Engineering Mathematics: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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→∞ (3x3 + 4x − 5)/(4x3 + x + 9).

f(x) = (3x3 + 4x − 5)/(4x3 + x + 9)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 4x)/sin(3x), with x in radians.

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

Let f(x) = (3x − 1)4. Find f′(0).

f(x) = (3x − 1)4x = 0
f′(0) =
4.DerivativesHard
xyx = 1

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

f(x) = x2 − x − 2x = 1
b =
5.IntegralsMedium
xy16y = x ln x

Evaluate I = ∫₁6 x·ln x dx.

∫₁6 x ln x dx
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
ReIm55−5−51010−10−10θz

Find the argument θ of z = −9 + j5 in degrees, as the principal value between −180° and 180°.

z = −9 + j5
θ = °
8.Complex numbersHard
ReIm11−1−122−2−2θz

Find the imaginary part of (−1 + j2)³.

z = −1 + j2n = 3
Im(z³) =
9.Differential equationsMedium
t (min)T (°C)29Ta​164 °C

A forged steel part at 460 °C is left to cool in a workshop at 29 °C. After 29 min its temperature is 164 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 460 °CTa = 29 °CT(29) = 164 °C
k = min⁻¹
10.Differential equationsHard
t (min)m (kg)t = ?19 kg

A tank holds 100 L of water in which 5 kg of salt is dissolved. Brine with 0.25 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 19 kg of salt?

V = 100 Lm0 = 5 kgc = 0.25 kg/Lr = 5 L/min
t = min
11.Matrices & determinantsMedium
A =5−1−11−25−45−4

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

A = [5 −1 −1; 1 −2 5; −4 5 −4]
det A =
12.Matrices & determinantsHard
A =6−3−1−2b =−10−3

Solve the system 6x − 3y = −10 and −x − 2y = −3. Find x.

6x − 3y = −10−x − 2y = −3
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (6, −3, −1) onto b = (−2, −5, 1), that is, the component of a in the direction of b.

a = (6, −3, −1)b = (−2, −5, 1)
compb a =
14.Vectors: dot & cross productHard
xyzab

The unit vector n̂ is perpendicular to both a = (−4, 0, 3) and b = (3, 4, 3) and has a positive z-component. Find its y-component n̂y.

a = (−4, 0, 3)b = (3, 4, 3)n̂z > 0
n̂y =
15.ProbabilityMedium
n = 13k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 13? The order does not matter.

n = 13k = 4
N =
16.ProbabilityHard
n = 8, p = 0.25: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 8p = 0.25
P =
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 20, 13, 47, 56, 13.

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

A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is below 90?

μ = 100σ = 20x = 90
P =
19.Normal distributionMedium
1360 kg1288 kgT = X₁ + … + X₁₆?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 16 of them exceeds 1288 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 16> 1288 kg
P =
20.Normal distributionHard
235240245250255260265gtop 90%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 90% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g90%
x = g

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 + 4/x2 + (−5)/x3)/(4 + 1/x2 + 9/x3)

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 = 1.333

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)]/[3·cos 3x]

Substitute x = 0

L = 4/3 = 1.333

Answer

L = 1.333
3.Derivatives

Answer: f′(0) = −12

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 − 1)3·3

Substitute

f′(0) = 4·(−1)3·3 = −12

Answer

f′(0) = −12
4.Derivatives

Answer: b = −3

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) = −2
f′(x) = 2x − 1 → f′(1) = 1

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −2 − 1·1 = −3

Answer

b = −3
5.Integrals

Answer: I = 23.5

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.

Choose u and dv

u = ln x, du = dx/x; dv = x dx, v = x²/2

Antiderivative

∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4

Limits

(18·ln 6 − 9) − (0 − 0.25) = 23.5

Answer

I = 23.5
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: θ = 150.9 °

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.

Quadrant

a = −9, b = 5 → II

Reference angle

α = atan(5/9) = 29.05°

Argument

θ = 180° − α = 150.9°

Answer

θ = 150.9°

In polar form z = 10.3∠150.9°.

8.Complex numbers

Answer: Im(z³) = −2

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.

Powers step by step

z² = (−1 + j2)(−1 + j2) = −3 − j4
z³ = (−3 − j4)(−1 + j2) = 11 − j2

Check with De Moivre

z = 2.236∠116.6° → z³ = 11.18∠−10.3°
Im(z³) = −2

Answer

Im(z³) = −2
9.Differential equations

Answer: k = 0.04003 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−29k = 135/431

Solve for k

k = ln(431/135)/29 = 0.04003 min⁻¹

Answer

k = 0.04003 min⁻¹
10.Differential equations

Answer: t = 24.08 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.

Result

m(t) = 25 + (−20)·e−t/20

Solve for t

e−t/20 = (25 − 19)/(25 − 5) = 0.3
t = 20·ln(3.333) = 24.08 min

Answer

t = 24.08 min
11.Matrices & determinants

Answer: det A = −66

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 = −17, M₁₂ = 16, M₁₃ = −3

Cofactor expansion

det A = 5·(−17) − (−1)·16 + (−1)·(−3) = −66

Answer

det A = −66
12.Matrices & determinants

Answer: x = −0.7333

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 = 6·(−2) − (−3)·(−1) = −15

Replace the x column

det Aₓ = (−10)·(−2) − (−3)·(−3) = 11
x = 11/(−15) = −0.7333

Answer

x = −0.7333
y = 1.867
13.Vectors: dot & cross product

Answer: compb a = 0.3651

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 = 6·(−2) + (−3)·(−5) + (−1)·1 = 2

Length of b

|b| = √((−2)² + (−5)² + 1²) = 5.477

Scalar projection

compb a = a·b/|b| = 2/5.477 = 0.3651

Answer

compb a = 0.3651
14.Vectors: dot & cross product

Answer: n̂y = −0.7241

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

(a × b)x = aybz − azby = 0·3 − 3·4 = −12
(a × b)y = azbx − axbz = 3·3 − (−4)·3 = 21
(a × b)z = axby − aybx = (−4)·4 − 0·3 = −16

Normalise

The z-component of a × b is negative, so use −(a × b).

|a × b| = √((−12)² + 21² + (−16)²) = 29
n̂ = −(−12, 21, −16)/29 = (0.4138, −0.7241, 0.5517)
n̂y = −0.7241

Answer

n̂y = −0.7241
15.Probability

Answer: N = 715

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.

Combinations

C(13, 4) = (13·12·11·10)/(4·3·2·1) = 715

Answer

N = 715
16.Probability

Answer: P = 0.3671

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) = 0.758 = 0.1001
P(X = 1) = 8·0.25·0.757 = 0.267

Add them

P = 0.1001 + 0.267 = 0.3671

Answer

P = 0.3671
17.Descriptive statistics

Answer: s = 20.27

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 = (−9.8)² + (−16.8)² + 17.2² + 26.2² + (−16.8)² = 1643

Standard deviation

s = √(1643/4) = 20.27

Answer

s = 20.27
18.Descriptive statistics

Answer: P = 0.3085

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 0.3085

Answer

P = 0.3085
19.Normal distribution

Answer: P = 0.9332

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 total

μT = 1360 kg, σT = 12·√16 = 48 kg

Standardise

z = (1288 − 1360)/48 = −1.5

Probability

P = 1 − Φ(−1.5) = 0.9332

Answer

P = 0.9332
20.Normal distribution

Answer: x = 243.6 g

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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 250 − 1.28·5 = 243.6 g

Answer

x = 243.6 g

ExamProblems · examproblems.com/courses/engineering-mathematics