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→∞ (√(9x2 − 5x) − 3x).

f(x) = √(9x2 − 5x) − 3xx → ∞
L =
2.LimitsHard
xyx = 0

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

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

Let f(x) = x·sin(3x), with x in radians. Find f′(1.5).

f(x) = x·sin(3x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
xyx = 1

The tangent line to f(x) = −x2 − 5x + 6 at x = 1 crosses the y-axis at y = b. Find b.

f(x) = −x2 − 5x + 6x = 1
b =
5.IntegralsMedium
xyπy = x·sin x

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

∫₀π x·sin x dx
I =
6.IntegralsHard
xy3y = 3xy = x²

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

y = 3xy = x²
V =
7.Complex numbersMedium
ReIm55−5−5z₁z₂w

Given z₁ = 5∠55° and z₂ = 1.5∠20°, find the argument θ of w = z₁·z₂ in degrees, between −180° and 180°.

z₁ = 5∠55°z₂ = 1.5∠20°
θ = °
8.Complex numbersHard
ReIm11−1−122−2−233−3−3ABCD

Exactly one of the points A–D is a root of z³ = 8∠0°. Click it.

w = 8∠0°n = 3
  1. 1A (3.6∠−120°)
  2. 2B (2∠−120°)
  3. 3C (2∠−90°)
  4. 4D (2∠−60°)
9.Differential equationsMedium
t (min)T (°C)58Ta​T = ?

A forged steel part at 340 °C is left to cool in a workshop at 20 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.01 min⁻¹, find its temperature T after 58 min.

T0 = 340 °CTa = 20 °Ck = 0.01 min⁻¹t = 58 min
T = °C
10.Differential equationsHard
xy12(1, 1)y(2) = ?

Solve the initial value problem y′ + y/x = 5x2 for x > 0, with y(1) = 1. Find y(2).

y′ + y/x = 5x2y(1) = 1
y(2) =
11.Matrices & determinantsMedium
A =−250−4−5−13−50

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

A = [−2 5 0; −4 −5 −1; 3 −5 0]
det A =
12.Matrices & determinantsHard
A =−321−2

Find the larger eigenvalue λ of the matrix A shown.

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

Find the scalar projection compb a of a = (5, −7, 2) onto b = (−2, 5, −4), that is, the component of a in the direction of b.

a = (5, −7, 2)b = (−2, 5, −4)
compb a =
14.Vectors: dot & cross productHard
xyzab

For a = (5, −3, 2) and b = (−1, −3, 5), find the x-component (a × b)x of the cross product a × b.

a = (5, −3, 2)b = (−1, −3, 5)
(a × b)x =
15.ProbabilityMedium
xp(x)00.110.1520.330.2540.2μ = 2.3

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

x: 0, 1, 2, 3, 4p(x): 0.1, 0.15, 0.3, 0.25, 0.2
σ =
16.ProbabilityHard
n = 10, p = 0.4: P(X ≤ 1) = ?k012345678910

10 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?

n = 10p = 0.4
P =
17.Descriptive statisticsMedium
152025303540

Find the sample standard deviation s of the data: 28, 12, 43, 40, 35.

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

A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 66?

μ = 70σ = 8x = 66
P =
19.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8690

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8690. Find z to two decimals.

A = 0.8690
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 = −0.8333

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) = −5x/(√(9x2 − 5x) + 3x)

Divide by x

f(x) = −5/(√(9 + (−5)/x) + 3)

Let x → ∞

L = −5/(3 + 3) = −0.8333

Answer

L = −0.8333
2.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).

First application

2·sin 2x/(2x) → 0/0

Second application

4·cos 2x/2

Substitute x = 0

L = 4/2 = 2

Answer

L = 2
3.Derivatives

Answer: f′(1.5) = −1.926

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 and chain rule

f′(x) = sin(3x) + 3x·cos(3x)

Substitute

f′(1.5) = −0.9775 + 4.5·(−0.2108) = −1.926

Answer

f′(1.5) = −1.926
4.Derivatives

Answer: b = 7

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) = 0
f′(x) = −2x − 5 → f′(1) = −7

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 0 − (−7)·1 = 7

Answer

b = 7
5.Integrals

Answer: I = 3.142

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·cos π + sin π) − (0 + 0) = 3.142

Answer

I = 3.142
6.Integrals

Answer: V = 101.8

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

3x = 1x² → x = 0, x = 3

Washer method

V = π∫₀3 ((3x)² − (1x²)²) dx

Integrate

V = π(9·3³/3 − 1·3⁵/5) = π·32.4 = 101.8

Answer

V = 101.8
7.Complex numbers

Answer: θ = 75 °

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 · 1.5 = 7.5
θ = 55° + 20° = 75°

Argument

θ = 75°

Answer

θ = 75°
8.Complex numbers

Answer: B (2∠−120°)

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 and arguments of the roots

|z| = 2
θ = 0°, 120°, −120°

The point

B = 2∠−120°

Answer

B (2∠−120°)
9.Differential equations

Answer: T = 199.2 °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) = 20 + 320·e−0.01t

Substitute t

T(58) = 20 + 320·0.5599 = 199.2 °C

Answer

T = 199.2 °C
10.Differential equations

Answer: y(2) = 9.875

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∫1/x dx = x1

Integrate

(x1y)′ = 5x3 → x1y = 5x4/4 + C

Initial condition

1 = 5/4 + C → C = −0.25

Evaluate

y(2) = (5·24/4 + (−0.25))/21 = 9.875

Answer

y(2) = 9.875
11.Matrices & determinants

Answer: det A = −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.

2×2 minors of the first row

M₁₁ = −5, M₁₂ = 3, M₁₃ = 35

Cofactor expansion

det A = (−2)·(−5) − 5·3 + 0·35 = −5

Answer

det A = −5
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 = −7.901

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 = 5·(−2) + (−7)·5 + 2·(−4) = −53

Length of b

|b| = √((−2)² + 5² + (−4)²) = 6.708

Scalar projection

compb a = a·b/|b| = −53/6.708 = −7.901

Answer

compb a = −7.901

Negative: a has a component pointing against b.

14.Vectors: dot & cross product

Answer: (a × b)x = −9

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 = (−3)·5 − 2·(−3) = −9
(a × b)y = azbx − axbz = 2·(−1) − 5·5 = −27
(a × b)z = axby − aybx = 5·(−3) − (−3)·(−1) = −18

Requested component

(a × b)x = −9

Answer

(a × b)x = −9

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

15.Probability

Answer: σ = 1.229

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 + 1·0.15 + 2·0.3 + 3·0.25 + 4·0.2 = 2.3

Mean of the squares

E(X²) = 0·0.1 + 1·0.15 + 4·0.3 + 9·0.25 + 16·0.2 = 6.8

Variance

Var(X) = 6.8 − 2.3² = 1.51
σ = √1.51 = 1.229

Answer

σ = 1.229
16.Probability

Answer: P = 0.04636

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) = 0.610 = 0.006047
P(X = 1) = 10·0.4·0.69 = 0.04031

Add them

P = 0.006047 + 0.04031 = 0.04636

Answer

P = 0.04636
17.Descriptive statistics

Answer: s = 12.34

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 = 158/5 = 31.6

Squared deviations

Σ(x − x̄)² = (−3.6)² + (−19.6)² + 11.4² + 8.4² + 3.4² = 609.2

Standard deviation

s = √(609.2/4) = 12.34

Answer

s = 12.34
18.Descriptive statistics

Answer: P = 0.3085

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 0.3085

Answer

P = 0.3085
19.Normal distribution

Answer: z = 1.51

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 + 0.8690)/2 = 0.9345

Read the z-table backwards

Find 0.9345 inside the table and read off z.

z = 1.51

Answer

z = 1.51
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

ExamProblems · examproblems.com/courses/engineering-mathematics