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 + 1/x)2x. Give a decimal value.

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

Find the limit L = limx→0 ln(1 + 3x)/sin(5x), with x in radians.

f(x) = ln(1 + 3x)/sin(5x)x → 0
L =
3.DerivativesMedium
xyx = 1.5

Let f(x) = x3·ex. Find f′(1.5).

f(x) = x3·exx = 1.5
f′(1.5) =
4.DerivativesHard
xyx = −1

The tangent line to f(x) = 0.5x2 + 2x + 5 at x = −1 crosses the y-axis at y = b. Find b.

f(x) = 0.5x2 + 2x + 5x = −1
b =
5.IntegralsMedium
xy12

Evaluate I = ∫12 2/x dx.

f(x) = 2/x[1, 2]
I =
6.IntegralsHard
xy4y = 2xy = 0.5x²

The region between the line y = 2x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.

y = 2xy = 0.5x²
V =
7.Complex numbersMedium
ReIm55−5−5z₁z₂w

Given z₁ = 5∠55° and z₂ = 1.5∠110°, find the argument θ of w = z₁·z₂ in degrees, between −180° and 180°.

z₁ = 5∠55°z₂ = 1.5∠110°
θ = °
8.Complex numbersHard
ReIm0.50.5−0.5−0.511−1−1θz

Use De Moivre's theorem to find the real part of z⁸ for z = −1 + j.

z = −1 + jn = 8
Re(z⁸) =
9.Differential equationsMedium
t (min)T (°C)33Ta​T = ?

A forged steel part at 380 °C is left to cool in a workshop at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 33 min.

T0 = 380 °CTa = 23 °Ck = 0.03 min⁻¹t = 33 min
T = °C
10.Differential equationsHard
tyy(0) = 1y(1.5) = ?

Solve y″ + 4y′ + 13y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.

y″ + 4y′ + 13y = 0y(0) = 1y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
A =−4−5−4−45−32−44

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

A = [−4 −5 −4; −4 5 −3; 2 −4 4]
det A =
12.Matrices & determinantsHard
A =5−441b =93

Solve the system 5x − 4y = 9 and 4x + y = 3. Find x.

5x − 4y = 94x + y = 3
x =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (0, −3, −1) and b = (3, 2, 5), in degrees.

a = (0, −3, −1)b = (3, 2, 5)
θ = °
14.Vectors: dot & cross productHard
xyzab

Find the area A of the parallelogram spanned by a = (−5, −2, −5) and b = (2, 0, 3).

a = (−5, −2, −5)b = (2, 0, 3)
A =
15.ProbabilityMedium
xp(x)00.2510.320.330.15μ = 1.35

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

x: 0, 1, 2, 3p(x): 0.25, 0.3, 0.3, 0.15
σ =
16.ProbabilityHard
xp(x)00.1510.0520.2530.240.35μ

X has the distribution shown, and Y = 0.5X + 4. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.15, 0.05, 0.25, 0.2, 0.35Y = 0.5X + 4
σY =
17.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 41, 50, 42, 37, 16.

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

A quantity is normally distributed with mean 100 and standard deviation 12. What proportion P of values is below 112?

μ = 100σ = 12x = 112
P =
19.Normal distributionMedium
460510P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 460 and 510 h? Give P as a decimal.

μ = 500σ = 40x1 = 460x2 = 510
P =
20.Normal distributionHard
μ = 32x = ?10%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 10% of them lie (the 10th percentile).

μ = 32σ = 6P = 10%
x = min

3Solutions

1.Limits

Answer: L = 7.389

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 + 1/x)

Evaluate

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

ln L = 2·1 = 2

Undo the logarithm

L = e2 = 7.389

Answer

L = 7.389
2.Limits

Answer: L = 0.6

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)]/[5·cos 5x]

Substitute x = 0

L = 3/5 = 0.6

Answer

L = 0.6
3.Derivatives

Answer: f′(1.5) = 45.38

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 rule

f′(x) = u′v + uv′ = ex·(3x2 + 1·x3)

Substitute

f′(1.5) = 4.482·(6.75 + 3.375) = 45.38

Answer

f′(1.5) = 45.38
4.Derivatives

Answer: b = 4.5

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) = 3.5
f′(x) = x + 2 → f′(−1) = 1

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 3.5 − 1·(−1) = 4.5

Answer

b = 4.5
5.Integrals

Answer: I = 1.386

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.

Antiderivative

F(x) = 2·ln x

Evaluate at the limits

I = 2·(ln 2 − ln 1) = 2·ln 2 = 1.386

Answer

I = 1.386
6.Integrals

Answer: V = 107.2

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.

Intersection points

2x = 0.5x² → x = 0, x = 4

Washer method

V = π∫₀4 ((2x)² − (0.5x²)²) dx

Integrate

V = π(4·4³/3 − 0.25·4⁵/5) = π·34.13 = 107.2

Answer

V = 107.2
7.Complex numbers

Answer: θ = 165 °

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| = 5 · 1.5 = 7.5
θ = 55° + 110° = 165°

Argument

θ = 165°

Answer

θ = 165°
8.Complex numbers

Answer: Re(z⁸) = 16

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.

Polar form

z = −1 + j = √2∠135°

De Moivre

z⁸ = (√2)⁸∠1080° = 16∠0°

Rectangular form

z⁸ = 16·(cos 0° + j sin 0°) = 16
Re(z⁸) = 16

Answer

Re(z⁸) = 16
9.Differential equations

Answer: T = 155.7 °C

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.

Solution of the cooling law

T(t) = 23 + 357·e−0.03t

Substitute t

T(33) = 23 + 357·0.3716 = 155.7 °C

Answer

T = 155.7 °C
10.Differential equations

Answer: y(1.5) = −0.04294

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.

Complex roots

r = −2 ± 3j

Constants

A = 1, B = 2·1/3 = 0.6667

Evaluate

y(1.5) = e−3(1·cos 4.5 + 0.6667·sin 4.5) = −0.04294

Answer

y(1.5) = −0.04294
11.Matrices & determinants

Answer: det A = −106

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 = 8, M₁₂ = −10, M₁₃ = 6

Cofactor expansion

det A = (−4)·8 − (−5)·(−10) + (−4)·6 = −106

Answer

det A = −106
12.Matrices & determinants

Answer: x = 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.

Determinant of the coefficient matrix

det A = 5·1 − (−4)·4 = 21

Replace the x column

det Aₓ = 9·1 − (−4)·3 = 21
x = 21/21 = 1

Answer

x = 1
y = −1
13.Vectors: dot & cross product

Answer: θ = 124.4 °

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 = 0·3 + (−3)·2 + (−1)·5 = −11

Magnitudes

|a| = √(0² + (−3)² + (−1)²) = 3.162
|b| = √(3² + 2² + 5²) = 6.164

Angle

cos θ = −11/(3.162·6.164) = −0.5643
θ = arccos(−0.5643) = 124.4°

Answer

θ = 124.4°
14.Vectors: dot & cross product

Answer: A = 8.775

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

Area

A = |a × b| = √((−6)² + 5² + 4²) = 8.775

Answer

A = 8.775
15.Probability

Answer: σ = 1.014

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 = 0·0.25 + 1·0.3 + 2·0.3 + 3·0.15 = 1.35

Mean of the squares

E(X²) = 0·0.25 + 1·0.3 + 4·0.3 + 9·0.15 = 2.85

Variance

Var(X) = 2.85 − 1.35² = 1.028
σ = √1.028 = 1.014

Answer

σ = 1.014
16.Probability

Answer: σY = 0.6978

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 = 2.55, Var(X) = 1.947, σX = 1.396

Linear change

σY = |0.5|·1.396 = 0.6978

Answer

σY = 0.6978

The mean does move: E(Y) = 0.5·2.55 + 4 = 5.275.

17.Descriptive statistics

Answer: s = 12.76

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 = 186/5 = 37.2

Squared deviations

Σ(x − x̄)² = 3.8² + 12.8² + 4.8² + (−0.2)² + (−21.2)² = 650.8

Standard deviation

s = √(650.8/4) = 12.76

Answer

s = 12.76
18.Descriptive statistics

Answer: P = 0.8413

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 = 1

Standard normal table

Φ(1) = 0.8413
P = 0.8413

Answer

P = 0.8413
19.Normal distribution

Answer: P = 0.4401

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 = (460 − 500)/40 = −1
z₂ = (510 − 500)/40 = 0.25

Area between

P = Φ(0.25) − Φ(−1) = 0.5987 − 0.1587 = 0.4401

Answer

P = 0.4401
20.Normal distribution

Answer: x = 24.31 min

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 = 32 + −1.282·6 = 24.31 min

Answer

x = 24.31 min

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