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 − 2/x)3x. Give a decimal value.

f(x) = (1 − 2/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) = x·sin(x), with x in radians. Find f′(2).

f(x) = x·sin(x)x = 2 rad
f′(2) =
4.DerivativesHard
xyx = −1

The tangent line to f(x) = 0.5x2 − 5x + 1 at x = −1 crosses the y-axis at y = b. Find b.

f(x) = 0.5x2 − 5x + 1x = −1
b =
5.IntegralsMedium
xyπ/2y = x·sin x

Evaluate I = ∫₀π/2 x·sin x dx (x in radians).

∫₀π/2 x·sin x dx
I =
6.IntegralsHard
xyy = 0.5x2​y = 2x

Find the area A of the region enclosed by y = 0.5x2 and y = 2x.

y = 0.5x2y = 2x
A =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6θz

Write z = −6 + j4 in exponential form r·ejθ. Give θ in radians, between −π and π.

z = −6 + j4
θ = rad
8.Complex numbersHard
ReIm22−2−244−4−4z₀z₁

The roots of zⁿ = w with n = 2 and w = 25∠0° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1. Find the real part of z₁.

w = 25∠0°n = 2k = 1
Re(z₁) =
9.Differential equationsMedium
t (min)T (°C)33Ta​94 °C

A forged steel part at 400 °C is left to cool in a workshop at 21 °C. After 33 min its temperature is 94 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 400 °CTa = 21 °CT(33) = 94 °C
k = min⁻¹
10.Differential equationsHard
t (min)m (kg)40 minm = ?

A tank holds 400 L of pure water. Brine with 0.1 kg of salt per litre flows in at 10 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 40 min?

V = 400 Lm0 = 0 kgc = 0.1 kg/Lr = 10 L/min
m = kg
11.Matrices & determinantsMedium
A =−2−235−1−50−5−5

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

A = [−2 −2 3; 5 −1 −5; 0 −5 −5]
det A =
12.Matrices & determinantsHard
A =−633−3b =2−7

Solve the system −6x + 3y = 2 and 3x − 3y = −7. Find x.

−6x + 3y = 23x − 3y = −7
x =
13.Vectors: dot & cross productMedium
xyzab

Find the dot product a·b of a = (−6, −4, −5) and b = (−6, −5, −3).

a = (−6, −4, −5)b = (−6, −5, −3)
a·b =
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (−30, −20, 20) N acts at point A(0, −3, 4). Coordinates are in metres. Find the y-component My of the moment of the force about the origin O(0, 0, 0).

F = (−30, −20, 20) NA(0, −3, 4) mO(0, 0, 0) m
My = N·m
15.ProbabilityMedium
n = 9k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 9? The order does not matter.

n = 9k = 4
N =
16.ProbabilityHard
xp(x)10.320.130.140.1550.35μ

X has the distribution shown, and Y = 5X + 4. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.3, 0.1, 0.1, 0.15, 0.35Y = 5X + 4
σY =
17.Descriptive statisticsMedium
101520253035404550

Find the sample standard deviation s of the data: 10, 38, 28, 46, 27, 27, 51.

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 155.

μ = 170σ = 10x = 155
z =
19.Normal distributionMedium
2035.3P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 20 and 35.3 min? Give P as a decimal.

μ = 32σ = 6x1 = 20x2 = 35.3
P =
20.Normal distributionHard
μ = 100x = ?25%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 25% of them lie (the 25th percentile).

μ = 100σ = 15P = 25%
x = points

3Solutions

1.Limits

Answer: L = 0.002479

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 − 2/x)

Evaluate

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

ln L = 3·(−2) = −6

Undo the logarithm

L = e−6 = 0.002479

Answer

L = 0.002479
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) = 0.077

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(x) + x·cos(x)

Substitute

f′(2) = 0.9093 + 2·(−0.4161) = 0.077

Answer

f′(2) = 0.077
4.Derivatives

Answer: b = 0.5

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.

Point and slope

f(−1) = 6.5
f′(x) = x − 5 → f′(−1) = −6

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 6.5 − (−6)·(−1) = 0.5

Answer

b = 0.5
5.Integrals

Answer: I = 1

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

(−π/2·cos π/2 + sin π/2) − (0 + 0) = 1

Answer

I = 1
6.Integrals

Answer: A = 5.333

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

0.5x2 = 2x → x = 0, x = 4

Integrate upper minus lower

A = [x2 − 0.1667x3]04 = 5.333

Answer

A = 5.333
7.Complex numbers

Answer: θ = 2.554 rad

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

r = √(36 + 16) = 7.211

Argument in radians

θ = 146.3° = 146.3·π/180 = 2.554 rad

Exponential form

z = 7.211·ej2.554

Answer

θ = 2.554 rad
8.Complex numbers

Answer: Re(z₁) = −5

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 of the roots

251/2 = 5

Argument of the root

θ₁ = (0° + 360°·1)/2 = 180°

Rectangular form

z₁ = 5·(cos 180° + j sin 180°) = −5
Re(z₁) = −5

Answer

Re(z₁) = −5

The 2 roots lie on the circle of radius 5, 180° apart.

9.Differential equations

Answer: k = 0.04991 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−33k = 73/379

Solve for k

k = ln(379/73)/33 = 0.04991 min⁻¹

Answer

k = 0.04991 min⁻¹
10.Differential equations

Answer: m = 25.28 kg

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.

Set up the equation

dm/dt = 10·0.1 − (10/400)·m = 1 − 0.025·m

Solve

The integrating factor is ert/V; the solution approaches the equilibrium c·V.

m(t) = 40 + (−40)·e−t/40

Substitute t

m(40) = 40 + (−40)·0.3679 = 25.28 kg

Answer

m = 25.28 kg
11.Matrices & determinants

Answer: det A = −85

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 = −20, M₁₂ = −25, M₁₃ = −25

Cofactor expansion

det A = (−2)·(−20) − (−2)·(−25) + 3·(−25) = −85

Answer

det A = −85
12.Matrices & determinants

Answer: x = 1.667

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 = (−6)·(−3) − 3·3 = 9

Replace the x column

det Aₓ = 2·(−3) − 3·(−7) = 15
x = 15/9 = 1.667

Answer

x = 1.667
y = 4
13.Vectors: dot & cross product

Answer: a·b = 71

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 = (−6)·(−6) + (−4)·(−5) + (−5)·(−3) = 36 + 20 + 15 = 71

Answer

a·b = 71

Positive: the angle between the vectors is less than 90°.

14.Vectors: dot & cross product

Answer: My = −120 N·m

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.

Position vector from O to A

r = A − O = (0, −3, 4) − (0, 0, 0) = (0, −3, 4) m

Moment M = r × F

(r × F)x = ryFz − rzFy = (−3)·20 − 4·(−20) = 20 N·m
(r × F)y = rzFx − rxFz = 4·(−30) − 0·20 = −120 N·m
(r × F)z = rxFy − ryFx = 0·(−20) − (−3)·(−30) = −90 N·m

Requested component

My = −120 N·m

Answer

My = −120 N·m
15.Probability

Answer: N = 126

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.

Combinations

C(9, 4) = (9·8·7·6)/(4·3·2·1) = 126

Answer

N = 126
16.Probability

Answer: σY = 8.408

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 = 3.15, Var(X) = 2.828, σX = 1.682

Linear change

σY = |5|·1.682 = 8.408

Answer

σY = 8.408

The mean does move: E(Y) = 5·3.15 + 4 = 19.75.

17.Descriptive statistics

Answer: s = 13.79

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 = 227/7 = 32.43

Squared deviations

Σ(x − x̄)² = (−22.43)² + 5.571² + (−4.429)² + 13.57² + (−5.429)² + (−5.429)² + 18.57² = 1142

Standard deviation

s = √(1142/6) = 13.79

Answer

s = 13.79
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 = (155 − 170)/10 = −1.5

Answer

z = −1.5
19.Normal distribution

Answer: P = 0.6861

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 = (20 − 32)/6 = −2
z₂ = (35.3 − 32)/6 = 0.55

Area between

P = Φ(0.55) − Φ(−2) = 0.7088 − 0.02275 = 0.6861

Answer

P = 0.6861
20.Normal distribution

Answer: x = 89.88 points

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 = 100 + −0.6745·15 = 89.88 points

Answer

x = 89.88 points

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