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→∞ (√(x2 − 10x) − x).

f(x) = √(x2 − 10x) − xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 4x)/sin(5x), with x in radians.

f(x) = ln(1 + 4x)/sin(5x)x → 0
L =
3.DerivativesMedium
xyx = 2

Let f(x) = ln(x2 + 1). Find f′(2).

f(x) = ln(x2 + 1)x = 2
f′(2) =
4.DerivativesHard
xyx = 3

The tangent line to f(x) = −x2 + 3x + 6 at x = 3 crosses the y-axis at y = b. Find b.

f(x) = −x2 + 3x + 6x = 3
b =
5.IntegralsMedium
xy14

Evaluate I = ∫14 5/x dx.

f(x) = 5/x[1, 4]
I =
6.IntegralsHard
xy2y = x²eˣ

Evaluate I = ∫₀2 x²·eˣ dx.

∫₀2 x²eˣ dx
I =
7.Complex numbersMedium
ReIm22−2−244−4−4θz

The number z = 4·e−j0.5 has its angle in radians. Find its imaginary part b.

r = 4θ = −0.5 rad
b =
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

Solve z² + 8z + 17 = 0. The roots are a ± jb; find the real part a.

p = 8q = 17
a =
9.Differential equationsMedium
t (min)T (°C)15Ta​327 °C159 °C

A forged steel part at 500 °C is left to cool in a workshop at 22 °C. After 15 min its temperature is 327 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 159 °C?

T0 = 500 °CTa = 22 °CT(15) = 327 °CT = 159 °C
t = min
10.Differential equationsHard
xy13(1, −2)y(3) = ?

Solve the initial value problem y′ + y/x = 4x for x > 0, with y(1) = −2. Find y(3).

y′ + y/x = 4xy(1) = −2
y(3) =
11.Matrices & determinantsMedium
A =5−1−3−143−20−3

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

A = [5 −1 −3; −1 4 3; −2 0 −3]
det A =
12.Matrices & determinantsHard
A =5−5−32b =−7−10

Solve the system 5x − 5y = −7 and −3x + 2y = −10. Find x.

5x − 5y = −7−3x + 2y = −10
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (−3, 6, −6) onto b = (2, −1, 6), that is, the component of a in the direction of b.

a = (−3, 6, −6)b = (2, −1, 6)
compb a =
14.Vectors: dot & cross productHard
xyzABC

Find the area A of the triangle with corners A(−3, 4, −2), B(−4, −2, 3) and C(3, −4, 4). Coordinates are in metres.

A(−3, 4, −2)B(−4, −2, 3)C(3, −4, 4)
A = m²
15.ProbabilityMedium
xp(x)100.1200.3300.35400.25μ = 27.5

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

x: 10, 20, 30, 40p(x): 0.1, 0.3, 0.35, 0.25
σ =
16.ProbabilityHard
xp(x)00.220.3540.3560.1μ

X has the distribution shown, and Y = 10X + 4. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.2, 0.35, 0.35, 0.1Y = 10X + 4
σY =
17.Descriptive statisticsMedium
1520253035

Find the sample standard deviation s of the data: 12, 36, 18, 15, 11, 23.

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

A quantity is normally distributed with mean 100 and standard deviation 8. What proportion P of values is above 92?

μ = 100σ = 8x = 92
P =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8230

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8230. Find z to two decimals.

A = 0.8230
z =
20.Normal distributionHard
557085100115130145pointsbottom 40%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points40%
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).

Multiply by the conjugate

f(x) = −10x/(√(x2 − 10x) + x)

Divide by x

f(x) = −10/(√(1 + (−10)/x) + 1)

Let x → ∞

L = −10/(1 + 1) = −5

Answer

L = −5
2.Limits

Answer: L = 0.8

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).

L'Hôpital's rule

L = limx→0 [4/(1 + 4x)]/[5·cos 5x]

Substitute x = 0

L = 4/5 = 0.8

Answer

L = 0.8
3.Derivatives

Answer: f′(2) = 0.8

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 + 1)

Substitute

f′(2) = 4/5 = 0.8

Answer

f′(2) = 0.8
4.Derivatives

Answer: b = 15

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) = 6
f′(x) = −2x + 3 → f′(3) = −3

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 6 − (−3)·3 = 15

Answer

b = 15
5.Integrals

Answer: I = 6.931

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 4 − ln 1) = 5·ln 4 = 6.931

Answer

I = 6.931
6.Integrals

Answer: I = 12.78

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

7.389·(2) − 1·2 = 12.78

Answer

I = 12.78
7.Complex numbers

Answer: b = −1.918

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 −0.5 + j sin −0.5)

Components (radians)

a = 4·cos −0.5 = 3.51
b = 4·sin −0.5 = −1.918
b = −1.918

Answer

b = −1.918
8.Complex numbers

Answer: a = −4

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.

Discriminant

p² − 4q = 64 − 68 = −4 < 0

Quadratic formula

z = (−8 ± j√4)/2 = −4 + j, −4 − j

Answer

a = −4

Answer

a = −4
9.Differential equations

Answer: t = 41.72 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(478/305)/15 = 0.02995 min⁻¹

Time to reach the target

t = ln(478/137)/0.02995 = 41.72 min

Answer

t = 41.72 min
10.Differential equations

Answer: y(3) = 10.89

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∫1/x dx = x1

Integrate

(x1y)′ = 4x2 → x1y = 4x3/3 + C

Initial condition

−2 = 4/3 + C → C = −3.333

Evaluate

y(3) = (4·33/3 + (−3.333))/31 = 10.89

Answer

y(3) = 10.89
11.Matrices & determinants

Answer: det A = −75

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 = −12, M₁₂ = 9, M₁₃ = 8

Cofactor expansion

det A = 5·(−12) − (−1)·9 + (−3)·8 = −75

Answer

det A = −75
12.Matrices & determinants

Answer: x = 12.8

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 = 5·2 − (−5)·(−3) = −5

Replace the x column

det Aₓ = (−7)·2 − (−5)·(−10) = −64
x = −64/(−5) = 12.8

Answer

x = 12.8
y = 14.2
13.Vectors: dot & cross product

Answer: compb a = −7.496

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 = (−3)·2 + 6·(−1) + (−6)·6 = −48

Length of b

|b| = √(2² + (−1)² + 6²) = 6.403

Scalar projection

compb a = a·b/|b| = −48/6.403 = −7.496

Answer

compb a = −7.496

Negative: a has a component pointing against b.

14.Vectors: dot & cross product

Answer: A = 28.5 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, −6, 5)
AC = C − A = (6, −8, 6)

Cross product

(AB × AC)x = AByACz − ABzACy = (−6)·6 − 5·(−8) = 4
(AB × AC)y = ABzACx − ABxACz = 5·6 − (−1)·6 = 36
(AB × AC)z = ABxACy − AByACx = (−1)·(−8) − (−6)·6 = 44
|AB × AC| = √(4² + 36² + 44²) = 56.99

Area of the triangle

A = ½·56.99 = 28.5 m²

Answer

A = 28.5 m²
15.Probability

Answer: σ = 9.421

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 = 10·0.1 + 20·0.3 + 30·0.35 + 40·0.25 = 27.5

Mean of the squares

E(X²) = 100·0.1 + 400·0.3 + 900·0.35 + 1600·0.25 = 845

Variance

Var(X) = 845 − 27.5² = 88.75
σ = √88.75 = 9.421

Answer

σ = 9.421
16.Probability

Answer: σY = 18.19

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 = 2.7, Var(X) = 3.31, σX = 1.819

Linear change

σY = |10|·1.819 = 18.19

Answer

σY = 18.19

The mean does move: E(Y) = 10·2.7 + 4 = 31.

17.Descriptive statistics

Answer: s = 9.326

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 = 115/6 = 19.17

Squared deviations

Σ(x − x̄)² = (−7.167)² + 16.83² + (−1.167)² + (−4.167)² + (−8.167)² + 3.833² = 434.8

Standard deviation

s = √(434.8/5) = 9.326

Answer

s = 9.326
18.Descriptive statistics

Answer: P = 0.8413

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

Standard normal table

Φ(−1) = 0.1587
P = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
19.Normal distribution

Answer: z = 1.35

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 of z

Φ(z) = (1 + 0.8230)/2 = 0.9115

Read the z-table backwards

Find 0.9115 inside the table and read off z.

z = 1.35

Answer

z = 1.35
20.Normal distribution

Answer: x = 96.3 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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 100 − 0.25·15 = 96.3 points

Answer

x = 96.3 points

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