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→∞ (2x3 − 9)/(2x3 + 3x + 1).

f(x) = (2x3 − 9)/(2x3 + 3x + 1)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 4x)/(3x²), with x in radians.

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

Let f(x) = x2·e0.5x. Find f′(2).

f(x) = x2·e0.5xx = 2
f′(2) =
4.DerivativesHard
xy

Let f(x) = x3 + 7.5x2 + 12x − 5. Find the x-coordinate x of the local minimum.

f(x) = x3 + 7.5x2 + 12x − 5
x =
5.IntegralsMedium
xy12y = x ln x

Evaluate I = ∫₁2 x·ln x dx.

∫₁2 x ln x dx
I =
6.IntegralsHard
xy0.5y = x²eˣ

Evaluate I = ∫₀0.5 x²·eˣ dx.

∫₀0.5 x²eˣ dx
I =
7.Complex numbersMedium
ReIm22−2−244−4−466−6−6θz

Write z = 8∠50° in rectangular form a + jb and give the imaginary part b.

r = 8θ = 50°
b =
8.Complex numbersHard
ReIm11−1−122−2−2z₁z₂

Solve z² + 3z + 8.5 = 0. The roots are a ± jb; find b > 0.

p = 3q = 8.5
b =
9.Differential equationsMedium
t (min)T (°C)45Ta​94 °C

A forged steel part at 300 °C is left to cool in a workshop at 22 °C. After 45 min its temperature is 94 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 300 °CTa = 22 °CT(45) = 94 °C
k = min⁻¹
10.Differential equationsHard
t (min)m (kg)t = ?53 kg

A tank holds 400 L of water in which 5 kg of salt is dissolved. 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. After how many minutes t does the tank contain 53 kg of salt?

V = 400 Lm0 = 5 kgc = 0.25 kg/Lr = 20 L/min
t = min
11.Matrices & determinantsMedium
A =550−5−12404

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

A = [5 5 0; −5 −1 2; 4 0 4]
det A =
12.Matrices & determinantsHard
A =−13−4−5b =24

Solve the system −x + 3y = 2 and −4x − 5y = 4. Find x.

−x + 3y = 2−4x − 5y = 4
x =
13.Vectors: dot & cross productMedium
xyzab

Find the scalar projection compb a of a = (−2, 0, −6) onto b = (−2, 3, −3), that is, the component of a in the direction of b.

a = (−2, 0, −6)b = (−2, 3, −3)
compb a =
14.Vectors: dot & cross productHard
xyzab

For a = (0, 6, 5) and b = (−1, 5, −5), find the y-component (a × b)y of the cross product a × b.

a = (0, 6, 5)b = (−1, 5, −5)
(a × b)y =
15.ProbabilityMedium
xp(x)00.1520.340.3560.2μ = 3.2

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

x: 0, 2, 4, 6p(x): 0.15, 0.3, 0.35, 0.2
Var(X) =
16.ProbabilityHard
xp(x)10.220.2530.240.350.05μ

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

x: 1, 2, 3, 4, 5p(x): 0.2, 0.25, 0.2, 0.3, 0.05Y = 3X − 3
σY =
17.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 35, 41, 56, 15, 35, 21.

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

A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is above 173.2?

μ = 170σ = 8x = 173.2
P =
19.Normal distributionMedium
250.25259.25P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 250.25 and 259.25 g? Give P as a decimal.

μ = 250σ = 5x1 = 250.25x2 = 259.25
P =
20.Normal distributionHard
35455565758595pointstop 30%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 30% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points30%
x = points

3Solutions

1.Limits

Answer: L = 1

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

(2 + (−9)/x3)/(2 + 3/x2 + 1/x3)

Let x → ∞

L = 2/2 = 1

Answer

L = 1

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

2.Limits

Answer: L = 2.667

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

4·sin 4x/(6x) → 0/0

Second application

16·cos 4x/6

Substitute x = 0

L = 16/6 = 2.667

Answer

L = 2.667
3.Derivatives

Answer: f′(2) = 16.31

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′ = e0.5x·(2x + 0.5·x2)

Substitute

f′(2) = 2.718·(4 + 2) = 16.31

Answer

f′(2) = 16.31
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 + 15x + 12 = 3·(x2 + 5x + 4)

Solve f′(x) = 0

x = −4 or x = −1

Which one is which

f″(x) = 6x + 15: negative at a maximum, positive at a minimum.

f″(−4) = −9, f″(−1) = 9
x = −1

Answer

x = −1
5.Integrals

Answer: I = 0.6363

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

(2·ln 2 − 1) − (0 − 0.25) = 0.6363

Answer

I = 0.6363
6.Integrals

Answer: I = 0.0609

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

1.649·(1.25) − 1·2 = 0.0609

Answer

I = 0.0609
7.Complex numbers

Answer: b = 6.128

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.

Components

a = 8·cos(50°) = 5.142
b = 8·sin(50°) = 6.128

Rectangular form

z = 5.142 + j6.128
b = 6.128

Answer

b = 6.128
8.Complex numbers

Answer: b = 2.5

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 = 9 − 34 = −25 < 0

Quadratic formula

z = (−3 ± j√25)/2 = −1.5 + j2.5, −1.5 − j2.5

Answer

b = 2.5

Answer

b = 2.5
9.Differential equations

Answer: k = 0.03002 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−45k = 72/278

Solve for k

k = ln(278/72)/45 = 0.03002 min⁻¹

Answer

k = 0.03002 min⁻¹
10.Differential equations

Answer: t = 14.07 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) = 100 + (−95)·e−t/20

Solve for t

e−t/20 = (100 − 53)/(100 − 5) = 0.4947
t = 20·ln(2.021) = 14.07 min

Answer

t = 14.07 min
11.Matrices & determinants

Answer: det A = 120

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 = −4, M₁₂ = −28, M₁₃ = 4

Cofactor expansion

det A = 5·(−4) − 5·(−28) + 0·4 = 120

Answer

det A = 120
12.Matrices & determinants

Answer: x = −1.294

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 = (−1)·(−5) − 3·(−4) = 17

Replace the x column

det Aₓ = 2·(−5) − 3·4 = −22
x = −22/17 = −1.294

Answer

x = −1.294
y = 0.2353
13.Vectors: dot & cross product

Answer: compb a = 4.69

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) + 0·3 + (−6)·(−3) = 22

Length of b

|b| = √((−2)² + 3² + (−3)²) = 4.69

Scalar projection

compb a = a·b/|b| = 22/4.69 = 4.69

Answer

compb a = 4.69
14.Vectors: dot & cross product

Answer: (a × b)y = −5

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 = 6·(−5) − 5·5 = −55
(a × b)y = azbx − axbz = 5·(−1) − 0·(−5) = −5
(a × b)z = axby − aybx = 0·5 − 6·(−1) = 6

Requested component

(a × b)y = −5

Answer

(a × b)y = −5

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

15.Probability

Answer: Var(X) = 3.76

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 = 0·0.15 + 2·0.3 + 4·0.35 + 6·0.2 = 3.2

Mean of the squares

E(X²) = 0·0.15 + 4·0.3 + 16·0.35 + 36·0.2 = 14

Variance

Var(X) = 14 − 3.2² = 3.76

Answer

Var(X) = 3.76
16.Probability

Answer: σY = 3.659

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 = 2.75, Var(X) = 1.488, σX = 1.22

Linear change

σY = |3|·1.22 = 3.659

Answer

σY = 3.659

The mean does move: E(Y) = 3·2.75 − 3 = 5.25.

17.Descriptive statistics

Answer: s = 14.59

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 = 203/6 = 33.83

Squared deviations

Σ(x − x̄)² = 1.167² + 7.167² + 22.17² + (−18.83)² + 1.167² + (−12.83)² = 1065

Standard deviation

s = √(1065/5) = 14.59

Answer

s = 14.59
18.Descriptive statistics

Answer: P = 0.3446

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446

Answer

P = 0.3446
19.Normal distribution

Answer: P = 0.4479

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z-scores

z₁ = (250.25 − 250)/5 = 0.05
z₂ = (259.25 − 250)/5 = 1.85

Area between

P = Φ(1.85) − Φ(0.05) = 0.9678 − 0.5199 = 0.4479

Answer

P = 0.4479
20.Normal distribution

Answer: x = 70.2 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) = 0.7

z from the table

z ≈ 0.52

Back to the original units

x = 65 + 0.52·10 = 70.2 points

Answer

x = 70.2 points

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