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)x. Give a decimal value.

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

Find the limit L = limx→0 (1 − cos 5x)/(4x²), with x in radians.

f(x) = (1 − cos 5x)/(4x²)x → 0
L =
3.DerivativesMedium
xyx = 0.5

Let f(x) = x3·e−x. Find f′(0.5).

f(x) = x3·e−xx = 0.5
f′(0.5) =
4.DerivativesHard
xy

Let f(x) = x3 + 3x2 − 9x − 1. Find the x-coordinate x of the local maximum.

f(x) = x3 + 3x2 − 9x − 1
x =
5.IntegralsMedium
xy4y = x

The region under y = x from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = x0 ≤ x ≤ 4
V =
6.IntegralsHard
xyy = x2​y = x

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

y = x2y = x
A =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6z₁z₂w

Given z₁ = 2∠125° and z₂ = 4∠20°, find the imaginary part of w = z₁·z₂.

z₁ = 2∠125°z₂ = 4∠20°
Im(w) =
8.Complex numbersHard
ReIm11−1−122−2−2θz

Use De Moivre's theorem to find the imaginary part of z⁵ for z = 2 + j2.

z = 2 + j2n = 5
Im(z⁵) =
9.Differential equationsMedium
tyy(0) = 3y(1.5) = ?

Solve y″ + 2y′ + y = 0 with y(0) = 3 and y′(0) = 4. Find y(1.5).

y″ + 2y′ + y = 0y(0) = 3y′(0) = 4
y(1.5) =
10.Differential equationsHard
tyy(0) = 10y(0.5) = ?

Solve y″ + 3y′ + 11.25y = 0 with y(0) = 10 and y′(0) = 0 (released from rest). Find y(0.5); angles in radians.

y″ + 3y′ + 11.25y = 0y(0) = 10y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
A =−4−210−5−205−4

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

A = [−4 −2 1; 0 −5 −2; 0 5 −4]
det A =
12.Matrices & determinantsHard
A =−2−356

Find the larger eigenvalue λ of the matrix A shown.

A = [−2 −3; 5 6]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (−5, 2, −6) and b = (−2, 5, −1), in degrees.

a = (−5, 2, −6)b = (−2, 5, −1)
θ = °
14.Vectors: dot & cross productHard
xyzab

For a = (−4, 6, −5) and b = (−4, 1, −2), find the z-component (a × b)z of the cross product a × b.

a = (−4, 6, −5)b = (−4, 1, −2)
(a × b)z =
15.ProbabilityMedium
xp(x)100.25200.3300.15400.3μ = 25

For the distribution shown, find the variance Var(X) of X.

x: 10, 20, 30, 40p(x): 0.25, 0.3, 0.15, 0.3
Var(X) =
16.ProbabilityHard
€p(x)−10.8840.190.02

A game costs €1 to play. You win €5 with probability 0.1 and €10 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€5: p = 0.1€10: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 23, 58, 51, 37, 35, 56, 29.

n = 7
s =
18.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is below 38?

μ = 50σ = 8x = 38
P =
19.Normal distributionMedium
242.25256.25P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 242.25 and 256.25 g? Give P as a decimal.

μ = 250σ = 5x1 = 242.25x2 = 256.25
P =
20.Normal distributionHard
380420460500540580620hbottom 30%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h30%
x = h

3Solutions

1.Limits

Answer: L = 0.1353

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→∞ x·ln(1 − 2/x)

Evaluate

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

ln L = 1·(−2) = −2

Undo the logarithm

L = e−2 = 0.1353

Answer

L = 0.1353
2.Limits

Answer: L = 3.125

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/(8x) → 0/0

Second application

25·cos 5x/8

Substitute x = 0

L = 25/8 = 3.125

Answer

L = 3.125
3.Derivatives

Answer: f′(0.5) = 0.3791

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′ = e−x·(3x2 + (−1)·x3)

Substitute

f′(0.5) = 0.6065·(0.75 + (−0.125)) = 0.3791

Answer

f′(0.5) = 0.3791
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 = −3 or x = 1

Which one is which

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

f″(−3) = −12, f″(1) = 12
x = −3

Answer

x = −3
5.Integrals

Answer: V = 67.02

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∫₀4 1x² dx

Integrate

V = π·1·4³/3 = 67.02

Answer

V = 67.02

Check: a cone of radius 4 and height 4 has (1/3)πr²h = 67.02.

6.Integrals

Answer: A = 0.1667

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

x2 = x → x = 0, x = 1

Integrate upper minus lower

A = [0.5x2 − 0.3333x3]01 = 0.1667

Answer

A = 0.1667
7.Complex numbers

Answer: Im(w) = 4.589

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 · 4 = 8
θ = 125° + 20° = 145°

Rectangular form

Re(w) = 8·cos(145°) = −6.553
Im(w) = 8·sin(145°) = 4.589
Im(w) = 4.589

Answer

Im(w) = 4.589
8.Complex numbers

Answer: Im(z⁵) = −128

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 = 2 + j2 = 2√2∠45°

De Moivre

z⁵ = (2√2)⁵∠225° = 181∠−135°

Rectangular form

z⁵ = 181·(cos −135° + j sin −135°) = −128 − j128
Im(z⁵) = −128

Answer

Im(z⁵) = −128
9.Differential equations

Answer: y(1.5) = 3.012

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 + 1)² = 0 → r = −1

Constants

A = y(0) = 3
y′(0) = B − 1A → B = 4 + 3 = 7

Evaluate

y(1.5) = (3 + 7·1.5)·e−1.5 = 3.012

Answer

y(1.5) = 3.012
10.Differential equations

Answer: y(0.5) = 2.69

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 = −1.5 ± 3j

Constants

A = 10, B = 1.5·10/3 = 5

Evaluate

y(0.5) = e−0.75(10·cos 1.5 + 5·sin 1.5) = 2.69

Answer

y(0.5) = 2.69
11.Matrices & determinants

Answer: det A = −120

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 = 30, M₁₂ = 0, M₁₃ = 0

Cofactor expansion

det A = (−4)·30 − (−2)·0 + 1·0 = −120

Answer

det A = −120
12.Matrices & determinants

Answer: λ = 3

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 − (4)λ + (3) = 0

Solve

λ = (4 ± √(4))/2 → λ = 1, 3
λ = 3

Answer

λ = 3
13.Vectors: dot & cross product

Answer: θ = 53.93 °

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 = (−5)·(−2) + 2·5 + (−6)·(−1) = 26

Magnitudes

|a| = √((−5)² + 2² + (−6)²) = 8.062
|b| = √((−2)² + 5² + (−1)²) = 5.477

Angle

cos θ = 26/(8.062·5.477) = 0.5888
θ = arccos(0.5888) = 53.93°

Answer

θ = 53.93°
14.Vectors: dot & cross product

Answer: (a × b)z = 20

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 = 6·(−2) − (−5)·1 = −7
(a × b)y = azbx − axbz = (−5)·(−4) − (−4)·(−2) = 12
(a × b)z = axby − aybx = (−4)·1 − 6·(−4) = 20

Requested component

(a × b)z = 20

Answer

(a × b)z = 20

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

15.Probability

Answer: Var(X) = 135

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 = 10·0.25 + 20·0.3 + 30·0.15 + 40·0.3 = 25

Mean of the squares

E(X²) = 100·0.25 + 400·0.3 + 900·0.15 + 1600·0.3 = 760

Variance

Var(X) = 760 − 25² = 135

Answer

Var(X) = 135
16.Probability

Answer: E(G) = −0.3 €

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.

Net outcomes

G: 4 (p = 0.1), 9 (p = 0.02), −1 (p = 0.88)

Expected value

E(G) = 4·0.1 + 9·0.02 − 1·0.88 = −0.3 €

Answer

E(G) = −0.3 €

A negative value means you lose money on average.

17.Descriptive statistics

Answer: s = 13.74

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 = 289/7 = 41.29

Squared deviations

Σ(x − x̄)² = (−18.29)² + 16.71² + 9.714² + (−4.286)² + (−6.286)² + 14.71² + (−12.29)² = 1133

Standard deviation

s = √(1133/6) = 13.74

Answer

s = 13.74
18.Descriptive statistics

Answer: P = 0.06681

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 0.06681

Answer

P = 0.06681
19.Normal distribution

Answer: P = 0.8338

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 = (242.25 − 250)/5 = −1.55
z₂ = (256.25 − 250)/5 = 1.25

Area between

P = Φ(1.25) − Φ(−1.55) = 0.8944 − 0.06057 = 0.8338

Answer

P = 0.8338
20.Normal distribution

Answer: x = 479 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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 500 − 0.52·40 = 479 h

Answer

x = 479 h

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