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 + 3/x)x. Give a decimal value.

f(x) = (1 + 3/x)xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (e−0.5x − 1 + 0.5x)/x².

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

Let f(x) = (3x + 5)4. Find f′(−1).

f(x) = (3x + 5)4x = −1
f′(−1) =
4.DerivativesHard
xy

Let f(x) = x3 + 4.5x2 + 6x − 2. Find the x-coordinate x of the local maximum.

f(x) = x3 + 4.5x2 + 6x − 2
x =
5.IntegralsMedium
xy13

Evaluate I = ∫13 5/x dx.

f(x) = 5/x[1, 3]
I =
6.IntegralsHard
xyy = 2x2​y = 3x

Find the area A of the region enclosed by y = 2x2 and y = 3x.

y = 2x2y = 3x
A =
7.Complex numbersMedium
ReIm55−5−51010−10−10θz

Write z = −3 − j9 in exponential form r·ejθ. Give θ in radians, between −π and π.

z = −3 − j9
θ = rad
8.Complex numbersHard
ReIm11−1−122−2−2z₀z₁z₂z₃z₄z₅

The roots of zⁿ = w with n = 6 and w = 64∠−120° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 5. Find the real part of z₂.

w = 64∠−120°n = 6k = 2
Re(z₂) =
9.Differential equationsMedium
t (min)T (°C)15Ta​15 °C12 °C

A can of soda at 6 °C is taken out of the fridge into a room at 26 °C. After 15 min its temperature is 15 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 12 °C?

T0 = 6 °CTa = 26 °CT(15) = 15 °CT = 12 °C
t = min
10.Differential equationsHard
tyK = 1y(1) = ?

A system obeys y″ + y′ + 25.25y = 25.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.

y″ + y′ + 25.25y = 25.25y(0) = 0y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
A =241−3B =56−52

Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).

A = [2 4; 1 −3]B = [5 6; −5 2]
c11 =
12.Matrices & determinantsHard
A =4312

Find the larger eigenvalue λ of the matrix A shown.

A = [4 3; 1 2]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (5, −5, −3) and b = (−6, 1, 4), in degrees.

a = (5, −5, −3)b = (−6, 1, 4)
θ = °
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (−60, 60, −10) N acts at point A(−1, −3, −3). Coordinates are in metres. Find the magnitude |M| of the moment of the force about the origin O(0, 0, 0).

F = (−60, 60, −10) NA(−1, −3, −3) mO(0, 0, 0) m
|M| = N·m
15.ProbabilityMedium
xp(x)10.0520.3530.340.3μ = 2.85

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

x: 1, 2, 3, 4p(x): 0.05, 0.35, 0.3, 0.3
Var(X) =
16.ProbabilityHard
xp(x)00.3510.320.35μ

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

x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 0.5X + 4
σY =
17.Descriptive statisticsMedium
202530354045

Find the sample standard deviation s of the data: 43, 18, 41, 17, 48, 17.

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

Exam scores are normally distributed with mean 100 and standard deviation 8. Find the z-score z of a score of 110.

μ = 100σ = 8x = 110
z =
19.Normal distributionMedium
88122.5P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 88 and 122.5 points? Give P as a decimal.

μ = 100σ = 15x1 = 88x2 = 122.5
P =
20.Normal distributionHard
μ = 250x = ?80%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 80% of them lie (the 80th percentile).

μ = 250σ = 5P = 80%
x = g

3Solutions

1.Limits

Answer: L = 20.09

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 + 3/x)

Evaluate

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

ln L = 1·3 = 3

Undo the logarithm

L = e3 = 20.09

Answer

L = 20.09
2.Limits

Answer: L = 0.125

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

(−0.5·e−0.5x + 0.5)/(2x) → 0/0

Second application

0.25·e−0.5x/2

Substitute x = 0

L = 0.25·1/2 = 0.125

Answer

L = 0.125
3.Derivatives

Answer: f′(−1) = 96

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.

Chain rule

f′(x) = 4·(3x + 5)3·3

Substitute

f′(−1) = 4·23·3 = 96

Answer

f′(−1) = 96
4.Derivatives

Answer: x = −2

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 + 6 = 3·(x2 + 3x + 2)

Solve f′(x) = 0

x = −2 or x = −1

Which one is which

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

f″(−2) = −3, f″(−1) = 3
x = −2

Answer

x = −2
5.Integrals

Answer: I = 5.493

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) = 5·ln x

Evaluate at the limits

I = 5·(ln 3 − ln 1) = 5·ln 3 = 5.493

Answer

I = 5.493
6.Integrals

Answer: A = 1.125

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

2x2 = 3x → x = 0, x = 1.5

Integrate upper minus lower

A = [1.5x2 − 0.6667x3]01.5 = 1.125

Answer

A = 1.125
7.Complex numbers

Answer: θ = −1.893 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 = √(9 + 81) = 9.487

Argument in radians

θ = −108.4° = −108.4·π/180 = −1.893 rad

Exponential form

z = 9.487·e−j1.893

Answer

θ = −1.893 rad
8.Complex numbers

Answer: Re(z₂) = −0.3473

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

641/6 = 2

Argument of the root

θ₂ = (−120° + 360°·2)/6 = 100°

Rectangular form

z₂ = 2·(cos 100° + j sin 100°) = −0.3473 + j1.97
Re(z₂) = −0.3473

Answer

Re(z₂) = −0.3473

The 6 roots lie on the circle of radius 2, 60° apart.

9.Differential equations

Answer: t = 8.949 min

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.

Cooling constant

k = ln(-20/-11)/15 = 0.03986 min⁻¹

Time to reach the target

t = ln(-20/(−14))/0.03986 = 8.949 min

Answer

t = 8.949 min
10.Differential equations

Answer: y(1) = 0.8861

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 = 1, r = −0.5 ± 5j

Constants from the initial values

y(0) = 1 + A = 0 → A = −1
y′(0) = −0.5A + 5B = 0 → B = −0.1

Evaluate

y(1) = 1·(1 − e−0.5(cos 5 + 0.1·sin 5)) = 0.8861

Answer

y(1) = 0.8861
11.Matrices & determinants

Answer: c11 = −10

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

c11 = 2·5 + 4·(−5) = −10

Answer

c11 = −10
12.Matrices & determinants

Answer: λ = 5

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 − (6)λ + (5) = 0

Solve

λ = (6 ± √(16))/2 → λ = 1, 5
λ = 5

Answer

λ = 5
13.Vectors: dot & cross product

Answer: θ = 147.2 °

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·(−6) + (−5)·1 + (−3)·4 = −47

Magnitudes

|a| = √(5² + (−5)² + (−3)²) = 7.681
|b| = √((−6)² + 1² + 4²) = 7.28

Angle

cos θ = −47/(7.681·7.28) = −0.8405
θ = arccos(−0.8405) = 147.2°

Answer

θ = 147.2°
14.Vectors: dot & cross product

Answer: |M| = 361.4 N·m

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.

Position vector from O to A

r = A − O = (−1, −3, −3) − (0, 0, 0) = (−1, −3, −3) m

Moment M = r × F

(r × F)x = ryFz − rzFy = (−3)·(−10) − (−3)·60 = 210 N·m
(r × F)y = rzFx − rxFz = (−3)·(−60) − (−1)·(−10) = 170 N·m
(r × F)z = rxFy − ryFx = (−1)·60 − (−3)·(−60) = −240 N·m

Magnitude

|M| = √(210² + 170² + (−240)²) = 361.4 N·m

Answer

|M| = 361.4 N·m
15.Probability

Answer: Var(X) = 0.8275

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 = 1·0.05 + 2·0.35 + 3·0.3 + 4·0.3 = 2.85

Mean of the squares

E(X²) = 1·0.05 + 4·0.35 + 9·0.3 + 16·0.3 = 8.95

Variance

Var(X) = 8.95 − 2.85² = 0.8275

Answer

Var(X) = 0.8275
16.Probability

Answer: σY = 0.4183

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 = 1, Var(X) = 0.7, σX = 0.8367

Linear change

σY = |0.5|·0.8367 = 0.4183

Answer

σY = 0.4183

The mean does move: E(Y) = 0.5·1 + 4 = 4.5.

17.Descriptive statistics

Answer: s = 14.79

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 = 184/6 = 30.67

Squared deviations

Σ(x − x̄)² = 12.33² + (−12.67)² + 10.33² + (−13.67)² + 17.33² + (−13.67)² = 1093

Standard deviation

s = √(1093/5) = 14.79

Answer

s = 14.79
18.Descriptive statistics

Answer: z = 1.25

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 = (110 − 100)/8 = 1.25

Answer

z = 1.25
19.Normal distribution

Answer: P = 0.7213

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 = (122.5 − 100)/15 = 1.5

Area between

P = Φ(1.5) − Φ(−0.8) = 0.9332 − 0.2119 = 0.7213

Answer

P = 0.7213
20.Normal distribution

Answer: x = 254.2 g

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) = 0.8 → z = 0.8416

Back to x

x = 250 + 0.8416·5 = 254.21 g

Answer

x = 254.21 g

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