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→∞ (−3x2 + x + 3)/(4x2 + 5x + 2).

f(x) = (−3x2 + x + 3)/(4x2 + 5x + 2)x → ∞
L =
2.LimitsHard
xyx = 0

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

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

Let f(x) = x·sin(x), with x in radians. Find f′(0.5).

f(x) = x·sin(x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
xy

Let f(x) = x3 + 9x2 + 24x + 5. Find the x-coordinate x of the local minimum.

f(x) = x3 + 9x2 + 24x + 5
x =
5.IntegralsMedium
xy210

Evaluate I = ∫210 4/x dx.

f(x) = 4/x[2, 10]
I =
6.IntegralsHard
xy−33

Find the area A of the region between the parabola y = 9 − x² and the x-axis.

y = 9 − x²
A =
7.Complex numbersMedium
ReIm55−5−51010−10−10zABCD

The diagram shows z = −1 + j4. With w = 2∠90°, click the point that represents z·w.

z = −1 + j4w = 2∠90°
  1. 1A (−4 − j)
  2. 2B (−2 + j8)
  3. 3C (8 + j2)
  4. 4D (−8 − j2)
8.Complex numbersHard
ReIm22−2−244−4−4ABCD

Exactly one of the points A–D is a root of z³ = 8∠120°. Click it.

w = 8∠120°n = 3
  1. 1A (2∠−40°)
  2. 2B (2∠40°)
  3. 3C (3.6∠100°)
  4. 4D (2∠−140°)
9.Differential equationsMedium
t (min)T (°C)29Ta​T = ?

A can of soda at 3 °C is taken out of the fridge into a room at 22 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 29 min.

T0 = 3 °CTa = 22 °Ck = 0.03 min⁻¹t = 29 min
T = °C
10.Differential equationsHard
t (min)m (kg)10 minm = ?

A tank holds 200 L of pure water. Brine with 0.25 kg of salt per litre flows in at 20 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 10 min?

V = 200 Lm0 = 0 kgc = 0.25 kg/Lr = 20 L/min
m = kg
11.Matrices & determinantsMedium
A =1−1530−1403

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

A = [1 −1 5; 3 0 −1; 4 0 3]
det A =
12.Matrices & determinantsHard
A =−2−2−16b =511

Solve the system −2x − 2y = 5 and −x + 6y = 11. Find x.

−2x − 2y = 5−x + 6y = 11
x =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (1, −6, 3) and b = (−1, 2, 4), in degrees.

a = (1, −6, 3)b = (−1, 2, 4)
θ = °
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (−5, 5, 3), b = (0, 2, −4) and c = (2, 3, 0).

a = (−5, 5, 3)b = (0, 2, −4)c = (2, 3, 0)
V =
15.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5P(six) = 1/6 on each roll

A fair die is rolled 5 times. What is the probability P of getting at least one six?

n = 5
P =
16.ProbabilityHard
€p(x)−20.840.15180.05

A game costs €2 to play. You win €6 with probability 0.15 and €20 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€6: p = 0.15€20: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
34363840424446

Find the sample standard deviation s of the data: 42, 34, 36, 46, 44.

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

A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 188?

μ = 170σ = 12x = 188
P =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.7775

Drag z so that the shaded area between −z and z under the standard normal curve is 0.7775. Find z to two decimals.

A = 0.7775
z =
20.Normal distributionHard
557085100115130145pointstop 70%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points70%
x = points

3Solutions

1.Limits

Answer: L = −0.75

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

(−3 + 1/x + 3/x2)/(4 + 5/x + 2/x2)

Let x → ∞

L = −3/4 = −0.75

Answer

L = −0.75

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

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·e−x + 1)/(2x) → 0/0

Second application

1·e−x/2

Substitute x = 0

L = 1·1/2 = 0.5

Answer

L = 0.5
3.Derivatives

Answer: f′(0.5) = 0.9182

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 and chain rule

f′(x) = sin(x) + x·cos(x)

Substitute

f′(0.5) = 0.4794 + 0.5·0.8776 = 0.9182

Answer

f′(0.5) = 0.9182
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 + 18x + 24 = 3·(x2 + 6x + 8)

Solve f′(x) = 0

x = −4 or x = −2

Which one is which

f″(x) = 6x + 18: negative at a maximum, positive at a minimum.

f″(−4) = −6, f″(−2) = 6
x = −2

Answer

x = −2
5.Integrals

Answer: I = 6.438

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) = 4·ln x

Evaluate at the limits

I = 4·(ln 10 − ln 2) = 4·ln 5 = 6.438

Answer

I = 6.438
6.Integrals

Answer: A = 36

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.

Zeros

9 − x² = 0 → x = ±3

Integrate

A = [9x − x³/3]−33 = 18 − (−18) = 36

Answer

A = 36
7.Complex numbers

Answer: D (−8 − j2)

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.

Effect of w

|z·w| = 2·|z|, arg(z·w) = arg z + 90°

The point

z·w = −8 − j2 → D

Answer

D (−8 − j2)
8.Complex numbers

Answer: B (2∠40°)

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 = 40°, 160°, −80°

The point

B = 2∠40°

Answer

B (2∠40°)
9.Differential equations

Answer: T = 14.04 °C

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.

Solution of the cooling law

T(t) = 22 + (−19)·e−0.03t

Substitute t

T(29) = 22 + (−19)·0.419 = 14.04 °C

Answer

T = 14.04 °C
10.Differential equations

Answer: m = 31.61 kg

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.

Set up the equation

dm/dt = 20·0.25 − (20/200)·m = 5 − 0.1·m

Solve

The integrating factor is ert/V; the solution approaches the equilibrium c·V.

m(t) = 50 + (−50)·e−t/10

Substitute t

m(10) = 50 + (−50)·0.3679 = 31.61 kg

Answer

m = 31.61 kg
11.Matrices & determinants

Answer: det A = 13

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 = 0, M₁₂ = 13, M₁₃ = 0

Cofactor expansion

det A = 1·0 − (−1)·13 + 5·0 = 13

Answer

det A = 13
12.Matrices & determinants

Answer: x = −3.714

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 = (−2)·6 − (−2)·(−1) = −14

Replace the x column

det Aₓ = 5·6 − (−2)·11 = 52
x = 52/(−14) = −3.714

Answer

x = −3.714
y = 1.214
13.Vectors: dot & cross product

Answer: θ = 91.84 °

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 = 1·(−1) + (−6)·2 + 3·4 = −1

Magnitudes

|a| = √(1² + (−6)² + 3²) = 6.782
|b| = √((−1)² + 2² + 4²) = 4.583

Angle

cos θ = −1/(6.782·4.583) = −0.03217
θ = arccos(−0.03217) = 91.84°

Answer

θ = 91.84°
14.Vectors: dot & cross product

Answer: V = 112

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 b × c

(b × c)x = bycz − bzcy = 2·0 − (−4)·3 = 12
(b × c)y = bzcx − bxcz = (−4)·2 − 0·0 = −8
(b × c)z = bxcy − bycx = 0·3 − 2·2 = −4

Dot product with a

a·(b × c) = (−5)·12 + 5·(−8) + 3·(−4) = −112
V = |−112| = 112

Answer

V = 112
15.Probability

Answer: P = 0.5981

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.

Complement

P(no six) = (5/6)5 = 0.4019
P = 1 − 0.4019 = 0.5981

Answer

P = 0.5981
16.Probability

Answer: E(G) = −0.1 €

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 = 0.15), 18 (p = 0.05), −2 (p = 0.8)

Expected value

E(G) = 4·0.15 + 18·0.05 − 2·0.8 = −0.1 €

Answer

E(G) = −0.1 €

A negative value means you lose money on average.

17.Descriptive statistics

Answer: s = 5.177

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 = 202/5 = 40.4

Squared deviations

Σ(x − x̄)² = 1.6² + (−6.4)² + (−4.4)² + 5.6² + 3.6² = 107.2

Standard deviation

s = √(107.2/4) = 5.177

Answer

s = 5.177
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: z = 1.22

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 + 0.7775)/2 = 0.8888

Read the z-table backwards

Find 0.8888 inside the table and read off z.

z = 1.22

Answer

z = 1.22
20.Normal distribution

Answer: x = 92.2 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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 100 − 0.52·15 = 92.2 points

Answer

x = 92.2 points

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