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→∞ (√(4x2 − 10x) − 2x).

f(x) = √(4x2 − 10x) − 2xx → ∞
L =
2.LimitsHard
xyx = 0

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

f(x) = (e−x − 1 + x)/x²x → 0
L =
3.DerivativesMedium
xyx = 1.5

Let f(x) = x·sin(3x), with x in radians. Find f′(1.5).

f(x) = x·sin(3x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
xyx = 2

The tangent line to f(x) = x2 + 5x − 3 at x = 2 crosses the y-axis at y = b. Find b.

f(x) = x2 + 5x − 3x = 2
b =
5.IntegralsMedium
xy2y = √(x)

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

y = √(x)0 ≤ x ≤ 2
V =
6.IntegralsHard
xy−11

Find the area A of the region between the parabola y = 1 − x² and the x-axis.

y = 1 − x²
A =
7.Complex numbersMedium
ReIm22−2−244−4−4θz

Write z = −2 − j5 in exponential form r·ejθ. Give θ in radians, between −π and π.

z = −2 − j5
θ = rad
8.Complex numbersHard
ReIm22−2−244−4−466−6−6ABCD

Exactly one of the points A–D is a root of z³ = 27∠60°. Click it.

w = 27∠60°n = 3
  1. 1A (3∠100°)
  2. 2B (3∠−70°)
  3. 3C (5.4∠−100°)
  4. 4D (3∠−100°)
9.Differential equationsMedium
t (min)T (°C)27Ta​T = ?

A can of soda at 3 °C is taken out of the fridge into a room at 30 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.06 min⁻¹, find its temperature T after 27 min.

T0 = 3 °CTa = 30 °Ck = 0.06 min⁻¹t = 27 min
T = °C
10.Differential equationsHard
tyy(0) = 1y(1.5) = ?

Solve y″ + 2y′ + 5y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.

y″ + 2y′ + 5y = 0y(0) = 1y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
A =5−5−5−55−454−1

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

A = [5 −5 −5; −5 5 −4; 5 4 −1]
det A =
12.Matrices & determinantsHard
A =−521−5b =113

Solve the system −5x + 2y = 11 and x − 5y = 3. Find x.

−5x + 2y = 11x − 5y = 3
x =
13.Vectors: dot & cross productMedium
xyzdFAB

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

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

For a = (6, −2, −5) and b = (−2, 3, 0), find the y-component (a × b)y of the cross product a × b.

a = (6, −2, −5)b = (−2, 3, 0)
(a × b)y =
15.ProbabilityMedium
red: 6blue: 4

A bag holds 6 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

6 / 4
P =
16.ProbabilityHard
xp(x)10.2520.330.140.35μ

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

x: 1, 2, 3, 4p(x): 0.25, 0.3, 0.1, 0.35Y = 5X + 4
σY =
17.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 25, 18, 21, 53, 39, 27.

n = 6
s =
18.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 170 and standard deviation 20. What proportion P of values is above 195?

μ = 170σ = 20x = 195
P =
19.Normal distributionMedium
144 kg142.6 kgT = X₁ + … + X₁₂?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 12 of them exceeds 142.6 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 12> 142.6 kg
P =
20.Normal distributionHard
380420460500540580620htop 80%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h80%
x = h

3Solutions

1.Limits

Answer: L = −2.5

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).

Multiply by the conjugate

f(x) = −10x/(√(4x2 − 10x) + 2x)

Divide by x

f(x) = −10/(√(4 + (−10)/x) + 2)

Let x → ∞

L = −10/(2 + 2) = −2.5

Answer

L = −2.5
2.Limits

Answer: L = 0.5

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

(−1·e−x + 1)/(2x) → 0/0

Second application

1·e−x/2

Substitute x = 0

L = 1·1/2 = 0.5

Answer

L = 0.5
3.Derivatives

Answer: f′(1.5) = −1.926

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′(1.5) = −0.9775 + 4.5·(−0.2108) = −1.926

Answer

f′(1.5) = −1.926
4.Derivatives

Answer: b = −7

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) = 11
f′(x) = 2x + 5 → f′(2) = 9

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 11 − 9·2 = −7

Answer

b = −7
5.Integrals

Answer: V = 6.283

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 x dx

Integrate

V = π·1·2²/2 = 6.283

Answer

V = 6.283
6.Integrals

Answer: A = 1.333

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.

Zeros

1 − x² = 0 → x = ±1

Integrate

A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333

Answer

A = 1.333
7.Complex numbers

Answer: θ = −1.951 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 = √(4 + 25) = 5.385

Argument in radians

θ = −111.8° = −111.8·π/180 = −1.951 rad

Exponential form

z = 5.385·e−j1.951

Answer

θ = −1.951 rad
8.Complex numbers

Answer: D (3∠−100°)

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 = 20°, 140°, −100°

The point

D = 3∠−100°

Answer

D (3∠−100°)
9.Differential equations

Answer: T = 24.66 °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) = 30 + (−27)·e−0.06t

Substitute t

T(27) = 30 + (−27)·0.1979 = 24.66 °C

Answer

T = 24.66 °C
10.Differential equations

Answer: y(1.5) = −0.2052

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 = −1 ± 2j

Constants

A = 1, B = 1·1/2 = 0.5

Evaluate

y(1.5) = e−1.5(1·cos 3 + 0.5·sin 3) = −0.2052

Answer

y(1.5) = −0.2052
11.Matrices & determinants

Answer: det A = 405

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 = 11, M₁₂ = 25, M₁₃ = −45

Cofactor expansion

det A = 5·11 − (−5)·25 + (−5)·(−45) = 405

Answer

det A = 405
12.Matrices & determinants

Answer: x = −2.652

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)·(−5) − 2·1 = 23

Replace the x column

det Aₓ = 11·(−5) − 2·3 = −61
x = −61/23 = −2.652

Answer

x = −2.652
y = −1.13
13.Vectors: dot & cross product

Answer: W = −120 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 = (2, −3, 3) − (4, 3, 3) = (−2, −6, 0) m

Work

W = F·d = 15·(−2) + 15·(−6) + 30·0 = −120 J

Answer

W = −120 J

Negative work: the force acts mostly against the motion.

14.Vectors: dot & cross product

Answer: (a × b)y = 10

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 = (−2)·0 − (−5)·3 = 15
(a × b)y = azbx − axbz = (−5)·(−2) − 6·0 = 10
(a × b)z = axby − aybx = 6·3 − (−2)·(−2) = 14

Requested component

(a × b)y = 10

Answer

(a × b)y = 10

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

15.Probability

Answer: P = 0.3333

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 = 6/10 · 5/9 = 30/90 = 0.3333

Answer

P = 0.3333
16.Probability

Answer: σY = 6.016

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) = 1.447, σX = 1.203

Linear change

σY = |5|·1.203 = 6.016

Answer

σY = 6.016

The mean does move: E(Y) = 5·2.55 + 4 = 16.75.

17.Descriptive statistics

Answer: s = 13.17

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 = 183/6 = 30.5

Squared deviations

Σ(x − x̄)² = (−5.5)² + (−12.5)² + (−9.5)² + 22.5² + 8.5² + (−3.5)² = 867.5

Standard deviation

s = √(867.5/5) = 13.17

Answer

s = 13.17
18.Descriptive statistics

Answer: P = 0.1056

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
19.Normal distribution

Answer: P = 0.695

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 total

μT = 144 kg, σT = 0.8·√12 = 2.771 kg

Standardise

z = (142.6 − 144)/2.771 = −0.51

Probability

P = 1 − Φ(−0.51) = 0.695

Answer

P = 0.695
20.Normal distribution

Answer: x = 466 h

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) = 0.2

z from the table

z ≈ −0.84

Back to the original units

x = 500 − 0.84·40 = 466 h

Answer

x = 466 h

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