
27.235P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 27.2 and 35 min? Give P as a decimal.

μ = 32σ = 6x1 = 27.2x2 = 35
P =
20.Normal distributionHard
235240245250255260265gtop 15%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 15% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g15%
x = g

3Solutions

1.Limits

Answer: L = 403.4

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→∞ 2·x·ln(1 + 3/x)

Evaluate

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

ln L = 2·3 = 6

Undo the logarithm

L = e6 = 403.4

Answer

L = 403.4
2.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).

L'Hôpital's rule

L = limx→0 [3/(1 + 3x)]/[4·cos 4x]

Substitute x = 0

L = 3/4 = 0.75

Answer

L = 0.75
3.Derivatives

Answer: f′(0) = 12

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) = 3·(4x − 1)2·4

Substitute

f′(0) = 3·(−1)2·4 = 12

Answer

f′(0) = 12
4.Derivatives

Answer: x = 5

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 − 21x + 30 = 3·(x2 − 7x + 10)

Solve f′(x) = 0

x = 2 or x = 5

Which one is which

f″(x) = 6x − 21: negative at a maximum, positive at a minimum.

f″(2) = −9, f″(5) = 9
x = 5

Answer

x = 5
5.Integrals

Answer: V = 25.13

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 4x dx

Integrate

V = π·4·2²/2 = 25.13

Answer

V = 25.13
6.Integrals

Answer: A = 1.125

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 = 3x → x = 0, x = 1.5

Integrate upper minus lower

A = [1.5x2 − 0.6667x3]01.5 = 1.125

Answer

A = 1.125
7.Complex numbers

Answer: Re(w) = −5.303

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| = 2.5 · 3 = 7.5
θ = 70° + 65° = 135°

Rectangular form

Re(w) = 7.5·cos(135°) = −5.303
Im(w) = 7.5·sin(135°) = 5.303
Re(w) = −5.303

Answer

Re(w) = −5.303
8.Complex numbers

Answer: Im(z⁴) = 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.

Powers step by step

z² = (2 − j3)(2 − j3) = −5 − j12
z³ = (−5 − j12)(2 − j3) = −46 − j9
z⁴ = (−46 − j9)(2 − j3) = −119 + j120

Check with De Moivre

z = 3.606∠−56.31° → z⁴ = 169∠134.8°
Im(z⁴) = 120

Answer

Im(z⁴) = 120
9.Differential equations

Answer: k = 0.04002 min⁻¹

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.

Ratio of temperature differences

e−28k = 136/417

Solve for k

k = ln(417/136)/28 = 0.04002 min⁻¹

Answer

k = 0.04002 min⁻¹
10.Differential equations

Answer: y(3) = 26.67

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 = e∫2/x dx = x2

Integrate

(x2y)′ = 5x4 → x2y = 5x5/5 + C

Initial condition

−2 = 5/5 + C → C = −3

Evaluate

y(3) = (5·35/5 + (−3))/32 = 26.67

Answer

y(3) = 26.67
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 = (−3)·2 + 4·5 = 14

Answer

c12 = 14
12.Matrices & determinants

Answer: λ = 1

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 + (−1) = 0

Solve

λ = (0 ± √(4))/2 → λ = −1, 1
λ = 1

Answer

λ = 1
13.Vectors: dot & cross product

Answer: θ = 86.1 °

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)·2 + (−5)·(−1) + 2·5 = 3

Magnitudes

|a| = √((−6)² + (−5)² + 2²) = 8.062
|b| = √(2² + (−1)² + 5²) = 5.477

Angle

cos θ = 3/(8.062·5.477) = 0.06794
θ = arccos(0.06794) = 86.1°

Answer

θ = 86.1°
14.Vectors: dot & cross product

Answer: n̂y = 0.4088

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

(a × b)x = aybz − azby = (−5)·5 − 2·3 = −31
(a × b)y = azbx − axbz = 2·(−2) − 2·5 = −14
(a × b)z = axby − aybx = 2·3 − (−5)·(−2) = −4

Normalise

The z-component of a × b is negative, so use −(a × b).

|a × b| = √((−31)² + (−14)² + (−4)²) = 34.25
n̂ = −(−31, −14, −4)/34.25 = (0.9051, 0.4088, 0.1168)
n̂y = 0.4088

Answer

n̂y = 0.4088
15.Probability

Answer: P = 0.4213

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)3 = 0.5787
P = 1 − 0.5787 = 0.4213

Answer

P = 0.4213
16.Probability

Answer: P = 0.04636

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) = 0.610 = 0.006047
P(X = 1) = 10·0.4·0.69 = 0.04031

Add them

P = 0.006047 + 0.04031 = 0.04636

Answer

P = 0.04636
17.Descriptive statistics

Answer: s = 16.78

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 = 199/5 = 39.8

Squared deviations

Σ(x − x̄)² = (−23.8)² + (−10.8)² + 6.2² + 15.2² + 13.2² = 1127

Standard deviation

s = √(1127/4) = 16.78

Answer

s = 16.78
18.Descriptive statistics

Answer: P = 0.6554

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 = 0.6554

Answer

P = 0.6554
19.Normal distribution

Answer: P = 0.4796

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 − 32)/6 = 0.5

Area between

P = Φ(0.5) − Φ(−0.8) = 0.6915 − 0.2119 = 0.4796

Answer

P = 0.4796
20.Normal distribution

Answer: x = 255.2 g

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) = 0.85

z from the table

z ≈ 1.04

Back to the original units

x = 250 + 1.04·5 = 255.2 g

Answer

x = 255.2 g

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