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 − 4x + 8)/(2x2 + x + 5).

f(x) = (−4x2 − 4x + 8)/(2x2 + x + 5)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (e−2x − 1 + 2x)/x².

f(x) = (e−2x − 1 + 2x)/x²x → 0
L =
3.DerivativesMedium
xyx = 0.5

Let f(x) = x2·e0.5x. Find f′(0.5).

f(x) = x2·e0.5xx = 0.5
f′(0.5) =
4.DerivativesHard
xy

Let f(x) = x3 − 4.5x2 − 4. Find the x-coordinate x of the local minimum.

f(x) = x3 − 4.5x2 − 4
x =
5.IntegralsMedium
xy03

Evaluate I = ∫03 ex dx.

f(x) = ex[0, 3]
I =
6.IntegralsHard
xy1.667y = 5xy = 3x²

The region between the line y = 5x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.

y = 5xy = 3x²
V =
7.Complex numbersMedium
ReIm11−1−122−2−233−3−3θz

Find the argument θ of z = 3 − j2 in degrees, as the principal value between −180° and 180°.

z = 3 − j2
θ = °
8.Complex numbersHard
ReIm11−1−122−2−2z₀z₁z₂z₃

The roots of zⁿ = w with n = 4 and w = 16∠−90° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2, 3. Find the real part of z₃.

w = 16∠−90°n = 4k = 3
Re(z₃) =
9.Differential equationsMedium
tyy(0) = 1y(0.5) = ?

Solve y″ + 7y′ + 10y = 0 with y(0) = 1 and y′(0) = 6. Find y(0.5).

y″ + 7y′ + 10y = 0y(0) = 1y′(0) = 6
y(0.5) =
10.Differential equationsHard
t (min)m (kg)25 minm = ?

A tank holds 100 L of water in which 20 kg of salt is dissolved. Brine with 0.05 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 25 min?

V = 100 Lm0 = 20 kgc = 0.05 kg/Lr = 5 L/min
m = kg
11.Matrices & determinantsMedium
A =60−60B =64−15

Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).

A = [6 0; −6 0]B = [6 4; −1 5]
c22 =
12.Matrices & determinantsHard
A =−5−4−63b =−10−12

Solve the system −5x − 4y = −10 and −6x + 3y = −12. Find x.

−5x − 4y = −10−6x + 3y = −12
x =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (−10, −30, 30) N acts on an object while it moves in a straight line from point A(−3, −1, 0) to point B(2, 4, −3). Coordinates are in metres. Find the work W done by the force.

F = (−10, −30, 30) NA(−3, −1, 0) mB(2, 4, −3) m
W = J
14.Vectors: dot & cross productHard
xyzab

For a = (0, −5, −5) and b = (4, 4, −6), find the x-component (a × b)x of the cross product a × b.

a = (0, −5, −5)b = (4, 4, −6)
(a × b)x =
15.ProbabilityMedium
red: 7blue: 9

A bag holds 7 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 9
P =
16.ProbabilityHard
n = 7, p = 0.6: P(X ≤ 1) = ?k01234567

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

n = 7p = 0.6
P =
17.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 38, 18, 28, 48, 55, 42.

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

A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is below 97.5?

μ = 100σ = 5x = 97.5
P =
19.Normal distributionMedium
88118.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 88 and 118.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 88x2 = 118.75
P =
20.Normal distributionHard
157164171178185192199cmbottom 75%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 75% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm75%
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

(−4 + (−4)/x + 8/x2)/(2 + 1/x + 5/x2)

Let x → ∞

L = −4/2 = −2

Answer

L = −2

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

2.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).

First application

(−2·e−2x + 2)/(2x) → 0/0

Second application

4·e−2x/2

Substitute x = 0

L = 4·1/2 = 2

Answer

L = 2
3.Derivatives

Answer: f′(0.5) = 1.445

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′ = e0.5x·(2x + 0.5·x2)

Substitute

f′(0.5) = 1.284·(1 + 0.125) = 1.445

Answer

f′(0.5) = 1.445
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 − 9x = 3·(x2 − 3x)

Solve f′(x) = 0

x = 0 or x = 3

Which one is which

f″(x) = 6x − 9: negative at a maximum, positive at a minimum.

f″(0) = −9, f″(3) = 9
x = 3

Answer

x = 3
5.Integrals

Answer: I = 19.09

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) = ex/1

Evaluate at the limits

I = (e3 − 1)/1 = (20.09 − 1)/1 = 19.09

Answer

I = 19.09
6.Integrals

Answer: V = 48.48

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

5x = 3x² → x = 0, x = 1.667

Washer method

V = π∫₀1.667 ((5x)² − (3x²)²) dx

Integrate

V = π(25·1.667³/3 − 9·1.667⁵/5) = π·15.43 = 48.48

Answer

V = 48.48
7.Complex numbers

Answer: θ = −33.69 °

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.

Quadrant

a = 3, b = −2 → IV

Reference angle

α = atan(2/3) = 33.69°

Argument

θ = −α = −33.69°

Answer

θ = −33.69°

In polar form z = 3.606∠−33.69°.

8.Complex numbers

Answer: Re(z₃) = −0.7654

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

161/4 = 2

Argument of the root

θ₃ = (−90° + 360°·3)/4 = 247.5° → −112.5°

Rectangular form

z₃ = 2·(cos −112.5° + j sin −112.5°) = −0.7654 − j1.848
Re(z₃) = −0.7654

Answer

Re(z₃) = −0.7654

The 4 roots lie on the circle of radius 2, 90° apart.

9.Differential equations

Answer: y(0.5) = 1.13

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.

Characteristic roots

(r + 2)(r + 5) = 0 → r = −2, −5

Constants from the initial values

A + B = 1, −2A − 5B = 6
A = 3.667, B = −2.667

Evaluate

y(0.5) = 3.667·e−1 + (−2.667)·e−2.5 = 1.13

Answer

y(0.5) = 1.13
10.Differential equations

Answer: m = 9.298 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 = 5·0.05 − (5/100)·m = 0.25 − 0.05·m

Solve

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

m(t) = 5 + 15·e−t/20

Substitute t

m(25) = 5 + 15·0.2865 = 9.298 kg

Answer

m = 9.298 kg
11.Matrices & determinants

Answer: c22 = −24

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

c22 = (−6)·4 + 0·5 = −24

Answer

c22 = −24
12.Matrices & determinants

Answer: x = 2

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)·3 − (−4)·(−6) = −39

Replace the x column

det Aₓ = (−10)·3 − (−4)·(−12) = −78
x = −78/(−39) = 2

Answer

x = 2
y = 0
13.Vectors: dot & cross product

Answer: W = −290 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, 4, −3) − (−3, −1, 0) = (5, 5, −3) m

Work

W = F·d = (−10)·5 + (−30)·5 + 30·(−3) = −290 J

Answer

W = −290 J

Negative work: the force acts mostly against the motion.

14.Vectors: dot & cross product

Answer: (a × b)x = 50

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 = (−5)·(−6) − (−5)·4 = 50
(a × b)y = azbx − axbz = (−5)·4 − 0·(−6) = −20
(a × b)z = axby − aybx = 0·4 − (−5)·4 = 20

Requested component

(a × b)x = 50

Answer

(a × b)x = 50

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

15.Probability

Answer: P = 0.175

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 = 7/16 · 6/15 = 42/240 = 0.175

Answer

P = 0.175
16.Probability

Answer: P = 0.01884

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) = 0.47 = 0.001638
P(X = 1) = 7·0.6·0.46 = 0.0172

Add them

P = 0.001638 + 0.0172 = 0.01884

Answer

P = 0.01884
17.Descriptive statistics

Answer: s = 13.45

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 = 229/6 = 38.17

Squared deviations

Σ(x − x̄)² = (−0.1667)² + (−20.17)² + (−10.17)² + 9.833² + 16.83² + 3.833² = 904.8

Standard deviation

s = √(904.8/5) = 13.45

Answer

s = 13.45
18.Descriptive statistics

Answer: P = 0.3085

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 0.3085

Answer

P = 0.3085
19.Normal distribution

Answer: P = 0.6825

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 = (88 − 100)/15 = −0.8
z₂ = (118.75 − 100)/15 = 1.25

Area between

P = Φ(1.25) − Φ(−0.8) = 0.8944 − 0.2119 = 0.6825

Answer

P = 0.6825
20.Normal distribution

Answer: x = 182.7 cm

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) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 178 + 0.67·7 = 182.7 cm

Answer

x = 182.7 cm

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