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→∞ (−3x3 − 3x + 6)/(2x3 + 3x + 8).

f(x) = (−3x3 − 3x + 6)/(2x3 + 3x + 8)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 3x)/sin(5x), with x in radians.

f(x) = ln(1 + 3x)/sin(5x)x → 0
L =
3.DerivativesMedium
xyx = 1.5

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

f(x) = x·sin(x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
xyx = −2

The tangent line to f(x) = x2 + 3x − 1 at x = −2 crosses the y-axis at y = b. Find b.

f(x) = x2 + 3x − 1x = −2
b =
5.IntegralsMedium
xy02

Evaluate I = ∫02 e2x dx.

f(x) = e2x[0, 2]
I =
6.IntegralsHard
xy−11

Find the area A of the region between the parabola y = 1 − x² and the x-axis.

y = 1 − x²
A =
7.Complex numbersMedium
ReIm11−1−122−2−233−3−3zABCD

The diagram shows z = 1 − j. With w = 0.5∠−90°, click the point that represents z/w.

z = 1 − jw = 0.5∠−90°
  1. 1A (1 + j)
  2. 2B (2 + j2)
  3. 3C (−2 − j2)
  4. 4D (2 − j2)
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

The equation z² − 4z + 20 = 0 has two complex roots. Find their modulus |z|.

p = −4q = 20
|z| =
9.Differential equationsMedium
t (min)T (°C)12Ta​T = ?

A cup of coffee at 91 °C is left in a room at 21 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.1 min⁻¹, find its temperature T after 12 min.

T0 = 91 °CTa = 21 °Ck = 0.1 min⁻¹t = 12 min
T = °C
10.Differential equationsHard
xy13(1, −2)y(3) = ?

Solve the initial value problem y′ + y/x = 1 for x > 0, with y(1) = −2. Find y(3).

y′ + y/x = 1y(1) = −2
y(3) =
11.Matrices & determinantsMedium
A =33−51−3−4−1−5−4

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

A = [3 3 −5; 1 −3 −4; −1 −5 −4]
det A =
12.Matrices & determinantsHard
A =3−46−7

Find the larger eigenvalue λ of the matrix A shown.

A = [3 −4; 6 −7]
λ =
13.Vectors: dot & cross productMedium
xyzdFAB

A constant force F = (−15, −30, 20) N acts on an object while it moves in a straight line from point A(5, 0, 0) to point B(−5, 0, 3). Coordinates are in metres. Find the work W done by the force.

F = (−15, −30, 20) NA(5, 0, 0) mB(−5, 0, 3) m
W = J
14.Vectors: dot & cross productHard
xyzab

For a = (4, 4, −2) and b = (5, 3, 5), find the x-component (a × b)x of the cross product a × b.

a = (4, 4, −2)b = (5, 3, 5)
(a × b)x =
15.ProbabilityMedium
xp(x)00.310.2520.130.35μ = 1.5

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

x: 0, 1, 2, 3p(x): 0.3, 0.25, 0.1, 0.35
Var(X) =
16.ProbabilityHard
xp(x)100.35200.25300.3400.1μ

X has the distribution shown, and Y = 0.5X + 5. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.35, 0.25, 0.3, 0.1Y = 0.5X + 5
σY =
17.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 45, 28, 37, 46, 44, 42, 57.

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

A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is below 175?

μ = 170σ = 5x = 175
P =
19.Normal distributionMedium
750 g764.8 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 764.8 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 764.8 g
P =
20.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =

3Solutions

1.Limits

Answer: L = −1.5

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).

Divide by the highest power

(−3 + (−3)/x2 + 6/x3)/(2 + 3/x2 + 8/x3)

Let x → ∞

L = −3/2 = −1.5

Answer

L = −1.5

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

2.Limits

Answer: L = 0.6

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).

L'Hôpital's rule

L = limx→0 [3/(1 + 3x)]/[5·cos 5x]

Substitute x = 0

L = 3/5 = 0.6

Answer

L = 0.6
3.Derivatives

Answer: f′(1.5) = 1.104

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(x) + x·cos(x)

Substitute

f′(1.5) = 0.9975 + 1.5·0.07074 = 1.104

Answer

f′(1.5) = 1.104
4.Derivatives

Answer: b = −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.

Point and slope

f(−2) = −3
f′(x) = 2x + 3 → f′(−2) = −1

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −3 − (−1)·(−2) = −5

Answer

b = −5
5.Integrals

Answer: I = 26.8

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.

Antiderivative

F(x) = e2x/2

Evaluate at the limits

I = (e4 − 1)/2 = (54.6 − 1)/2 = 26.8

Answer

I = 26.8
6.Integrals

Answer: A = 1.333

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

1 − x² = 0 → x = ±1

Integrate

A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333

Answer

A = 1.333
7.Complex numbers

Answer: B (2 + j2)

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| = |z|/0.5, arg(z/w) = arg z − (−90)°

The point

z/w = 2 + j2 → B

Answer

B (2 + j2)
8.Complex numbers

Answer: |z| = 4.472

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 = 16 − 80 = −64 < 0

Quadratic formula

z = (4 ± j√64)/2 = 2 + j4, 2 − j4

Answer

|z| = √(2² + 4²) = √20 = 4.472

Answer

|z| = 4.472
9.Differential equations

Answer: T = 42.08 °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) = 21 + 70·e−0.1t

Substitute t

T(12) = 21 + 70·0.3012 = 42.08 °C

Answer

T = 42.08 °C
10.Differential equations

Answer: y(3) = 0.6667

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)′ = x → x1y = 1x2/2 + C

Initial condition

−2 = 1/2 + C → C = −2.5

Evaluate

y(3) = (1·32/2 + (−2.5))/31 = 0.6667

Answer

y(3) = 0.6667
11.Matrices & determinants

Answer: det A = 40

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 = −8, M₁₂ = −8, M₁₃ = −8

Cofactor expansion

det A = 3·(−8) − 3·(−8) + (−5)·(−8) = 40

Answer

det A = 40
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 − (−4)λ + (3) = 0

Solve

λ = (−4 ± √(4))/2 → λ = −3, −1
λ = −1

Answer

λ = −1
13.Vectors: dot & cross product

Answer: W = 210 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 = (−5, 0, 3) − (5, 0, 0) = (−10, 0, 3) m

Work

W = F·d = (−15)·(−10) + (−30)·0 + 20·3 = 210 J

Answer

W = 210 J
14.Vectors: dot & cross product

Answer: (a × b)x = 26

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 = 4·5 − (−2)·3 = 26
(a × b)y = azbx − axbz = (−2)·5 − 4·5 = −30
(a × b)z = axby − aybx = 4·3 − 4·5 = −8

Requested component

(a × b)x = 26

Answer

(a × b)x = 26

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

15.Probability

Answer: Var(X) = 1.55

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 = 0·0.3 + 1·0.25 + 2·0.1 + 3·0.35 = 1.5

Mean of the squares

E(X²) = 0·0.3 + 1·0.25 + 4·0.1 + 9·0.35 = 3.8

Variance

Var(X) = 3.8 − 1.5² = 1.55

Answer

Var(X) = 1.55
16.Probability

Answer: σY = 5.068

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 = 21.5, Var(X) = 102.8, σX = 10.14

Linear change

σY = |0.5|·10.14 = 5.068

Answer

σY = 5.068

The mean does move: E(Y) = 0.5·21.5 + 5 = 15.75.

17.Descriptive statistics

Answer: s = 8.864

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 = 299/7 = 42.71

Squared deviations

Σ(x − x̄)² = 2.286² + (−14.71)² + (−5.714)² + 3.286² + 1.286² + (−0.7143)² + 14.29² = 471.4

Standard deviation

s = √(471.4/6) = 8.864

Answer

s = 8.864
18.Descriptive statistics

Answer: P = 0.8413

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 = 1

Standard normal table

Φ(1) = 0.8413
P = 0.8413

Answer

P = 0.8413
19.Normal distribution

Answer: P = 0.06681

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 sum

μ = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (764.8 − 750)/9.849 = 1.5

Upper tail

P = 1 − Φ(1.5) = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
20.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389

ExamProblems · examproblems.com/courses/engineering-mathematics