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→∞ (√(9x2 + 9x) − 3x).

f(x) = √(9x2 + 9x) − 3xx → ∞
L =
2.LimitsHard
xyx = 0

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

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

Let f(x) = x·e−x. Find f′(2).

f(x) = x·e−xx = 2
f′(2) =
4.DerivativesHard
xyx = −1

The tangent line to f(x) = 3x2 + 4x + 2 at x = −1 crosses the y-axis at y = b. Find b.

f(x) = 3x2 + 4x + 2x = −1
b =
5.IntegralsMedium
xy13

Evaluate I = ∫13 2/x dx.

f(x) = 2/x[1, 3]
I =
6.IntegralsHard
xyy = x2​y = 4x

Find the area A of the region enclosed by y = x2 and y = 4x.

y = x2y = 4x
A =
7.Complex numbersMedium
ReIm22−2−244−4−4θz

Write z = −5 − j5 in exponential form r·ejθ. Give θ in radians, between −π and π.

z = −5 − j5
θ = rad
8.Complex numbersHard
ReIm22−2−244−4−4ABCD

Exactly one of the points A–D is a root of z³ = 8∠−90°. Click it.

w = 8∠−90°n = 3
  1. 1A (2∠60°)
  2. 2B (3.6∠90°)
  3. 3C (2∠90°)
  4. 4D (2∠−90°)
9.Differential equationsMedium
tyy(0) = 3y(2) = ?

Solve y″ + 4y′ + 3y = 0 with y(0) = 3 and y′(0) = 1. Find y(2).

y″ + 4y′ + 3y = 0y(0) = 3y′(0) = 1
y(2) =
10.Differential equationsHard
xy13(1, 3)y(3) = ?

Solve the initial value problem y′ + 2y/x = 2x2 for x > 0, with y(1) = 3. Find y(3).

y′ + 2y/x = 2x2y(1) = 3
y(3) =
11.Matrices & determinantsMedium
A =3−52−6B =−6−2−34

Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).

A = [3 −5; 2 −6]B = [−6 −2; −3 4]
c12 =
12.Matrices & determinantsHard
A =−12−98

Find the larger eigenvalue λ of the matrix A shown.

A = [−1 2; −9 8]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the value of k for which the vectors a = (−1, 3, −1) and b = (4, −4, k) are perpendicular.

a = (−1, 3, −1)b = (4, −4, k)
k =
14.Vectors: dot & cross productHard
xyzABC

Find the area A of the triangle with corners A(−3, −4, 1), B(−4, −1, 0) and C(3, 1, −1). Coordinates are in metres.

A(−3, −4, 1)B(−4, −1, 0)C(3, 1, −1)
A = m²
15.ProbabilityMedium
xp(x)00.210.3520.0530.1540.25μ = 1.9

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

x: 0, 1, 2, 3, 4p(x): 0.2, 0.35, 0.05, 0.15, 0.25
σ =
16.ProbabilityHard
n = 6, p = 0.7: P(X ≤ 1) = ?k0123456

6 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?

n = 6p = 0.7
P =
17.Descriptive statisticsMedium
2025303540

Find the sample standard deviation s of the data: 41, 25, 31, 21, 40, 39, 20.

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

A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is above 77.2?

μ = 70σ = 12x = 77.2
P =
19.Normal distributionMedium
200 mm199.6 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.6 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.6 mm
P =
20.Normal distributionHard
557085100115130145pointstop 75%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 75% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points75%
x = points

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) = 9x/(√(9x2 + 9x) + 3x)

Divide by x

f(x) = 9/(√(9 + 9/x) + 3)

Let x → ∞

L = 9/(3 + 3) = 1.5

Answer

L = 1.5
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·ex − 1)/(2x) → 0/0

Second application

1·ex/2

Substitute x = 0

L = 1·1/2 = 0.5

Answer

L = 0.5
3.Derivatives

Answer: f′(2) = −0.1353

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·(1 + (−1)·x)

Substitute

f′(2) = 0.1353·(1 + (−2)) = −0.1353

Answer

f′(2) = −0.1353
4.Derivatives

Answer: b = −1

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(−1) = 1
f′(x) = 6x + 4 → f′(−1) = −2

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 1 − (−2)·(−1) = −1

Answer

b = −1
5.Integrals

Answer: I = 2.197

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) = 2·ln x

Evaluate at the limits

I = 2·(ln 3 − ln 1) = 2·ln 3 = 2.197

Answer

I = 2.197
6.Integrals

Answer: A = 10.67

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 = 4x → x = 0, x = 4

Integrate upper minus lower

A = [2x2 − 0.3333x3]04 = 10.67

Answer

A = 10.67
7.Complex numbers

Answer: θ = −2.356 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 = √(25 + 25) = 7.071

Argument in radians

θ = −135° = −135·π/180 = −2.356 rad

Exponential form

z = 7.071·e−j2.356

Answer

θ = −2.356 rad
8.Complex numbers

Answer: C (2∠90°)

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| = 2
θ = −30°, 90°, −150°

The point

C = 2∠90°

Answer

C (2∠90°)
9.Differential equations

Answer: y(2) = 0.6717

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 + 1)(r + 3) = 0 → r = −1, −3

Constants from the initial values

A + B = 3, −1A − 3B = 1
A = 5, B = −2

Evaluate

y(2) = 5·e−2 + (−2)·e−6 = 0.6717

Answer

y(2) = 0.6717
10.Differential equations

Answer: y(3) = 11.09

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.

Integrating factor

μ = e∫2/x dx = x2

Integrate

(x2y)′ = 2x4 → x2y = 2x5/5 + C

Initial condition

3 = 2/5 + C → C = 2.6

Evaluate

y(3) = (2·35/5 + 2.6)/32 = 11.09

Answer

y(3) = 11.09
11.Matrices & determinants

Answer: c12 = −26

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

c12 = 3·(−2) + (−5)·4 = −26

Answer

c12 = −26
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 − (7)λ + (10) = 0

Solve

λ = (7 ± √(9))/2 → λ = 2, 5
λ = 5

Answer

λ = 5
13.Vectors: dot & cross product

Answer: k = −16

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 equals zero

a·b = (−1)·4 + 3·(−4) + (−1)·k = 0
−16 + (−1)·k = 0

Solve for k

k = −(−16)/(−1) = −16

Answer

k = −16
14.Vectors: dot & cross product

Answer: A = 12.19 m²

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.

Edge vectors

AB = B − A = (−1, 3, −1)
AC = C − A = (6, 5, −2)

Cross product

(AB × AC)x = AByACz − ABzACy = 3·(−2) − (−1)·5 = −1
(AB × AC)y = ABzACx − ABxACz = (−1)·6 − (−1)·(−2) = −8
(AB × AC)z = ABxACy − AByACx = (−1)·5 − 3·6 = −23
|AB × AC| = √((−1)² + (−8)² + (−23)²) = 24.37

Area of the triangle

A = ½·24.37 = 12.19 m²

Answer

A = 12.19 m²
15.Probability

Answer: σ = 1.513

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 = 0·0.2 + 1·0.35 + 2·0.05 + 3·0.15 + 4·0.25 = 1.9

Mean of the squares

E(X²) = 0·0.2 + 1·0.35 + 4·0.05 + 9·0.15 + 16·0.25 = 5.9

Variance

Var(X) = 5.9 − 1.9² = 2.29
σ = √2.29 = 1.513

Answer

σ = 1.513
16.Probability

Answer: P = 0.01094

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) = 0.36 = 0.000729
P(X = 1) = 6·0.7·0.35 = 0.01021

Add them

P = 0.000729 + 0.01021 = 0.01094

Answer

P = 0.01094
17.Descriptive statistics

Answer: s = 9.147

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 = 217/7 = 31

Squared deviations

Σ(x − x̄)² = 10² + (−6)² + 0² + (−10)² + 9² + 8² + (−11)² = 502

Standard deviation

s = √(502/6) = 9.147

Answer

s = 9.147
18.Descriptive statistics

Answer: P = 0.2743

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 = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
19.Normal distribution

Answer: P = 0.7881

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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (199.6 − 200)/0.5 = −0.8

Upper tail

P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
20.Normal distribution

Answer: x = 90 points

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) = 0.25

z from the table

z ≈ −0.67

Back to the original units

x = 100 − 0.67·15 = 90 points

Answer

x = 90 points

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