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→∞ (√(4x2 − 7x) − 2x).

f(x) = √(4x2 − 7x) − 2xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 2x)/(4x²), with x in radians.

f(x) = (1 − cos 2x)/(4x²)x → 0
L =
3.DerivativesMedium
xyx = 1.5

Let f(x) = x3·e2x. Find f′(1.5).

f(x) = x3·e2xx = 1.5
f′(1.5) =
4.DerivativesHard
xyx = 1

The tangent line to f(x) = 0.5x2 + 5x at x = 1 crosses the y-axis at y = b. Find b.

f(x) = 0.5x2 + 5xx = 1
b =
5.IntegralsMedium
xy02

Evaluate I = ∫02 2x·(x² + 1)2 dx.

f(x) = 2x·(x² + 1)2[0, 2]
I =
6.IntegralsHard
xy1.333y = 4xy = 3x²

The region between the line y = 4x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.

y = 4xy = 3x²
V =
7.Complex numbersMedium
ReIm55−5−51010−10−10zABCD

The diagram shows z = −4 − j4. With w = 2∠−90°, click the point that represents z·w.

z = −4 − j4w = 2∠−90°
  1. 1A (−8 + j8)
  2. 2B (−8 − j8)
  3. 3C (−2 + j2)
  4. 4D (−4 + j4)
8.Complex numbersHard
ReIm11−1−122−2−2θz

Find the real part of (1 + j2)³.

z = 1 + j2n = 3
Re(z³) =
9.Differential equationsMedium
t (min)T (°C)15Ta​16 °C23 °C

A can of soda at 5 °C is taken out of the fridge into a room at 26 °C. After 15 min its temperature is 16 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 23 °C?

T0 = 5 °CTa = 26 °CT(15) = 16 °CT = 23 °C
t = min
10.Differential equationsHard
t (min)m (kg)45 minm = ?

A tank holds 100 L of pure water. Brine with 0.1 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 45 min?

V = 100 Lm0 = 0 kgc = 0.1 kg/Lr = 2 L/min
m = kg
11.Matrices & determinantsMedium
A =−6−426B =342−2

Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).

A = [−6 −4; 2 6]B = [3 4; 2 −2]
c21 =
12.Matrices & determinantsHard
A =−221−3

Find the larger eigenvalue λ of the matrix A shown.

A = [−2 2; 1 −3]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (−1, −7, −5) onto b = (3, −6, 1), that is, the component of a in the direction of b.

a = (−1, −7, −5)b = (3, −6, 1)
compb a =
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (−5, −3, −1), b = (2, −4, 1) and c = (5, −1, 5).

a = (−5, −3, −1)b = (2, −4, 1)c = (5, −1, 5)
V =
15.ProbabilityMedium
xp(x)00.320.240.2560.25μ = 2.9

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

x: 0, 2, 4, 6p(x): 0.3, 0.2, 0.25, 0.25
Var(X) =
16.ProbabilityHard
xp(x)00.310.3520.35μ

X has the distribution shown, and Y = 3X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 3X + 1
σY =
17.Descriptive statisticsMedium
202530354045505560

Find the sample standard deviation s of the data: 25, 36, 60, 19, 52.

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

Exam scores are normally distributed with mean 500 and standard deviation 100. Find the z-score z of a score of 380.

μ = 500σ = 100x = 380
z =
19.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.7291

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7291. Find z to two decimals.

A = 0.7291
z =
20.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =

3Solutions

1.Limits

Answer: L = −1.75

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).

Multiply by the conjugate

f(x) = −7x/(√(4x2 − 7x) + 2x)

Divide by x

f(x) = −7/(√(4 + (−7)/x) + 2)

Let x → ∞

L = −7/(2 + 2) = −1.75

Answer

L = −1.75
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

2·sin 2x/(8x) → 0/0

Second application

4·cos 2x/8

Substitute x = 0

L = 4/8 = 0.5

Answer

L = 0.5
3.Derivatives

Answer: f′(1.5) = 271.2

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.

Product rule

f′(x) = u′v + uv′ = e2x·(3x2 + 2·x3)

Substitute

f′(1.5) = 20.09·(6.75 + 6.75) = 271.2

Answer

f′(1.5) = 271.2
4.Derivatives

Answer: b = −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.

Point and slope

f(1) = 5.5
f′(x) = x + 5 → f′(1) = 6

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 5.5 − 6·1 = −0.5

Answer

b = −0.5
5.Integrals

Answer: I = 41.33

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 u2 du

Integrate and evaluate

I = [u3/3]15 = (125 − 1)/3 = 41.33

Answer

I = 41.33
6.Integrals

Answer: V = 15.89

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.

Intersection points

4x = 3x² → x = 0, x = 1.333

Washer method

V = π∫₀1.333 ((4x)² − (3x²)²) dx

Integrate

V = π(16·1.333³/3 − 9·1.333⁵/5) = π·5.057 = 15.89

Answer

V = 15.89
7.Complex numbers

Answer: A (−8 + j8)

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 + j8 → A

Answer

A (−8 + j8)
8.Complex numbers

Answer: Re(z³) = −11

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∠63.43° → z³ = 11.18∠−169.7°
Re(z³) = −11

Answer

Re(z³) = −11
9.Differential equations

Answer: t = 39.34 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(-21/-10)/15 = 0.04946 min⁻¹

Time to reach the target

t = ln(-21/(−3))/0.04946 = 39.34 min

Answer

t = 39.34 min
10.Differential equations

Answer: m = 5.934 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 = 2·0.1 − (2/100)·m = 0.2 − 0.02·m

Solve

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

m(t) = 10 + (−10)·e−t/50

Substitute t

m(45) = 10 + (−10)·0.4066 = 5.934 kg

Answer

m = 5.934 kg
11.Matrices & determinants

Answer: c21 = 18

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

c21 = 2·3 + 6·2 = 18

Answer

c21 = 18
12.Matrices & determinants

Answer: λ = −1

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 − (−5)λ + (4) = 0

Solve

λ = (−5 ± √(9))/2 → λ = −4, −1
λ = −1

Answer

λ = −1
13.Vectors: dot & cross product

Answer: compb a = 5.013

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)·3 + (−7)·(−6) + (−5)·1 = 34

Length of b

|b| = √(3² + (−6)² + 1²) = 6.782

Scalar projection

compb a = a·b/|b| = 34/6.782 = 5.013

Answer

compb a = 5.013
14.Vectors: dot & cross product

Answer: V = 92

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 = (−4)·5 − 1·(−1) = −19
(b × c)y = bzcx − bxcz = 1·5 − 2·5 = −5
(b × c)z = bxcy − bycx = 2·(−1) − (−4)·5 = 18

Dot product with a

a·(b × c) = (−5)·(−19) + (−3)·(−5) + (−1)·18 = 92
V = |92| = 92

Answer

V = 92
15.Probability

Answer: Var(X) = 5.39

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 = 0·0.3 + 2·0.2 + 4·0.25 + 6·0.25 = 2.9

Mean of the squares

E(X²) = 0·0.3 + 4·0.2 + 16·0.25 + 36·0.25 = 13.8

Variance

Var(X) = 13.8 − 2.9² = 5.39

Answer

Var(X) = 5.39
16.Probability

Answer: σY = 2.414

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 = 1.05, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |3|·0.8047 = 2.414

Answer

σY = 2.414

The mean does move: E(Y) = 3·1.05 + 1 = 4.15.

17.Descriptive statistics

Answer: s = 17.42

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 = 192/5 = 38.4

Squared deviations

Σ(x − x̄)² = (−13.4)² + (−2.4)² + 21.6² + (−19.4)² + 13.6² = 1213

Standard deviation

s = √(1213/4) = 17.42

Answer

s = 17.42
18.Descriptive statistics

Answer: z = −1.2

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 = (380 − 500)/100 = −1.2

Answer

z = −1.2
19.Normal distribution

Answer: z = −0.61

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 − 0.7291 = 0.2709

Read the z-table backwards

Find 0.2709 inside the table and read off z.

z = −0.61

Answer

z = −0.61
20.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881

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