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→∞ (6x2 + 4x − 6)/(3x2 + 6x + 7).

f(x) = (6x2 + 4x − 6)/(3x2 + 6x + 7)x → ∞
L =
2.LimitsHard
xyx = 0

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

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

Let f(x) = (−2x + 5)5. Find f′(−1).

f(x) = (−2x + 5)5x = −1
f′(−1) =
4.DerivativesHard
xy

Let f(x) = x3 − 1.5x2 − 6x − 3. Find the x-coordinate x of the local maximum.

f(x) = x3 − 1.5x2 − 6x − 3
x =
5.IntegralsMedium
xy02

Evaluate I = ∫02 2x·(x² + 1)3 dx.

f(x) = 2x·(x² + 1)3[0, 2]
I =
6.IntegralsHard
xyy = x2​y = x

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

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

The number z = 5·e−j0.5 has its angle in radians. Find its real part a.

r = 5θ = −0.5 rad
a =
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

One root of z³ − 9z² + 43z − 75 = 0 is z₁ = 3 + j4. The coefficients are real. Find the real root r.

z₁ = 3 + j4
r =
9.Differential equationsMedium
t (min)T (°C)25Ta​37 °C31 °C

A cup of coffee at 84 °C is left in a room at 23 °C. After 25 min its temperature is 37 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 31 °C?

T0 = 84 °CTa = 23 °CT(25) = 37 °CT = 31 °C
t = min
10.Differential equationsHard
ty2y(2) = ?

Solve y′ + 2y = 7e−t with y(0) = 5. Find y(2).

y′ + 2y = 7e−ty(0) = 5
y(2) =
11.Matrices & determinantsMedium
A =−210−25−3414

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

A = [−2 1 0; −2 5 −3; 4 1 4]
det A =
12.Matrices & determinantsHard
A =66−3−3

Find the larger eigenvalue λ of the matrix A shown.

A = [6 6; −3 −3]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (2, 3, 6) and b = (−2, 5, −5), in degrees.

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

For a = (−5, −1, −2) and b = (−1, 1, 4), find the z-component (a × b)z of the cross product a × b.

a = (−5, −1, −2)b = (−1, 1, 4)
(a × b)z =
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)−30.78120.21470.02

A game costs €3 to play. You win €15 with probability 0.2 and €150 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€15: p = 0.2€150: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
1015202530354045

Find the sample standard deviation s of the data: 38, 48, 34, 10, 12, 35.

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

Exam scores are normally distributed with mean 100 and standard deviation 15. Find the z-score z of a score of 92.5.

μ = 100σ = 15x = 92.5
z =
19.Normal distributionMedium
240 g241.5 gT = X₁ + … + X₆?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 6 of them exceeds 241.5 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 6> 241.5 g
P =
20.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =

3Solutions

1.Limits

Answer: L = 2

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

(6 + 4/x + (−6)/x2)/(3 + 6/x + 7/x2)

Let x → ∞

L = 6/3 = 2

Answer

L = 2

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

2.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).

L'Hôpital's rule

L = limx→0 [3/(1 + 3x)]/[4·cos 4x]

Substitute x = 0

L = 3/4 = 0.75

Answer

L = 0.75
3.Derivatives

Answer: f′(−1) = −24010

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) = 5·(−2x + 5)4·(−2)

Substitute

f′(−1) = 5·74·(−2) = −24010

Answer

f′(−1) = −24010
4.Derivatives

Answer: x = −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.

Derivative

f′(x) = 3x2 − 3x − 6 = 3·(x2 − x − 2)

Solve f′(x) = 0

x = −1 or x = 2

Which one is which

f″(x) = 6x − 3: negative at a maximum, positive at a minimum.

f″(−1) = −9, f″(2) = 9
x = −1

Answer

x = −1
5.Integrals

Answer: I = 156

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.

Substitution

u = x² + 1, du = 2x dx
I = ∫15 u3 du

Integrate and evaluate

I = [u4/4]15 = (625 − 1)/4 = 156

Answer

I = 156
6.Integrals

Answer: A = 0.1667

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 = x → x = 0, x = 1

Integrate upper minus lower

A = [0.5x2 − 0.3333x3]01 = 0.1667

Answer

A = 0.1667
7.Complex numbers

Answer: a = 4.388

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 = 5·(cos −0.5 + j sin −0.5)

Components (radians)

a = 5·cos −0.5 = 4.388
b = 5·sin −0.5 = −2.397
a = 4.388

Answer

a = 4.388
8.Complex numbers

Answer: r = 3

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.

Conjugate pair

z₂ = 3 − j4
(z − z₁)(z − z₂) = z² − 2·3·z + (3² + 4²) = z² − 6z + 25

Factor out the quadratic

z³ − 9z² + 43z − 75 = (z² − 6z + 25)(z − 3)

Real root

r = 3

Answer

r = 3

Check with the sum of the roots: z₁ + z₂ + r = 6 + 3 = 9 = −(−9).

9.Differential equations

Answer: t = 34.51 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(61/14)/25 = 0.05887 min⁻¹

Time to reach the target

t = ln(61/8)/0.05887 = 34.51 min

Answer

t = 34.51 min
10.Differential equations

Answer: y(2) = 0.9107

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 = e2t
(e2ty)′ = 7et

General solution

y = 7/1·e−t + C·e−2t = 7·e−t + C·e−2t

Initial condition

5 = 7 + C → C = −2

Evaluate

y(2) = 7·0.1353 + (−2)·0.01832 = 0.9107

Answer

y(2) = 0.9107
11.Matrices & determinants

Answer: det A = −50

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 = 23, M₁₂ = 4, M₁₃ = −22

Cofactor expansion

det A = (−2)·23 − 1·4 + 0·(−22) = −50

Answer

det A = −50
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 − (3)λ + (0) = 0

Solve

λ = (3 ± √(9))/2 → λ = 0, 3
λ = 3

Answer

λ = 3
13.Vectors: dot & cross product

Answer: θ = 111.7 °

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

a·b = 2·(−2) + 3·5 + 6·(−5) = −19

Magnitudes

|a| = √(2² + 3² + 6²) = 7
|b| = √((−2)² + 5² + (−5)²) = 7.348

Angle

cos θ = −19/(7·7.348) = −0.3694
θ = arccos(−0.3694) = 111.7°

Answer

θ = 111.7°
14.Vectors: dot & cross product

Answer: (a × b)z = −6

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, component by component

(a × b)x = aybz − azby = (−1)·4 − (−2)·1 = −2
(a × b)y = azbx − axbz = (−2)·(−1) − (−5)·4 = 22
(a × b)z = axby − aybx = (−5)·1 − (−1)·(−1) = −6

Requested component

(a × b)z = −6

Answer

(a × b)z = −6

Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.

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) = 3 €

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: 12 (p = 0.2), 147 (p = 0.02), −3 (p = 0.78)

Expected value

E(G) = 12·0.2 + 147·0.02 − 3·0.78 = 3 €

Answer

E(G) = 3 €

A positive value means you win money on average.

17.Descriptive statistics

Answer: s = 15.18

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 = 177/6 = 29.5

Squared deviations

Σ(x − x̄)² = 8.5² + 18.5² + 4.5² + (−19.5)² + (−17.5)² + 5.5² = 1152

Standard deviation

s = √(1152/5) = 15.18

Answer

s = 15.18
18.Descriptive statistics

Answer: z = −0.5

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 = (92.5 − 100)/15 = −0.5

Answer

z = −0.5
19.Normal distribution

Answer: P = 0.305

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 = 240 g, σT = 1.2·√6 = 2.939 g

Standardise

z = (241.5 − 240)/2.939 = 0.51

Probability

P = 1 − Φ(0.51) = 0.305

Answer

P = 0.305
20.Normal distribution

Answer: P = 0.8413

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 difference

μ = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413

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