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→∞ (−2x3 − 6x + 2)/(x3 + 5x + 1).

f(x) = (−2x3 − 6x + 2)/(x3 + 5x + 1)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 2x)/(3x²), with x in radians.

f(x) = (1 − cos 2x)/(3x²)x → 0
L =
3.DerivativesMedium
xyx = 0.5

Let f(x) = x·sin(2x), with x in radians. Find f′(0.5).

f(x) = x·sin(2x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
xyx = −2

The tangent line to f(x) = 0.5x2 + 3x + 1 at x = −2 crosses the y-axis at y = b. Find b.

f(x) = 0.5x2 + 3x + 1x = −2
b =
5.IntegralsMedium
xy12y = x ln x

Evaluate I = ∫₁2 x·ln x dx.

∫₁2 x ln x dx
I =
6.IntegralsHard
xy1.5y = x²eˣ

Evaluate I = ∫₀1.5 x²·eˣ dx.

∫₀1.5 x²eˣ dx
I =
7.Complex numbersMedium
ReIm0.50.5−0.5−0.511−1−11.51.5−1.5−1.5z₁z₂w

Given z₁ = 1.5∠−60° and z₂ = 2∠−140°, find the real part of w = z₁/z₂.

z₁ = 1.5∠−60°z₂ = 2∠−140°
Re(w) =
8.Complex numbersHard
ReIm22−2−244−4−4ABCD

Exactly one of the points A–D is a root of z⁶ = 729∠180°. Click it.

w = 729∠180°n = 6
  1. 1A (3∠60°)
  2. 2B (3∠30°)
  3. 3C (5.4∠60°)
  4. 4D (3∠180°)
9.Differential equationsMedium
t (min)T (°C)18Ta​33 °C27 °C

A cup of coffee at 80 °C is left in a room at 19 °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 27 °C?

T0 = 80 °CTa = 19 °CT(18) = 33 °CT = 27 °C
t = min
10.Differential equationsHard
tyK = 10ymax​ = ?

A system obeys y″ + y′ + 16.25y = 162.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + y′ + 16.25y = 162.5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
A =33−23−3−2−1−24

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

A = [3 3 −2; 3 −3 −2; −1 −2 4]
det A =
12.Matrices & determinantsHard
A =1−34−2b =−912

Solve the system x − 3y = −9 and 4x − 2y = 12. Find x.

x − 3y = −94x − 2y = 12
x =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (0, 15, 30) N acts on an object while it moves in a straight line from point A(0, 0, 4) to point B(0, 3, −2). Coordinates are in metres. Find the work W done by the force.

F = (0, 15, 30) NA(0, 0, 4) mB(0, 3, −2) m
W = J
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (5, −3, 1), b = (−2, 1, −4) and c = (−4, −5, −1).

a = (5, −3, 1)b = (−2, 1, −4)c = (−4, −5, −1)
V =
15.ProbabilityMedium
xp(x)00.3520.240.360.15μ = 2.5

For the distribution shown, find the standard deviation σ of X.

x: 0, 2, 4, 6p(x): 0.35, 0.2, 0.3, 0.15
σ =
16.ProbabilityHard
n = 5, p = 0.6: P(X ≤ 1) = ?k012345

5 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 5p = 0.6
P =
17.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 53, 41, 33, 12, 48.

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

Exam scores are normally distributed with mean 70 and standard deviation 10. Find the z-score z of a score of 58.

μ = 70σ = 10x = 58
z =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.6629

Drag z so that the shaded area between −z and z under the standard normal curve is 0.6629. Find z to two decimals.

A = 0.6629
z =
20.Normal distributionHard
μ = 178x = ?10%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 10% of them lie (the 10th percentile).

μ = 178σ = 7P = 10%
x = cm

3Solutions

1.Limits

Answer: L = −2

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).

Divide by the highest power

(−2 + (−6)/x2 + 2/x3)/(1 + 5/x2 + 1/x3)

Let x → ∞

L = −2/1 = −2

Answer

L = −2

For equal degrees the limit is the ratio of the leading coefficients.

2.Limits

Answer: L = 0.6667

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

2·sin 2x/(6x) → 0/0

Second application

4·cos 2x/6

Substitute x = 0

L = 4/6 = 0.6667

Answer

L = 0.6667
3.Derivatives

Answer: f′(0.5) = 1.382

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(2x) + 2x·cos(2x)

Substitute

f′(0.5) = 0.8415 + 1·0.5403 = 1.382

Answer

f′(0.5) = 1.382
4.Derivatives

Answer: b = −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.

Point and slope

f(−2) = −3
f′(x) = x + 3 → f′(−2) = 1

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −3 − 1·(−2) = −1

Answer

b = −1
5.Integrals

Answer: I = 0.6363

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 = ln x, du = dx/x; dv = x dx, v = x²/2

Antiderivative

∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4

Limits

(2·ln 2 − 1) − (0 − 0.25) = 0.6363

Answer

I = 0.6363
6.Integrals

Answer: I = 3.602

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.

First round

∫x²eˣ dx = x²eˣ − ∫2x eˣ dx

Second round

∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)

Limits

4.482·(1.25) − 1·2 = 3.602

Answer

I = 3.602
7.Complex numbers

Answer: Re(w) = 0.1302

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 / 2 = 0.75
θ = −60° − (−140)° = 80°

Rectangular form

Re(w) = 0.75·cos(80°) = 0.1302
Im(w) = 0.75·sin(80°) = 0.7386
Re(w) = 0.1302

Answer

Re(w) = 0.1302
8.Complex numbers

Answer: B (3∠30°)

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 = 30°, 90°, 150°, −150°, −90°, −30°

The point

B = 3∠30°

Answer

B (3∠30°)
9.Differential equations

Answer: t = 24.84 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(61/14)/18 = 0.08177 min⁻¹

Time to reach the target

t = ln(61/8)/0.08177 = 24.84 min

Answer

t = 24.84 min
10.Differential equations

Answer: ymax = 16.75

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 = −0.5 ± 4j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10
y′(0) = −0.5A + 4B = 0 → B = −1.25

Peak

t = π/4 = 0.7854
ymax = 10·(1 + e−0.5·π/4) = 16.75

Answer

ymax = 16.75
11.Matrices & determinants

Answer: det A = −60

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 = −16, M₁₂ = 10, M₁₃ = −9

Cofactor expansion

det A = 3·(−16) − 3·10 + (−2)·(−9) = −60

Answer

det A = −60
12.Matrices & determinants

Answer: x = 5.4

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 = 1·(−2) − (−3)·4 = 10

Replace the x column

det Aₓ = (−9)·(−2) − (−3)·12 = 54
x = 54/10 = 5.4

Answer

x = 5.4
y = 4.8
13.Vectors: dot & cross product

Answer: W = −135 J

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.

Displacement

d = B − A = (0, 3, −2) − (0, 0, 4) = (0, 3, −6) m

Work

W = F·d = 0·0 + 15·3 + 30·(−6) = −135 J

Answer

W = −135 J

Negative work: the force acts mostly against the motion.

14.Vectors: dot & cross product

Answer: V = 133

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 b × c

(b × c)x = bycz − bzcy = 1·(−1) − (−4)·(−5) = −21
(b × c)y = bzcx − bxcz = (−4)·(−4) − (−2)·(−1) = 14
(b × c)z = bxcy − bycx = (−2)·(−5) − 1·(−4) = 14

Dot product with a

a·(b × c) = 5·(−21) + (−3)·14 + 1·14 = −133
V = |−133| = 133

Answer

V = 133
15.Probability

Answer: σ = 2.179

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 = 0·0.35 + 2·0.2 + 4·0.3 + 6·0.15 = 2.5

Mean of the squares

E(X²) = 0·0.35 + 4·0.2 + 16·0.3 + 36·0.15 = 11

Variance

Var(X) = 11 − 2.5² = 4.75
σ = √4.75 = 2.179

Answer

σ = 2.179
16.Probability

Answer: P = 0.08704

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) = 0.45 = 0.01024
P(X = 1) = 5·0.6·0.44 = 0.0768

Add them

P = 0.01024 + 0.0768 = 0.08704

Answer

P = 0.08704
17.Descriptive statistics

Answer: s = 16.07

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 = 187/5 = 37.4

Squared deviations

Σ(x − x̄)² = 15.6² + 3.6² + (−4.4)² + (−25.4)² + 10.6² = 1033

Standard deviation

s = √(1033/4) = 16.07

Answer

s = 16.07
18.Descriptive statistics

Answer: z = −1.2

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 = (58 − 70)/10 = −1.2

Answer

z = −1.2
19.Normal distribution

Answer: z = 0.96

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 of z

Φ(z) = (1 + 0.6629)/2 = 0.8315

Read the z-table backwards

Find 0.8315 inside the table and read off z.

z = 0.96

Answer

z = 0.96
20.Normal distribution

Answer: x = 169 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) = 0.1 → z = −1.282

Back to x

x = 178 + −1.282·7 = 169.03 cm

Answer

x = 169.03 cm

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