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 + 6x) − 2x).

f(x) = √(4x2 + 6x) − 2xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 5x)/(2x²), with x in radians.

f(x) = (1 − cos 5x)/(2x²)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
xy

Let f(x) = x3 − 3x2 − 9x + 5. Find the x-coordinate x of the local minimum.

f(x) = x3 − 3x2 − 9x + 5
x =
5.IntegralsMedium
xy02

Evaluate I = ∫02 2x·(x² + 1)2 dx.

f(x) = 2x·(x² + 1)2[0, 2]
I =
6.IntegralsHard
xyy = 0.5x2​y = 3x

Find the area A of the region enclosed by y = 0.5x2 and y = 3x.

y = 0.5x2y = 3x
A =
7.Complex numbersMedium
ReIm22−2−244−4−4z₁z₂w

Given z₁ = 2.5∠150° and z₂ = 0.5∠45°, find the argument θ of w = z₁/z₂ in degrees, between −180° and 180°.

z₁ = 2.5∠150°z₂ = 0.5∠45°
θ = °
8.Complex numbersHard
ReIm11−1−122−2−233−3−3z₁z₂

One root of z³ − 10z² + 34z − 40 = 0 is z₁ = 3 + j. The coefficients are real. Find the real root r.

z₁ = 3 + j
r =
9.Differential equationsMedium
tyy(0) = 1y(1.5) = ?

Solve y″ + 8y′ + 16y = 0 with y(0) = 1 and y′(0) = 0. Find y(1.5).

y″ + 8y′ + 16y = 0y(0) = 1y′(0) = 0
y(1.5) =
10.Differential equationsHard
tyK = 10ymax​ = ?

A system obeys y″ + 2y′ + 26y = 260 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + 2y′ + 26y = 260y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
A =15−55B =60−44

Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).

A = [1 5; −5 5]B = [6 0; −4 4]
c21 =
12.Matrices & determinantsHard
A =5−45−4

Find the larger eigenvalue λ of the matrix A shown.

A = [5 −4; 5 −4]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (−6, −4, −7) onto b = (−2, −5, 6), that is, the component of a in the direction of b.

a = (−6, −4, −7)b = (−2, −5, 6)
compb a =
14.Vectors: dot & cross productHard
xyzab

The unit vector n̂ is perpendicular to both a = (−1, −2, 5) and b = (−1, −4, 4) and has a positive z-component. Find its y-component n̂y.

a = (−1, −2, 5)b = (−1, −4, 4)n̂z > 0
n̂y =
15.ProbabilityMedium
xp(x)10.320.2530.0540.1550.25μ = 2.8

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

x: 1, 2, 3, 4, 5p(x): 0.3, 0.25, 0.05, 0.15, 0.25
σ =
16.ProbabilityHard
n = 7, p = 0.1: P(X = 3) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 3 correct answers?

n = 7p = 0.1k = 3
P =
17.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 16, 13, 60, 38, 51.

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

A quantity is normally distributed with mean 100 and standard deviation 8. What proportion P of values is below 104.8?

μ = 100σ = 8x = 104.8
P =
19.Normal distributionMedium
1360 kg1336 kgT = X₁ + … + X₁₆?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 16 of them is below 1336 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 16< 1336 kg
P =
20.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =

3Solutions

1.Limits

Answer: L = 1.5

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) = 6x/(√(4x2 + 6x) + 2x)

Divide by x

f(x) = 6/(√(4 + 6/x) + 2)

Let x → ∞

L = 6/(2 + 2) = 1.5

Answer

L = 1.5
2.Limits

Answer: L = 6.25

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

5·sin 5x/(4x) → 0/0

Second application

25·cos 5x/4

Substitute x = 0

L = 25/4 = 6.25

Answer

L = 6.25
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: x = 3

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 − 6x − 9 = 3·(x2 − 2x − 3)

Solve f′(x) = 0

x = −1 or x = 3

Which one is which

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

f″(−1) = −12, f″(3) = 12
x = 3

Answer

x = 3
5.Integrals

Answer: I = 41.33

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.

Substitution

u = x² + 1, du = 2x dx
I = ∫15 u2 du

Integrate and evaluate

I = [u3/3]15 = (125 − 1)/3 = 41.33

Answer

I = 41.33
6.Integrals

Answer: A = 18

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 = 3x → x = 0, x = 6

Integrate upper minus lower

A = [1.5x2 − 0.1667x3]06 = 18

Answer

A = 18
7.Complex numbers

Answer: θ = 105 °

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 / 0.5 = 5
θ = 150° − 45° = 105°

Argument

θ = 105°

Answer

θ = 105°
8.Complex numbers

Answer: r = 4

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.

Conjugate pair

z₂ = 3 − j
(z − z₁)(z − z₂) = z² − 2·3·z + (3² + 1²) = z² − 6z + 10

Factor out the quadratic

z³ − 10z² + 34z − 40 = (z² − 6z + 10)(z − 4)

Real root

r = 4

Answer

r = 4

Check with the sum of the roots: z₁ + z₂ + r = 6 + 4 = 10 = −(−10).

9.Differential equations

Answer: y(1.5) = 0.01735

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.

Double root

(r + 4)² = 0 → r = −4

Constants

A = y(0) = 1
y′(0) = B − 4A → B = 0 + 4 = 4

Evaluate

y(1.5) = (1 + 4·1.5)·e−6 = 0.01735

Answer

y(1.5) = 0.01735
10.Differential equations

Answer: ymax = 15.33

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 = −1 ± 5j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10
y′(0) = −1A + 5B = 0 → B = −2

Peak

t = π/5 = 0.6283
ymax = 10·(1 + e−1·π/5) = 15.33

Answer

ymax = 15.33
11.Matrices & determinants

Answer: c21 = −50

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

c21 = (−5)·6 + 5·(−4) = −50

Answer

c21 = −50
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 − (1)λ + (0) = 0

Solve

λ = (1 ± √(1))/2 → λ = 0, 1
λ = 1

Answer

λ = 1
13.Vectors: dot & cross product

Answer: compb a = −1.24

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) + (−4)·(−5) + (−7)·6 = −10

Length of b

|b| = √((−2)² + (−5)² + 6²) = 8.062

Scalar projection

compb a = a·b/|b| = −10/8.062 = −1.24

Answer

compb a = −1.24

Negative: a has a component pointing against b.

14.Vectors: dot & cross product

Answer: n̂y = −0.08192

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)·4 − 5·(−4) = 12
(a × b)y = azbx − axbz = 5·(−1) − (−1)·4 = −1
(a × b)z = axby − aybx = (−1)·(−4) − (−2)·(−1) = 2

Normalise

The z-component of a × b is already positive.

|a × b| = √(12² + (−1)² + 2²) = 12.21
n̂ = (12, −1, 2)/12.21 = (0.9831, −0.08192, 0.1638)
n̂y = −0.08192

Answer

n̂y = −0.08192
15.Probability

Answer: σ = 1.6

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 = 1·0.3 + 2·0.25 + 3·0.05 + 4·0.15 + 5·0.25 = 2.8

Mean of the squares

E(X²) = 1·0.3 + 4·0.25 + 9·0.05 + 16·0.15 + 25·0.25 = 10.4

Variance

Var(X) = 10.4 − 2.8² = 2.56
σ = √2.56 = 1.6

Answer

σ = 1.6
16.Probability

Answer: P = 0.02296

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.

Binomial formula

P(X = 3) = 35·0.13·0.94 = 35·0.001·0.6561 = 0.02296

Answer

P = 0.02296
17.Descriptive statistics

Answer: s = 20.82

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 = 178/5 = 35.6

Squared deviations

Σ(x − x̄)² = (−19.6)² + (−22.6)² + 24.4² + 2.4² + 15.4² = 1733

Standard deviation

s = √(1733/4) = 20.82

Answer

s = 20.82
18.Descriptive statistics

Answer: P = 0.7257

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 0.7257

Answer

P = 0.7257
19.Normal distribution

Answer: P = 0.3085

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 = 1360 kg, σT = 12·√16 = 48 kg

Standardise

z = (1336 − 1360)/48 = −0.5

Probability

P = Φ(−0.5) = 0.3085

Answer

P = 0.3085
20.Normal distribution

Answer: P = 0.8413

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 difference

μ = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413

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