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)x. Give a decimal value.

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

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

f(x) = (1 − cos 3x)/(x²)x → 0
L =
3.DerivativesMedium
xyx = 0.5

Let f(x) = x·ex. Find f′(0.5).

f(x) = x·exx = 0.5
f′(0.5) =
4.DerivativesHard
xy

Let f(x) = x3 + 4.5x2 − 5. Find the x-coordinate x of the local maximum.

f(x) = x3 + 4.5x2 − 5
x =
5.IntegralsMedium
xy1y = x·sin x

Evaluate I = ∫₀1 x·sin x dx (x in radians).

∫₀1 x·sin x dx
I =
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
ReIm11−1−122−2−2θz

Write z = 2.5∠−20° in rectangular form a + jb and give the imaginary part b.

r = 2.5θ = −20°
b =
8.Complex numbersHard
ReIm11−1−122−2−233−3−3ABCD

Exactly one of the points A–D is a root of z³ = 27∠120°. Click it.

w = 27∠120°n = 3
  1. 1A (1.5∠40°)
  2. 2B (3∠40°)
  3. 3C (3∠−140°)
  4. 4D (3∠10°)
9.Differential equationsMedium
t (min)T (°C)18Ta​328 °C206 °C

A forged steel part at 460 °C is left to cool in a workshop at 23 °C. After 18 min its temperature is 328 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 206 °C?

T0 = 460 °CTa = 23 °CT(18) = 328 °CT = 206 °C
t = min
10.Differential equationsHard
tyK = 10y(1.5) = ?

A system obeys y″ + 2y′ + 26y = 260 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.

y″ + 2y′ + 26y = 260y(0) = 0y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
A =−541−1B =5−1−6−1

Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).

A = [−5 4; 1 −1]B = [5 −1; −6 −1]
c11 =
12.Matrices & determinantsHard
A =466−2b =48

Solve the system 4x + 6y = 4 and 6x − 2y = 8. Find x.

4x + 6y = 46x − 2y = 8
x =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (30, 10, 25) N acts on an object while it moves in a straight line from point A(4, 1, 3) to point B(−1, −4, −4). Coordinates are in metres. Find the work W done by the force.

F = (30, 10, 25) NA(4, 1, 3) mB(−1, −4, −4) m
W = J
14.Vectors: dot & cross productHard
xyzab

The unit vector n̂ is perpendicular to both a = (−4, 4, −2) and b = (−2, −1, 1) and has a positive z-component. Find its x-component n̂x.

a = (−4, 4, −2)b = (−2, −1, 1)n̂z > 0
n̂x =
15.ProbabilityMedium
P(at least one six) = ??1?2?3?4P(six) = 1/6 on each roll

A fair die is rolled 4 times. What is the probability P of getting at least one six?

n = 4
P =
16.ProbabilityHard
n = 11, p = 0.7: P(X ≤ 1) = ?k01234567891011

11 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?

n = 11p = 0.7
P =
17.Descriptive statisticsMedium
1015202530354045505560

Find the sample standard deviation s of the data: 37, 60, 10, 54, 24, 22, 37.

n = 7
s =
18.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 12. What proportion P of values is below 84?

μ = 60σ = 12x = 84
P =
19.Normal distributionMedium
680 kg629.1 kgT = X₁ + … + X₈?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 8 of them exceeds 629.1 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 8> 629.1 kg
P =
20.Normal distributionHard
235240245250255260265gbottom 90%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g90%
x = g

3Solutions

1.Limits

Answer: L = 0.1353

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→∞ x·ln(1 − 2/x)

Evaluate

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

ln L = 1·(−2) = −2

Undo the logarithm

L = e−2 = 0.1353

Answer

L = 0.1353
2.Limits

Answer: L = 4.5

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

3·sin 3x/(2x) → 0/0

Second application

9·cos 3x/2

Substitute x = 0

L = 9/2 = 4.5

Answer

L = 4.5
3.Derivatives

Answer: f′(0.5) = 2.473

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′ = ex·(1 + 1·x)

Substitute

f′(0.5) = 1.649·(1 + 0.5) = 2.473

Answer

f′(0.5) = 2.473
4.Derivatives

Answer: x = −3

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 + 9x = 3·(x2 + 3x)

Solve f′(x) = 0

x = −3 or x = 0

Which one is which

f″(x) = 6x + 9: negative at a maximum, positive at a minimum.

f″(−3) = −9, f″(0) = 9
x = −3

Answer

x = −3
5.Integrals

Answer: I = 0.3012

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

(−1·cos 1 + sin 1) − (0 + 0) = 0.3012

Answer

I = 0.3012
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: b = −0.8551

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 = 2.5·cos(−20°) = 2.349
b = 2.5·sin(−20°) = −0.8551

Rectangular form

z = 2.349 − j0.8551
b = −0.8551

Answer

b = −0.8551
8.Complex numbers

Answer: B (3∠40°)

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| = 3
θ = 40°, 160°, −80°

The point

B = 3∠40°

Answer

B (3∠40°)
9.Differential equations

Answer: t = 43.57 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(437/305)/18 = 0.01998 min⁻¹

Time to reach the target

t = ln(437/183)/0.01998 = 43.57 min

Answer

t = 43.57 min
10.Differential equations

Answer: y(1.5) = 8.808

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.

Steady value and roots

yss = 10, r = −1 ± 5j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10
y′(0) = −1A + 5B = 0 → B = −2

Evaluate

y(1.5) = 10·(1 − e−1.5(cos 7.5 + 0.2·sin 7.5)) = 8.808

Answer

y(1.5) = 8.808
11.Matrices & determinants

Answer: c11 = −49

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

c11 = (−5)·5 + 4·(−6) = −49

Answer

c11 = −49
12.Matrices & determinants

Answer: x = 1.273

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 = 4·(−2) − 6·6 = −44

Replace the x column

det Aₓ = 4·(−2) − 6·8 = −56
x = −56/(−44) = 1.273

Answer

x = 1.273
y = −0.1818
13.Vectors: dot & cross product

Answer: W = −375 J

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.

Displacement

d = B − A = (−1, −4, −4) − (4, 1, 3) = (−5, −5, −7) m

Work

W = F·d = 30·(−5) + 10·(−5) + 25·(−7) = −375 J

Answer

W = −375 J

Negative work: the force acts mostly against the motion.

14.Vectors: dot & cross product

Answer: n̂x = 0.1374

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

(a × b)x = aybz − azby = 4·1 − (−2)·(−1) = 2
(a × b)y = azbx − axbz = (−2)·(−2) − (−4)·1 = 8
(a × b)z = axby − aybx = (−4)·(−1) − 4·(−2) = 12

Normalise

The z-component of a × b is already positive.

|a × b| = √(2² + 8² + 12²) = 14.56
n̂ = (2, 8, 12)/14.56 = (0.1374, 0.5494, 0.8242)
n̂x = 0.1374

Answer

n̂x = 0.1374
15.Probability

Answer: P = 0.5177

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.

Complement

P(no six) = (5/6)4 = 0.4823
P = 1 − 0.4823 = 0.5177

Answer

P = 0.5177
16.Probability

Answer: P = 4.724 × 10⁻⁵

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) = 0.311 = 1.771 × 10⁻⁶
P(X = 1) = 11·0.7·0.310 = 4.547 × 10⁻⁵

Add them

P = 1.771 × 10⁻⁶ + 4.547 × 10⁻⁵ = 4.724 × 10⁻⁵

Answer

P = 4.724 × 10⁻⁵
17.Descriptive statistics

Answer: s = 17.84

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 = 244/7 = 34.86

Squared deviations

Σ(x − x̄)² = 2.143² + 25.14² + (−24.86)² + 19.14² + (−10.86)² + (−12.86)² + 2.143² = 1909

Standard deviation

s = √(1909/6) = 17.84

Answer

s = 17.84
18.Descriptive statistics

Answer: P = 0.9772

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 0.9772

Answer

P = 0.9772
19.Normal distribution

Answer: P = 0.9332

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 = 680 kg, σT = 12·√8 = 33.94 kg

Standardise

z = (629.1 − 680)/33.94 = −1.5

Probability

P = 1 − Φ(−1.5) = 0.9332

Answer

P = 0.9332
20.Normal distribution

Answer: x = 256.4 g

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 = 250 + 1.28·5 = 256.4 g

Answer

x = 256.4 g

ExamProblems · examproblems.com/courses/engineering-mathematics