Flashcards · IB Diploma Programme Mathematics: Applications and Interpretation · Topic 5

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5.1 Introduction to the concept of a limit; derivative as gradient function and rate of change

Limit: The value a function approaches as x approaches a number. At this level you estimate it from a table or graph only.
Derivative: The gradient function of a curve, and the rate of change of one variable with respect to another. Notation: dy/dx, f′(x), dV/dr, ds/dt.
Gradient of a curve at a point: The limit of the gradients of chords through that point as the chords shrink.
Limit vs value: A function may have a limit at a point where it cannot be evaluated (for example, a chord gradient with h = 0).
V is in cm³ and r is in cm. What are the units of dV/dr?: cm³ per cm. A derivative is a rate of change, so its units are (units of V) per (unit of r).

5.2 Increasing and decreasing functions; graphical interpretation of f′(x)

Increasing and decreasing functions: Increasing: f′(x) > 0. Decreasing: f′(x) < 0.
For f(x) = x² − 8x + 3, state where f is increasing: f′(x) = 2x − 8 > 0, so x > 4.
Reading the graph of y = f′(x): what does it tell you about f?: Where f′ is above the x-axis, f is increasing; below, f is decreasing. Where f′ crosses zero, f has a stationary point.
Common error when stating where a function is increasing: Giving points instead of intervals, or the wrong inequality direction. Answer with an interval such as x > 4.

5.3 Derivative of f(x) = ax^n and sums of such terms

Derivative of f(x) = axⁿ: f′(x) = anxⁿ⁻¹, for integer n. The derivative of a constant is 0, and sums are differentiated term by term.
Differentiate y = 3x² + 6/x: Rewrite as 3x² + 6x⁻¹, then dy/dx = 6x − 6x⁻² = 6x − 6/x².
Find f′(x) when f(x) = 3/x: f(x) = 3x⁻¹, so f′(x) = −3x⁻² = −3/x². The power drops from −1 to −2.
Differentiate y = 5x⁴ − 3x + 2: dy/dx = 20x³ − 3.

5.4 Tangents and normals at a given point

Tangent and normal to a curve: Tangent: line touching the curve with the same gradient as the curve at that point. Normal: line through the same point, perpendicular to the tangent.
How do you find the tangent and normal at x = a?: 1. Find y = f(a). 2. Find m = f′(a). 3. Tangent: y − y₁ = m(x − x₁). 4. Normal: gradient −1/m through the same point.
f(a) vs f′(a): f(a) is a y-coordinate; f′(a) is a gradient.
Find the tangent and normal to y = x² − 3x + 2/x at x = 2: y = 4 − 6 + 1 = −1. dy/dx = 2x − 3 − 2x⁻² = 0.5. Tangent: y = 0.5x − 2. Normal: gradient −2, so y = −2x + 3.

5.5 Introduction to integration as anti-differentiation; definite integrals and area using technology

Anti-derivative: A function whose derivative is the given function. Always includes + C.
Integral of axⁿ: ∫ axⁿ dx = (a/(n + 1))xⁿ⁺¹ + C, for integer n ≠ −1. Example: ∫ 6x⁻² dx = −6x⁻¹ + C = −6/x + C.
Boundary condition: A known point (x, y) used to find C.
dy/dx = 2x + 1 and y = 7 when x = 2. Find y: y = x² + x + C. 4 + 2 + C = 7, so C = 1 and y = x² + x + 1.
Indefinite vs definite integral: Indefinite gives a function + C; definite gives a number (found with technology at SL).
Area under a curve at SL: A = ∫ₐᵇ f(x) dx when f(x) > 0 on a ≤ x ≤ b. Write the integral with its limits, then evaluate on the GDC. Example: ∫₁³ 3x² dx = 26.

5.6 Values where the gradient is zero; local maximum and minimum points

Stationary point: Where f′(x) = 0 (gradient zero).
Local maximum/minimum: A stationary point that is higher/lower than the points close to it. It need not be the greatest/least value on the whole domain.
How do you classify a stationary point at SL?: Test the sign of f′(x) either side (or use the GDC graph). + to − is a local maximum; − to + is a local minimum. The second derivative test is HL only.
f(x) = x³ − 12x. Find and classify the stationary points: f′(x) = 3x² − 12 = 0 gives x = ±2. f′(−3) = 15, f′(0) = −12, f′(3) = 15, so x = −2 is a local maximum and x = 2 a local minimum.
Local vs global maximum: A local maximum can be beaten by an end-point value. f(x) = x³ − 6x² + 9x + 1 has a local maximum at (1, 5), but on 0 ≤ x ≤ 5 the greatest value is f(5) = 21.

5.7 Optimisation problems in context

How do you solve an optimisation problem?: 1. Write the quantity to optimise. 2. Use the constraint to reduce it to one variable. 3. Differentiate, set equal to 0, solve. 4. Justify max or min. 5. Answer in context, with units.
P(x) = −2x² + 120x − 1000. Find the maximum profit: P′(x) = −4x + 120 = 0 gives x = 30. P′ changes from + to −, so it is a maximum: P(30) = 800.
Optimisation: two checks before you give the answer: Justify that the stationary point is a maximum or minimum (sign of the derivative or a graph), and compare with the end points of the domain if one is given.
Optimisation: when should you round?: Only at the end. Keep x unrounded (store it in the GDC) when substituting to find the optimal value.

5.8 Approximating areas using the trapezoidal rule

Trapezoidal rule: Area ≈ (h/2)[y₀ + yₙ + 2(y₁ + … + yₙ₋₁)], where h = (b − a)/n and n is the number of equal strips.
Strips vs points: 6 x-values means 5 strips in the trapezoidal rule.
Use the trapezoidal rule with h = 1 and y-values 2, 3, 5, 6: (1/2)[2 + 6 + 2(3 + 5)] = (1/2)(24) = 12.
Widths of a pond are 0, 6.2, 8.4, 9.0, 7.1, 0 m at 5 m spacing. Estimate its area: 6 values, so 5 strips and h = 5. Area ≈ 2.5[0 + 0 + 2(30.7)] = 153.5 m².

5.9 Derivatives of sin x, cos x, tan x, e^x, ln x, x^n (n rational); chain, product and quotient rules; related rates of change (AHL only)

Derivatives of sin x, cos x, tan x, eˣ and ln xHL: sin x → cos x cos x → −sin x tan x → 1/cos²x eˣ → eˣ ln x → 1/x (x in radians)
Chain ruleHL: dy/dx = (dy/du)(du/dx). Example: y = ln(3x² + 1) gives dy/dx = 6x/(3x² + 1). Do not forget the inner derivative.
Product ruleHL: d/dx(uv) = u(dv/dx) + v(du/dx). Example: y = x²e⁻ˣ gives 2xe⁻ˣ − x²e⁻ˣ = xe⁻ˣ(2 − x).
Quotient ruleHL: d/dx(u/v) = (v(du/dx) − u(dv/dx))/v². The order in the numerator matters: swapping it flips the sign.
How do you solve a related rates problem?HL: 1. Write the equation linking the quantities. 2. Differentiate with respect to t (chain rule). 3. Substitute the known rate and values at that instant (only after differentiating). 4. Solve, with units. Example: dA/dt = 2πr(dr/dt).
Differentiate y = 5x^(2/5)HL: dy/dx = 5 × (2/5)x^(−3/5) = 2x^(−3/5). The power rule holds for rational n at HL.

5.10 The second derivative; concavity and points of inflexion (AHL only)

Second derivative and concavityHL: f′′(x) or d²y/dx², the derivative of the derivative. Concave-up: f′′(x) > 0. Concave-down: f′′(x) < 0.
Point of inflexion, and how to show oneHL: A point where the concavity changes. Solve f′′(x) = 0 and show that f′′ changes sign there; f′′ = 0 alone is not proof.
Second derivative test for a stationary pointHL: At f′(x) = 0: f′′ < 0 → maximum; f′′ > 0 → minimum; f′′ = 0 → test fails, so check the sign of f′ either side.
Find the x-coordinates of the points of inflexion of f(x) = x⁴ − 4x³HL: f′′(x) = 12x² − 24x = 12x(x − 2). f′′ changes sign at both roots, so x = 0 and x = 2.
Can a stationary point also be a point of inflexion?HL: Yes. On y = x³, (0, 0) has f′(0) = 0 and the concavity changes there.

5.11 Definite and indefinite integration of further functions; integration by inspection or substitution (AHL only)

Standard integrals of 1/x, sin x, cos x, 1/cos²x and eˣHL: ∫ 1/x dx = ln|x| + C ∫ sin x dx = −cos x + C ∫ cos x dx = sin x + C ∫ 1/cos²x dx = tan x + C ∫ eˣ dx = eˣ + C
Integrate sin(5x + 2) by inspectionHL: −(1/5) cos(5x + 2) + C. Divide by the coefficient of x.
∫ 3x²/(x³ + 4) dxHL: ln|x³ + 4| + C, because the numerator is the derivative of the denominator.
How do you integrate by substitution?HL: 1. Spot g(x) whose derivative g′(x) appears (up to a constant). 2. Let u = g(x), du = g′(x) dx. 3. Rewrite everything in u, including the limits if definite. 4. Integrate, then substitute back or use the u-limits.
Find ∫ x e^(x²) dxHL: u = x², du = 2x dx, so the integral is (1/2)∫ eᵘ du = (1/2)e^(x²) + C.

5.12 Area enclosed by a curve and an axis; volumes of revolution about the x- or y-axis (AHL only)

Definite integral vs area when the curve goes below the x-axisHL: A definite integral is negative where the curve is below the axis; an area is always positive. Integrate each piece separately and add the absolute values, or use ∫|f(x)| dx on the GDC.
Find ∫₀² (x − 1)³ dx and the area between y = (x − 1)³ and the x-axis for 0 ≤ x ≤ 2HL: The integral is 0, because the two halves cancel. The area is 1/4 + 1/4 = 1/2.
Volume of revolution about the x-axisHL: V = ∫ₐᵇ πy² dx, with x-limits. Example: y = √x for 0 ≤ x ≤ 4 gives V = π∫₀⁴ x dx = 8π.
Volume of revolution about the y-axisHL: V = ∫ₐᵇ πx² dy, with y-limits. Example: y = x² (so x² = y) for 0 ≤ y ≤ 4 gives V = π∫₀⁴ y dy = 8π.
Area between a curve and the y-axisHL: Rearrange to x in terms of y, convert the limits to y-values, and integrate x with respect to y: A = ∫ x dy (taking |x| where the curve is left of the y-axis).

5.13 Kinematic problems involving displacement, velocity and acceleration (AHL only)

Displacement, velocity and acceleration: how are they linked?HL: v = ds/dt and a = dv/dt = d²s/dt². Displacement is position from a fixed origin and can be negative. Notation: ẋ = dx/dt, ẍ = d²x/dt².
Acceleration when v is given in terms of sHL: a = v(dv/ds). Example: v = 2s + 3 at s = 1 gives a = 5 × 2 = 10 m s⁻².
Displacement vs total distance travelledHL: Displacement = ∫ v dt from t₁ to t₂ (can be negative or zero). Distance = ∫ |v| dt (never negative): find where v = 0 and add the absolute values of each piece.
v = 4 − t² m s⁻¹ for 0 ≤ t ≤ 3. Find the displacement and the distance travelledHL: Displacement = ∫₀³ (4 − t²) dt = 3 m. v = 0 at t = 2, so distance = 16/3 + 7/3 = 23/3 ≈ 7.67 m.
Speeding up or slowing down?HL: Same signs of v and a: speeding up. Opposite signs: slowing down. Speed is |v|, never negative.

5.14 Setting up and solving first order differential equations by separation of variables (AHL only)

Differential equationHL: An equation linking a function and its derivatives, for example dy/dx = ky.
General vs particular solutionHL: The general solution keeps the constant; the particular solution does not (the constant is fixed by an initial condition).
How do you solve dy/dx = g(x)h(y) by separating variables?HL: 1. Rearrange to (1/h(y)) dy = g(x) dx. 2. Integrate both sides; add one constant. 3. Use the initial condition to find it. 4. Rearrange to y = … if asked.
Solve dy/dx = 2y/x for x > 0, given y = 3 when x = 1HL: ∫ (1/y) dy = ∫ (2/x) dx, so ln y = 2 ln x + c and y = Ax². y(1) = 3 gives y = 3x².
dy/dx = ky: general solution, and the common slipHL: y = Ae^(kx). From ln y = kx + c, y = e^c × e^(kx) = Ae^(kx), not e^(kx) + c.

5.15 Slope fields and their diagrams (AHL only)

Slope fieldHL: Short segments at grid points, each with gradient equal to dy/dx there.
How does a solution curve relate to a slope field?HL: It follows the segments and never crosses them at an angle. Different starting points give different curves; to sketch one, start at the given point and follow the segments both ways.
Find the gradient of the slope field of dy/dx = xy − 1 at (2, 1)HL: (2)(1) − 1 = 1.
Slope field of dy/dx = x − y: where are segments horizontal, and which line is a solution?HL: Horizontal on y = x (where x − y = 0). On y = x − 1 every segment has gradient 1, the line's own gradient, so y = x − 1 is itself a solution curve.

5.16 Euler’s method for numerical solutions of first order and coupled differential equations (AHL only)

Euler's method for dy/dx = f(x, y)HL: A step-by-step numerical approximation using the gradient at the start of each step: xₙ₊₁ = xₙ + h, yₙ₊₁ = yₙ + h f(xₙ, yₙ).
One Euler step with h = 0.1: dy/dx = x + y, y(0) = 1HL: y₁ = 1 + 0.1(0 + 1) = 1.1.
Coupled systemHL: Two equations dx/dt and dy/dt that each depend on x and y.
Euler's method for a coupled system: the key ruleHL: xₙ₊₁ = xₙ + h f₁(xₙ, yₙ, tₙ), yₙ₊₁ = yₙ + h f₂(xₙ, yₙ, tₙ), tₙ₊₁ = tₙ + h. Both updates use the old values; never update y with the new x.
Euler estimate vs exact valueHL: Euler is an approximation; a smaller h usually reduces the error.
Euler's method on a GDC or spreadsheet: what must you write?HL: The formula, h and starting values, and at least the first step by hand. Keep full accuracy in the table and round only the final answer to 3 s.f.

5.17 Phase portraits for coupled linear differential equations; qualitative analysis by eigenvalue type (AHL only)

Phase portraitHL: A sketch of trajectories in the (x, y) plane, with arrows showing the direction of motion as t increases.
Equilibrium pointHL: A point where dx/dt = 0 and dy/dt = 0. For dx/dt = ax + by, dy/dt = cx + dy it is the origin.
Real eigenvalues of M: what does the phase portrait look like?HL: Both positive: all trajectories move away from the origin. Both negative: all move towards it. Opposite signs: a saddle point at the origin.
Complex vs imaginary eigenvaluesHL: Complex with non-zero real part gives a spiral (outwards if the real part is positive, inwards if negative); purely imaginary gives a closed orbit (circles or ellipses).
How do you find the direction of rotation of a spiral or ellipse?HL: Use a test point such as (1, 0): if dy/dt > 0 there, the motion is anticlockwise; if dy/dt < 0, clockwise. The real part only decides in or out.
Exact solution of a coupled linear system with real distinct eigenvaluesHL: (x, y) = Ae^(λ₁t)p₁ + Be^(λ₂t)p₂, where p₁, p₂ are eigenvectors. Put t = 0 to find A and B.
Characteristic polynomial for M with rows (a, b) and (c, d)HL: λ² − (a + d)λ + (ad − bc) = 0. Example: rows (1, 2), (2, 1) gives λ² − 2λ − 3 = 0, λ = 3 or −1: a saddle.

5.18 Second order differential equations by Euler’s method and via phase portraits (AHL only)

How do you write d²x/dt² = f(x, dx/dt, t) as a coupled system?HL: Let y = dx/dt. Then dx/dt = y and dy/dt = f(x, y, t). Solve with the coupled Euler method.
y in a second order systemHL: y means dx/dt, the velocity, not a second unknown function.
Matrix for d²x/dt² + a dx/dt + bx = 0HL: M = [0 1; −b −a], with x across and y = dx/dt up. Above the x-axis dx/dt > 0, so trajectories move right.
d²x/dt² + 4x = 0: eigenvalues and phase portraitHL: M = [0 1; −4 0], λ = ±2i: ellipses (undamped oscillation). At (1, 0), dy/dt = −4, so the motion is clockwise.
Write d²x/dt² + 4 dx/dt − 5x = 0 as a system and find the eigenvaluesHL: dx/dt = y, dy/dt = 5x − 4y. M = [0 1; 5 −4], λ² + 4λ − 5 = 0, so λ = 1 or −5: a saddle.