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→∞ (5x2 − 2x + 9)/(x2 + x + 1).

f(x) = (5x2 − 2x + 9)/(x2 + x + 1)x → ∞
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 = 2

Let f(x) = ln(x2 + 4). Find f′(2).

f(x) = ln(x2 + 4)x = 2
f′(2) =
4.DerivativesHard
xyx = 3

The tangent line to f(x) = −x2 + 5 at x = 3 crosses the y-axis at y = b. Find b.

f(x) = −x2 + 5x = 3
b =
5.IntegralsMedium
xy1y = 2x

The region under y = 2x from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 2x0 ≤ x ≤ 1
V =
6.IntegralsHard
xy3y = x²eˣ

Evaluate I = ∫₀3 x²·eˣ dx.

∫₀3 x²eˣ dx
I =
7.Complex numbersMedium
ReIm11−1−122−2−233−3−3θz

The number z = 4·ej0.6 has its angle in radians. Find its real part a.

r = 4θ = 0.6 rad
a =
8.Complex numbersHard
ReIm11−1−122−2−233−3−3ABCD

Exactly one of the points A–D is a root of z⁴ = 81∠180°. Click it.

w = 81∠180°n = 4
  1. 1A (3∠90°)
  2. 2B (3∠180°)
  3. 3C (3∠45°)
  4. 4D (1.5∠45°)
9.Differential equationsMedium
t (min)T (°C)20Ta​21 °C

A can of soda at 5 °C is taken out of the fridge into a room at 26 °C. After 20 min its temperature is 21 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 5 °CTa = 26 °CT(20) = 21 °C
k = min⁻¹
10.Differential equationsHard
ty0.5y(0.5) = ?

Solve y′ + 3y = 8e−0.5t with y(0) = 5. Find y(0.5).

y′ + 3y = 8e−0.5ty(0) = 5
y(0.5) =
11.Matrices & determinantsMedium
A =6−1−65B =−6−15−2

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

A = [6 −1; −6 5]B = [−6 −1; 5 −2]
c12 =
12.Matrices & determinantsHard
A =4−17−4

Find the larger eigenvalue λ of the matrix A shown.

A = [4 −1; 7 −4]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the value of k for which the vectors a = (2, −2, −6) and b = (4, k, −1) are perpendicular.

a = (2, −2, −6)b = (4, k, −1)
k =
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (−40, 50, −40) N acts at point A(4, 4, 4). Coordinates are in metres. Find the magnitude |M| of the moment of the force about point O(3, 3, −2).

F = (−40, 50, −40) NA(4, 4, 4) mO(3, 3, −2) m
|M| = N·m
15.ProbabilityMedium
xp(x)10.3520.330.0540.3μ = 2.3

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

x: 1, 2, 3, 4p(x): 0.35, 0.3, 0.05, 0.3
Var(X) =
16.ProbabilityHard
xp(x)00.0510.3520.330.3μ

X has the distribution shown, and Y = 5X + 10. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.05, 0.35, 0.3, 0.3Y = 5X + 10
σY =
17.Descriptive statisticsMedium
15202530354045

Find the sample standard deviation s of the data: 13, 16, 27, 47, 46.

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

A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is below 192.5?

μ = 170σ = 15x = 192.5
P =
19.Normal distributionMedium
65.580.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 65.5 and 80.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 65.5x2 = 80.5
P =
20.Normal distributionHard
557085100115130145pointstop 85%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 85% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points85%
x = points

3Solutions

1.Limits

Answer: L = 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).

Divide by the highest power

(5 + (−2)/x + 9/x2)/(1 + 1/x + 1/x2)

Let x → ∞

L = 5/1 = 5

Answer

L = 5

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

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′(2) = 0.5

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 for the logarithm

f′(x) = 2x/(x2 + 4)

Substitute

f′(2) = 4/8 = 0.5

Answer

f′(2) = 0.5
4.Derivatives

Answer: b = 14

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(3) = −4
f′(x) = −2x → f′(3) = −6

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −4 − (−6)·3 = 14

Answer

b = 14
5.Integrals

Answer: V = 4.189

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∫₀1 4x² dx

Integrate

V = π·4·1³/3 = 4.189

Answer

V = 4.189

Check: a cone of radius 2 and height 1 has (1/3)πr²h = 4.189.

6.Integrals

Answer: I = 98.43

Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.

First round

∫x²eˣ dx = x²eˣ − ∫2x eˣ dx

Second round

∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)

Limits

20.09·(5) − 1·2 = 98.43

Answer

I = 98.43
7.Complex numbers

Answer: a = 3.301

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.

Euler's formula

z = 4·(cos 0.6 + j sin 0.6)

Components (radians)

a = 4·cos 0.6 = 3.301
b = 4·sin 0.6 = 2.259
a = 3.301

Answer

a = 3.301
8.Complex numbers

Answer: C (3∠45°)

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| = 3
θ = 45°, 135°, −135°, −45°

The point

C = 3∠45°

Answer

C (3∠45°)
9.Differential equations

Answer: k = 0.07175 min⁻¹

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.

Ratio of temperature differences

e−20k = -5/-21

Solve for k

k = ln(-21/-5)/20 = 0.07175 min⁻¹

Answer

k = 0.07175 min⁻¹
10.Differential equations

Answer: y(0.5) = 2.894

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 = e3t
(e3ty)′ = 8e2.5t

General solution

y = 8/2.5·e−0.5t + C·e−3t = 3.2·e−0.5t + C·e−3t

Initial condition

5 = 3.2 + C → C = 1.8

Evaluate

y(0.5) = 3.2·0.7788 + 1.8·0.2231 = 2.894

Answer

y(0.5) = 2.894
11.Matrices & determinants

Answer: c12 = −4

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 = 6·(−1) + (−1)·(−2) = −4

Answer

c12 = −4
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 − (0)λ + (−9) = 0

Solve

λ = (0 ± √(36))/2 → λ = −3, 3
λ = 3

Answer

λ = 3
13.Vectors: dot & cross product

Answer: k = 7

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 = 2·4 + (−2)·k + (−6)·(−1) = 0
14 + (−2)·k = 0

Solve for k

k = −14/(−2) = 7

Answer

k = 7
14.Vectors: dot & cross product

Answer: |M| = 404.6 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 = (4, 4, 4) − (3, 3, −2) = (1, 1, 6) m

Moment M = r × F

(r × F)x = ryFz − rzFy = 1·(−40) − 6·50 = −340 N·m
(r × F)y = rzFx − rxFz = 6·(−40) − 1·(−40) = −200 N·m
(r × F)z = rxFy − ryFx = 1·50 − 1·(−40) = 90 N·m

Magnitude

|M| = √((−340)² + (−200)² + 90²) = 404.6 N·m

Answer

|M| = 404.6 N·m
15.Probability

Answer: Var(X) = 1.51

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 = 1·0.35 + 2·0.3 + 3·0.05 + 4·0.3 = 2.3

Mean of the squares

E(X²) = 1·0.35 + 4·0.3 + 9·0.05 + 16·0.3 = 6.8

Variance

Var(X) = 6.8 − 2.3² = 1.51

Answer

Var(X) = 1.51
16.Probability

Answer: σY = 4.548

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 = 1.85, Var(X) = 0.8275, σX = 0.9097

Linear change

σY = |5|·0.9097 = 4.548

Answer

σY = 4.548

The mean does move: E(Y) = 5·1.85 + 10 = 19.25.

17.Descriptive statistics

Answer: s = 16.12

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 = 149/5 = 29.8

Squared deviations

Σ(x − x̄)² = (−16.8)² + (−13.8)² + (−2.8)² + 17.2² + 16.2² = 1039

Standard deviation

s = √(1039/4) = 16.12

Answer

s = 16.12
18.Descriptive statistics

Answer: P = 0.9332

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) = 0.9332
P = 0.9332

Answer

P = 0.9332
19.Normal distribution

Answer: P = 0.4195

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 = (65.5 − 65)/10 = 0.05
z₂ = (80.5 − 65)/10 = 1.55

Area between

P = Φ(1.55) − Φ(0.05) = 0.9394 − 0.5199 = 0.4195

Answer

P = 0.4195
20.Normal distribution

Answer: x = 84.4 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) = 0.15

z from the table

z ≈ −1.04

Back to the original units

x = 100 − 1.04·15 = 84.4 points

Answer

x = 84.4 points

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