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→∞ (1 + 1/x)2x. Give a decimal value.

f(x) = (1 + 1/x)2xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 5x)/(3x²), with x in radians.

f(x) = (1 − cos 5x)/(3x²)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = x3·e−x. Find f′(1).

f(x) = x3·e−xx = 1
f′(1) =
4.DerivativesHard
xyx = −1

The tangent line to f(x) = 2x2 − 4 at x = −1 crosses the y-axis at y = b. Find b.

f(x) = 2x2 − 4x = −1
b =
5.IntegralsMedium
xy4y = x

The region under y = x from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = x0 ≤ x ≤ 4
V =
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
ReIm22−2−244−4−4z₁z₂w

Given z₁ = 5∠−60° and z₂ = 2∠−25°, find the real part of w = z₁/z₂.

z₁ = 5∠−60°z₂ = 2∠−25°
Re(w) =
8.Complex numbersHard
ReIm22−2−244−4−4z₀z₁z₂

The roots of zⁿ = w with n = 3 and w = 64∠45° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the argument of z₂ in degrees, between −180° and 180°.

w = 64∠45°n = 3k = 2
θ₂ = °
9.Differential equationsMedium
t (min)T (°C)27Ta​221 °C291 °C

A forged steel part at 360 °C is left to cool in a workshop at 27 °C. After 27 min its temperature is 221 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 291 °C?

T0 = 360 °CTa = 27 °CT(27) = 221 °CT = 291 °C
t = min
10.Differential equationsHard
t (min)m (kg)55 minm = ?

A tank holds 200 L of water in which 20 kg of salt is dissolved. Brine with 0.05 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 55 min?

V = 200 Lm0 = 20 kgc = 0.05 kg/Lr = 5 L/min
m = kg
11.Matrices & determinantsMedium
A =3−4−16B =−22−24

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

A = [3 −4; −1 6]B = [−2 2; −2 4]
c21 =
12.Matrices & determinantsHard
A =−5−352b =11−2

Solve the system −5x − 3y = 11 and 5x + 2y = −2. Find x.

−5x − 3y = 115x + 2y = −2
x =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (1, 4, 4) and b = (5, 3, 2), in degrees.

a = (1, 4, 4)b = (5, 3, 2)
θ = °
14.Vectors: dot & cross productHard
xyzrFOA

A force F = (−60, −50, −30) N acts at point A(3, 3, 3). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(−1, 0, 3).

F = (−60, −50, −30) NA(3, 3, 3) mO(−1, 0, 3) m
Mz = N·m
15.ProbabilityMedium
xp(x)00.110.3520.3530.2μ = 1.65

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

x: 0, 1, 2, 3p(x): 0.1, 0.35, 0.35, 0.2
Var(X) =
16.ProbabilityHard
xp(x)100.2200.15300.3400.35μ

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

x: 10, 20, 30, 40p(x): 0.2, 0.15, 0.3, 0.35Y = 3X + 1
σY =
17.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 12, 39, 54, 52, 15.

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

Exam scores are normally distributed with mean 70 and standard deviation 15. Find the z-score z of a score of 58.

μ = 70σ = 15x = 58
z =
19.Normal distributionMedium
125 kg127.2 kgT = X₁ + … + X₅?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 5 of them exceeds 127.2 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 5> 127.2 kg
P =
20.Normal distributionHard
557085100115130145pointsbottom 90%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points90%
x = points

3Solutions

1.Limits

Answer: L = 7.389

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).

Take the logarithm

ln L = limx→∞ 2·x·ln(1 + 1/x)

Evaluate

ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.

ln L = 2·1 = 2

Undo the logarithm

L = e2 = 7.389

Answer

L = 7.389
2.Limits

Answer: L = 4.167

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

5·sin 5x/(6x) → 0/0

Second application

25·cos 5x/6

Substitute x = 0

L = 25/6 = 4.167

Answer

L = 4.167
3.Derivatives

Answer: f′(1) = 0.7358

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′ = e−x·(3x2 + (−1)·x3)

Substitute

f′(1) = 0.3679·(3 + (−1)) = 0.7358

Answer

f′(1) = 0.7358
4.Derivatives

Answer: b = −6

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) = 4x → f′(−1) = −4

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −2 − (−4)·(−1) = −6

Answer

b = −6
5.Integrals

Answer: V = 67.02

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.

Disc method

V = π∫₀4 1x² dx

Integrate

V = π·1·4³/3 = 67.02

Answer

V = 67.02

Check: a cone of radius 4 and height 4 has (1/3)πr²h = 67.02.

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: Re(w) = 2.048

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.

Modulus and argument of w

|w| = 5 / 2 = 2.5
θ = −60° − (−25)° = −35°

Rectangular form

Re(w) = 2.5·cos(−35°) = 2.048
Im(w) = 2.5·sin(−35°) = −1.434
Re(w) = 2.048

Answer

Re(w) = 2.048
8.Complex numbers

Answer: θ₂ = −105 °

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 of the roots

641/3 = 4

Argument of the root

θ₂ = (45° + 360°·2)/3 = 255° → −105°

Argument

θ₂ = −105°

Answer

θ₂ = −105°

The 3 roots lie on the circle of radius 4, 120° apart.

9.Differential equations

Answer: t = 11.6 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(333/194)/27 = 0.02001 min⁻¹

Time to reach the target

t = ln(333/264)/0.02001 = 11.6 min

Answer

t = 11.6 min
10.Differential equations

Answer: m = 12.53 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 = 5·0.05 − (5/200)·m = 0.25 − 0.025·m

Solve

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

m(t) = 10 + 10·e−t/40

Substitute t

m(55) = 10 + 10·0.2528 = 12.53 kg

Answer

m = 12.53 kg
11.Matrices & determinants

Answer: c21 = −10

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 = (−1)·(−2) + 6·(−2) = −10

Answer

c21 = −10
12.Matrices & determinants

Answer: x = 3.2

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 = (−5)·2 − (−3)·5 = 5

Replace the x column

det Aₓ = 11·2 − (−3)·(−2) = 16
x = 16/5 = 3.2

Answer

x = 3.2
y = −9
13.Vectors: dot & cross product

Answer: θ = 45.09 °

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 = 1·5 + 4·3 + 4·2 = 25

Magnitudes

|a| = √(1² + 4² + 4²) = 5.745
|b| = √(5² + 3² + 2²) = 6.164

Angle

cos θ = 25/(5.745·6.164) = 0.706
θ = arccos(0.706) = 45.09°

Answer

θ = 45.09°
14.Vectors: dot & cross product

Answer: Mz = −20 N·m

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.

Position vector from O to A

r = A − O = (3, 3, 3) − (−1, 0, 3) = (4, 3, 0) m

Moment M = r × F

(r × F)x = ryFz − rzFy = 3·(−30) − 0·(−50) = −90 N·m
(r × F)y = rzFx − rxFz = 0·(−60) − 4·(−30) = 120 N·m
(r × F)z = rxFy − ryFx = 4·(−50) − 3·(−60) = −20 N·m

Requested component

Mz = −20 N·m

Answer

Mz = −20 N·m
15.Probability

Answer: Var(X) = 0.8275

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 = 0·0.1 + 1·0.35 + 2·0.35 + 3·0.2 = 1.65

Mean of the squares

E(X²) = 0·0.1 + 1·0.35 + 4·0.35 + 9·0.2 = 3.55

Variance

Var(X) = 3.55 − 1.65² = 0.8275

Answer

Var(X) = 0.8275
16.Probability

Answer: σY = 33.67

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 = 28, Var(X) = 126, σX = 11.22

Linear change

σY = |3|·11.22 = 33.67

Answer

σY = 33.67

The mean does move: E(Y) = 3·28 + 1 = 85.

17.Descriptive statistics

Answer: s = 19.96

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 = 172/5 = 34.4

Squared deviations

Σ(x − x̄)² = (−22.4)² + 4.6² + 19.6² + 17.6² + (−19.4)² = 1593

Standard deviation

s = √(1593/4) = 19.96

Answer

s = 19.96
18.Descriptive statistics

Answer: z = −0.8

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 = (58 − 70)/15 = −0.8

Answer

z = −0.8
19.Normal distribution

Answer: P = 0.02442

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 = 125 kg, σT = 0.5·√5 = 1.118 kg

Standardise

z = (127.2 − 125)/1.118 = 1.97

Probability

P = 1 − Φ(1.97) = 0.02442

Answer

P = 0.02442
20.Normal distribution

Answer: x = 119.2 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) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 100 + 1.28·15 = 119.2 points

Answer

x = 119.2 points

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