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→∞ (1 − 2/x)2x. Give a decimal value.

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

Find the limit L = limx→0 (e1.5x − 1 − 1.5x)/x².

f(x) = (e1.5x − 1 − 1.5x)/x²x → 0
L =
3.DerivativesMedium
xyx = −1

Let f(x) = ln(3x2 + 1). Find f′(−1).

f(x) = ln(3x2 + 1)x = −1
f′(−1) =
4.DerivativesHard
xyx = −2

The tangent line to f(x) = 0.5x2 − 5x − 4 at x = −2 crosses the y-axis at y = b. Find b.

f(x) = 0.5x2 − 5x − 4x = −2
b =
5.IntegralsMedium
xyπ/3y = x·sin x

Evaluate I = ∫₀π/3 x·sin x dx (x in radians).

∫₀π/3 x·sin x dx
I =
6.IntegralsHard
xy2.5y = x²eˣ

Evaluate I = ∫₀2.5 x²·eˣ dx.

∫₀2.5 x²eˣ dx
I =
7.Complex numbersMedium
ReIm55−5−51010−10−10zABCD

The diagram shows z = 4 + j2. With w = 2∠90°, click the point that represents z·w.

z = 4 + j2w = 2∠90°
  1. 1A (−1 + j2)
  2. 2B (8 + j4)
  3. 3C (4 − j8)
  4. 4D (−4 + j8)
8.Complex numbersHard
ReIm11−1−122−2−2z₀z₁z₂

The roots of zⁿ = w with n = 3 and w = 8∠−60° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the argument of z₀ in degrees, between −180° and 180°.

w = 8∠−60°n = 3k = 0
θ₀ = °
9.Differential equationsMedium
t (min)T (°C)6Ta​T = ?

A can of soda at 6 °C is taken out of the fridge into a room at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.07 min⁻¹, find its temperature T after 6 min.

T0 = 6 °CTa = 23 °Ck = 0.07 min⁻¹t = 6 min
T = °C
10.Differential equationsHard
xy12(1, −2)y(2) = ?

Solve the initial value problem y′ + 2y/x = 6 for x > 0, with y(1) = −2. Find y(2).

y′ + 2y/x = 6y(1) = −2
y(2) =
11.Matrices & determinantsMedium
A =410−1−5−133−5

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

A = [4 1 0; −1 −5 −1; 3 3 −5]
det A =
12.Matrices & determinantsHard
A =−2174

Find the larger eigenvalue λ of the matrix A shown.

A = [−2 1; 7 4]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the dot product a·b of a = (1, −6, −1) and b = (−5, 0, −7).

a = (1, −6, −1)b = (−5, 0, −7)
a·b =
14.Vectors: dot & cross productHard
xyzabc

Find the volume V of the parallelepiped whose edges from one corner are a = (−1, 4, −2), b = (−5, −5, −3) and c = (0, −5, 2).

a = (−1, 4, −2)b = (−5, −5, −3)c = (0, −5, 2)
V =
15.ProbabilityMedium
xp(x)10.0520.330.340.250.15μ = 3.1

For the distribution shown, find the standard deviation σ of X.

x: 1, 2, 3, 4, 5p(x): 0.05, 0.3, 0.3, 0.2, 0.15
σ =
16.ProbabilityHard
n = 7, p = 0.5: P(X ≤ 1) = ?k01234567

7 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?

n = 7p = 0.5
P =
17.Descriptive statisticsMedium
202530354045

Find the sample standard deviation s of the data: 46, 48, 25, 20, 17.

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

A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 160.4?

μ = 170σ = 12x = 160.4
P =
19.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9793

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9793. Find z to two decimals.

A = 0.9793
z =
20.Normal distributionHard
μ = 65x = ?90%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 90% of them lie (the 90th percentile).

μ = 65σ = 10P = 90%
x = points

3Solutions

1.Limits

Answer: L = 0.01832

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 − 2/x)

Evaluate

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

ln L = 2·(−2) = −4

Undo the logarithm

L = e−4 = 0.01832

Answer

L = 0.01832
2.Limits

Answer: L = 1.125

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

(1.5·e1.5x − 1.5)/(2x) → 0/0

Second application

2.25·e1.5x/2

Substitute x = 0

L = 2.25·1/2 = 1.125

Answer

L = 1.125
3.Derivatives

Answer: f′(−1) = −1.5

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.

Chain rule for the logarithm

f′(x) = 6x/(3x2 + 1)

Substitute

f′(−1) = −6/4 = −1.5

Answer

f′(−1) = −1.5
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(−2) = 8
f′(x) = x − 5 → f′(−2) = −7

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 8 − (−7)·(−2) = −6

Answer

b = −6
5.Integrals

Answer: I = 0.3424

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 = x, du = dx; dv = sin x dx, v = −cos x

Antiderivative

∫x sin x dx = −x cos x + ∫cos x dx = −x cos x + sin x

Limits

(−π/3·cos π/3 + sin π/3) − (0 + 0) = 0.3424

Answer

I = 0.3424
6.Integrals

Answer: I = 37.59

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

12.18·(3.25) − 1·2 = 37.59

Answer

I = 37.59
7.Complex numbers

Answer: D (−4 + 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 = −4 + j8 → D

Answer

D (−4 + j8)
8.Complex numbers

Answer: θ₀ = −20 °

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

81/3 = 2

Argument of the root

θ₀ = (−60° + 360°·0)/3 = −20°

Argument

θ₀ = −20°

Answer

θ₀ = −20°

The 3 roots lie on the circle of radius 2, 120° apart.

9.Differential equations

Answer: T = 11.83 °C

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.

Solution of the cooling law

T(t) = 23 + (−17)·e−0.07t

Substitute t

T(6) = 23 + (−17)·0.657 = 11.83 °C

Answer

T = 11.83 °C
10.Differential equations

Answer: y(2) = 3

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 = e∫2/x dx = x2

Integrate

(x2y)′ = 6x2 → x2y = 6x3/3 + C

Initial condition

−2 = 6/3 + C → C = −4

Evaluate

y(2) = (6·23/3 + (−4))/22 = 3

Answer

y(2) = 3
11.Matrices & determinants

Answer: det A = 104

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 = 28, M₁₂ = 8, M₁₃ = 12

Cofactor expansion

det A = 4·28 − 1·8 + 0·12 = 104

Answer

det A = 104
12.Matrices & determinants

Answer: λ = 5

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 − (2)λ + (−15) = 0

Solve

λ = (2 ± √(64))/2 → λ = −3, 5
λ = 5

Answer

λ = 5
13.Vectors: dot & cross product

Answer: a·b = 2

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.

Multiply matching components and add

a·b = 1·(−5) + (−6)·0 + (−1)·(−7) = (−5) + 0 + 7 = 2

Answer

a·b = 2

Positive: the angle between the vectors is less than 90°.

14.Vectors: dot & cross product

Answer: V = 15

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 = (−5)·2 − (−3)·(−5) = −25
(b × c)y = bzcx − bxcz = (−3)·0 − (−5)·2 = 10
(b × c)z = bxcy − bycx = (−5)·(−5) − (−5)·0 = 25

Dot product with a

a·(b × c) = (−1)·(−25) + 4·10 + (−2)·25 = 15
V = |15| = 15

Answer

V = 15
15.Probability

Answer: σ = 1.136

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 = 1·0.05 + 2·0.3 + 3·0.3 + 4·0.2 + 5·0.15 = 3.1

Mean of the squares

E(X²) = 1·0.05 + 4·0.3 + 9·0.3 + 16·0.2 + 25·0.15 = 10.9

Variance

Var(X) = 10.9 − 3.1² = 1.29
σ = √1.29 = 1.136

Answer

σ = 1.136
16.Probability

Answer: P = 0.0625

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) = 0.57 = 0.007813
P(X = 1) = 7·0.5·0.56 = 0.05469

Add them

P = 0.007813 + 0.05469 = 0.0625

Answer

P = 0.0625
17.Descriptive statistics

Answer: s = 14.72

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 = 156/5 = 31.2

Squared deviations

Σ(x − x̄)² = 14.8² + 16.8² + (−6.2)² + (−11.2)² + (−14.2)² = 866.8

Standard deviation

s = √(866.8/4) = 14.72

Answer

s = 14.72
18.Descriptive statistics

Answer: P = 0.2119

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 0.2119

Answer

P = 0.2119
19.Normal distribution

Answer: z = −2.04

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 − 0.9793 = 0.0207

Read the z-table backwards

Find 0.0207 inside the table and read off z.

z = −2.04

Answer

z = −2.04
20.Normal distribution

Answer: x = 77.82 points

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 for the percentile

Φ(z) = 0.9 → z = 1.282

Back to x

x = 65 + 1.282·10 = 77.82 points

Answer

x = 77.82 points

ExamProblems · examproblems.com/courses/engineering-mathematics