Flashcards · IB Diploma Programme Mathematics: Analysis and Approaches · Topic 5

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94 cards on Topic 5, subtopics 5.1 to 5.19. Pick a subtopic or study them all.

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

Limit (informal, SL): The value f(x) approaches as x approaches a given value. At SL, estimate it from a table or a graph; formal methods are not required.
Derivative as a gradient function: f′(x) is the gradient function. f′(a) is the gradient of the tangent at x = a, found as the limit of the gradients of chords as they shrink towards that point.
Table: (x⁴ − 1)/(x − 1) is 3.940399 at x = 0.99 and 4.060401 at x = 1.01. Estimate the limit as x → 1.: The values approach 4 from both sides, so the limit is 4.
The derivative as a rate of change: what do dV/dr and ds/dt mean?: dV/dr is the rate of change of V with respect to r; ds/dt is the rate of change of s with respect to time t.

5.2 Increasing and decreasing functions

Increasing and decreasing functions in terms of f′(x): Increasing on an interval where f′(x) > 0; decreasing where f′(x) < 0.
How do you find the intervals where a function is increasing or decreasing?: 1. Find f′(x) and factorise it. 2. Solve f′(x) = 0. 3. Test the sign of f′ in each interval. 4. State the intervals with strict inequalities.
For what values of x is g(x) = 2x³ − 9x² + 12x decreasing?: g′(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2) < 0 for 1 < x < 2.
Reading a graph of y = f′(x): where is f increasing, and where are its stationary points?: f is increasing where the graph of f′ is above the x-axis and decreasing where it is below. The zeros of f′ are the stationary points of f.

5.3 Derivative of f(x) = ax^n

Derivative of f(x) = axⁿ: f′(x) = anxⁿ⁻¹. The derivative of a constant is 0, and a sum is differentiated term by term.
Differentiate y = 7/x³ − 2√x: Rewrite first: y = 7x⁻³ − 2x^(1/2). dy/dx = −21x⁻⁴ − x^(−1/2)
Differentiate y = 6x⁻² + x³/3: dy/dx = −12x⁻³ + x²
Exam trap: differentiating 3/x or 1/√x: Rewrite as 3x⁻¹ or x^(−1/2) before using the power rule: d/dx(3x⁻¹) = −3x⁻², d/dx(x^(−1/2)) = −½x^(−3/2).

5.4 Tangents and normals at a given point

How do you find the tangent and normal to y = f(x) at x = a?: 1. Point: y₁ = f(a). 2. Tangent gradient m = f′(a). 3. Tangent: y − y₁ = m(x − a). 4. Normal gradient −1/m; normal: y − y₁ = (−1/m)(x − a).
Find the tangent and normal to y = e²ˣ at x = 0: Point (0, 1); dy/dx = 2e²ˣ, so m = 2. Tangent y = 2x + 1. Normal y = −x/2 + 1.
Find the equation of the tangent to y = √x at x = 9: Point (9, 3). dy/dx = 1/(2√x) = 1/6 at x = 9. y − 3 = (1/6)(x − 9), so y = x/6 + 3/2.
Exam trap: the y-coordinate of the point of contact: Substitute into f(x), not f′(x). And the normal gradient is −1/m, not 1/m or m.

5.5 Introduction to integration as anti-differentiation

Anti-derivative and indefinite integral: F is an anti-derivative of f if F′(x) = f(x). ∫ f(x) dx = F(x) + C: a family of functions, so include + C.
Integral of xⁿ: ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C, for n ≠ −1.
dy/dx = 3x² + 2 and y = 1 when x = 1. Find y.: y = x³ + 2x + C. Substitute: 1 = 1 + 2 + C, so C = −2. y = x³ + 2x − 2 (write the full function).
How do you find the area between a curve and the x-axis with technology, where f(x) > 0?: Area = ∫ₐᵇ f(x) dx. Write the integral with its limits on the page, then evaluate it with the GDC and give 3 s.f.

5.6 Derivatives of x^n, sin x, cos x, e^x and ln x; chain, product and quotient rules

Derivatives of sin x, cos x, eˣ and ln x: sin x → cos x cos x → −sin x eˣ → eˣ ln x → 1/x (Trig results need x in radians.)
Chain rule: If y = f(u) and u = g(x), dy/dx = (dy/du)(du/dx). Example: d/dx(x² + 1)⁵ = 5(x² + 1)⁴ × 2x = 10x(x² + 1)⁴.
Product rule: For y = uv: dy/dx = u(dv/dx) + v(du/dx). Example: d/dx(x ln x) = ln x + 1.
Quotient rule: For y = u/v: dy/dx = (v(du/dx) − u(dv/dx))/v². Example: d/dx(x/(x + 2)) = ((x + 2) − x)/(x + 2)² = 2/(x + 2)².
Derivatives of e^(kx), sin(kx + b) and ln(g(x)): d/dx e^(kx) = ke^(kx) d/dx sin(kx + b) = k cos(kx + b) d/dx ln(g(x)) = g′(x)/g(x), so d/dx ln(3x − 2) = 3/(3x − 2)
Product or composite? x² sin x and sin(x²): x² sin x is a product: use the product rule. sin(x²) is a composite: use the chain rule, giving 2x cos(x²).

5.7 The second derivative and graphical behaviour of functions

Second derivative: d²y/dx² or f″(x): the derivative of f′(x), the rate of change of the gradient.
At a local maximum of f, what does the graph of f′ do? And at a local minimum?: Local maximum: f′ crosses the x-axis from + to −. Local minimum: f′ crosses from − to +.
Concave-up and concave-down: what do f′ and f″ do?: Concave-up: f′ increasing, f″(x) > 0. Concave-down: f′ decreasing, f″(x) < 0.
A point of inflexion on f: what does it look like on the graphs of f′ and f″?: f′ has a turning point there, and f″ crosses the x-axis.

5.8 Local maximum and minimum points, optimization and points of inflexion

How do you find and classify stationary points?: 1. Solve f′(x) = 0. 2. Find y from f(x), not f′(x). 3. f″(x) > 0: minimum; f″(x) < 0: maximum; f″(x) = 0: use the sign of f′ either side.
Point of inflexion: the full condition: f″(x) = 0 AND f″ changes sign. Its gradient can be zero (horizontal inflexion, like (0, 0) on y = x³) or non-zero.
Exam trap: y = x⁴ has f″(0) = 0. Is (0, 0) a point of inflexion?: No. f″ does not change sign, and (0, 0) is a minimum. f″(x) = 0 is necessary but not sufficient.
Find the point of inflexion of y = x³ − 6x² + 5: y″ = 6x − 12 = 0 at x = 2, and y″ changes from − to +. y = 8 − 24 + 5 = −11, so the point is (2, −11).
f′(x) = (x − 1)²(x + 2). Classify the stationary points of f.: x = −2: f′ goes from − to +, a minimum. x = 1: f′ is positive on both sides, so a point of inflexion with zero gradient.
How do you solve an optimisation problem?: 1. Write the quantity Q in terms of the variables. 2. Use the constraint to make Q a function of one variable. 3. Solve dQ/dx = 0. 4. Justify max or min; check end points if the domain is restricted. 5. Answer the question, with units.

5.9 Kinematic problems: displacement, velocity, acceleration and distance

Links between displacement s, velocity v and acceleration a: v = ds/dt and a = dv/dt = d²s/dt². Going back: integrate a to get v, and v to get s, adding + C each time and using known values such as v(0) or s(0).
Velocity vs speed: Velocity has a sign (direction). Speed = |v|, never negative.
Displacement vs total distance travelled between t₁ and t₂: Displacement = ∫ v(t) dt (can be negative). Distance = ∫ |v(t)| dt (never negative). They differ when the particle turns round.
v(t) = 4 − 2t for 0 ≤ t ≤ 3. Find the displacement and the total distance.: Displacement = [4t − t²]₀³ = 3 m. v = 0 at t = 2: ∫₀² v dt = 4, ∫₂³ v dt = −1, so distance = 4 + 1 = 5 m.

5.10 Indefinite integration by inspection or substitution

Standard integrals of 1/x, sin x, cos x and eˣ: ∫ 1/x dx = ln x + C ∫ sin x dx = −cos x + C ∫ cos x dx = sin x + C ∫ eˣ dx = eˣ + C
Integrating a linear composite f(ax + b): ∫ f(ax + b) dx = (1/a)F(ax + b) + C. Divide by a, never multiply. Example: ∫ 3 sin(2x) dx = −(3/2) cos(2x) + C.
Reverse chain rule (integration by inspection): ∫ k g′(x) f(g(x)) dx = k F(g(x)) + C. Find the inner function g, check g′(x) is a factor (up to a constant), guess, differentiate to adjust the constant, add + C.
Find ∫ x² e^(x³) dx: Inner function x³ with derivative 3x², so ∫ x² e^(x³) dx = (1/3)e^(x³) + C.
Find ∫ x(x² − 1)³ dx: Inner function x² − 1 with derivative 2x: (1/8)(x² − 1)⁴ + C.
Find ∫ 1/(5x + 2) dx: (1/5) ln(5x + 2) + C. Do not forget the 1/a factor.

5.11 Definite integrals and areas between curves

Definite integral: ∫ₐᵇ f(x) dx = F(b) − F(a). It is a number, so there is no + C.
Area between two curves: Area = ∫ₐᵇ (top − bottom) dx. Solve f(x) = g(x) for the limits, test which curve is on top, and split where the curves cross.
How do you find the area of a region that lies partly below the x-axis (no GDC)?: 1. Find the roots in the interval. 2. Split the integral at each root. 3. Evaluate each part. 4. Add the absolute values. Example: y = x² − 4x + 3 on [0, 3] has integral 0 but area 8/3.
Find the area enclosed by y = x² and y = 2x: They meet at x = 0 and x = 2, with 2x on top. ∫₀² (2x − x²) dx = 4 − 8/3 = 4/3
Evaluate ∫₀^(ln 2) e²ˣ dx: [½e²ˣ] from 0 to ln 2 = ½(e^(2 ln 2) − 1) = ½(4 − 1) = 3/2

5.12 Continuity, differentiability and derivative from first principles (AHL only)

Continuity and differentiability (informal)HL: Continuous: no break in the graph. Differentiable: no break and no sharp corner. The guide says you will not be asked to test for either.
Derivative from first principlesHL: f′(x) = lim (h→0) [f(x + h) − f(x)]/h In this course, for polynomials only.
Find f′(x) from first principles for f(x) = 2x² − 3xHL: f(x + h) − f(x) = 4xh + 2h² − 3h. Divide by h: 4x + 2h − 3. Let h → 0: f′(x) = 4x − 3. (Divide by h before letting h → 0.)
Convergence and divergenceHL: A sequence or expression converges if it tends to a finite limit; otherwise it diverges. Example: rⁿ → 0 when |r| < 1.
Higher derivatives: notation, and the pattern for y = xeˣHL: dⁿy/dxⁿ or f⁽ⁿ⁾(x). For y = xeˣ: y′ = (x + 1)eˣ, y″ = (x + 2)eˣ, suggesting f⁽ⁿ⁾(x) = (x + n)eˣ, which is proved by induction.

5.13 Evaluation of limits using l’Hôpital’s rule or Maclaurin series (AHL only)

l’Hôpital’s ruleHL: If lim f(x)/g(x) is of the form 0/0 or ∞/∞, then lim f(x)/g(x) = lim f′(x)/g′(x). Differentiate top and bottom separately (not the quotient rule).
How do you apply l’Hôpital’s rule correctly?HL: 1. Substitute to confirm 0/0 or ∞/∞, and say which. 2. Differentiate numerator and denominator separately. 3. Substitute again; if still indeterminate, repeat.
Find lim (x→0) (1 − cos 2x)/x²HL: 0/0. Differentiate: (2 sin 2x)/(2x), still 0/0. Again: (4 cos 2x)/2 → 4/2 = 2.
Find lim (x→∞) (2x² − x)/(5x² + 3)HL: ∞/∞. l’Hôpital twice: (4x − 1)/(10x), then 4/10 = 2/5.
Standard limit for sin θ/θ, and lim (x→0) (sin 3x)/xHL: lim (θ→0) (sin θ)/θ = 1. (sin 3x)/x = 3 × (sin 3x)/(3x) → 3.
Use a Maclaurin series to find lim (x→0) (eˣ − 1 − x)/x²HL: eˣ − 1 − x = x²/2 + x³/6 + …, so the expression = 1/2 + x/6 + … → 1/2.

5.14 Implicit differentiation, related rates of change and optimisation (AHL only)

How do you differentiate implicitly?HL: 1. Differentiate each term with respect to x; terms in y pick up a factor dy/dx. 2. Use the product rule on mixed terms such as xy or x²y. 3. Collect the dy/dx terms and factorise.
Find dy/dx for x³ + y³ = 9 at (1, 2)HL: 3x² + 3y² dy/dx = 0, so dy/dx = −x²/y² = −1/4.
Find dy/dx for x² + xy + y² = 7 at (1, 2)HL: 2x + y + x dy/dx + 2y dy/dx = 0, so dy/dx = −(2x + y)/(x + 2y) = −4/5.
How do you solve a related rates problem?HL: 1. Write the formula linking the quantities. 2. Differentiate with respect to t, or use the chain rule, e.g. dA/dt = (dA/dr)(dr/dt). 3. Substitute the values at the instant LAST (not before differentiating).
Optimisation on a closed interval: where can the optimum be?HL: At a stationary point inside the interval or at an end point. Keep only solutions inside the interval, evaluate at them and at both end points, and compare.

5.15 Derivatives and indefinite integrals of further functions (AHL only)

Derivatives of tan x, sec x, cosec x and cot xHL: tan x → sec²x sec x → sec x tan x cosec x → −cosec x cot x cot x → −cosec²x
Derivatives of aˣ and logₐ xHL: d/dx aˣ = aˣ ln a (so 3ˣ → 3ˣ ln 3) d/dx logₐ x = 1/(x ln a)
Derivatives of arcsin x, arccos x and arctan xHL: arcsin x → 1/√(1 − x²) arccos x → −1/√(1 − x²) arctan x → 1/(1 + x²)
Standard integrals giving arctan and arcsinHL: ∫ 1/(a² + x²) dx = (1/a) arctan(x/a) + C ∫ 1/√(a² − x²) dx = arcsin(x/a) + C
Find ∫ 1/(x² + 6x + 13) dxHL: Complete the square: x² + 6x + 13 = (x + 3)² + 4. ∫ = (1/2) arctan((x + 3)/2) + C
Find ∫ (x + 7)/((x − 2)(x + 1)) dxHL: Partial fractions: 3/(x − 2) − 2/(x + 1). ∫ = 3 ln|x − 2| − 2 ln|x + 1| + C (modulus signs needed).

5.16 Integration by substitution and by parts (AHL only)

Integration by partsHL: ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx Choose u to get simpler when differentiated: ln x, arcsin x, arctan x first, then polynomials.
Find ∫ ln x dxHL: Parts with u = ln x, dv/dx = 1: x ln x − ∫ x(1/x) dx = x ln x − x + C.
Find ∫ x e³ˣ dxHL: u = x, dv/dx = e³ˣ, v = (1/3)e³ˣ: (x/3)e³ˣ − ∫ (1/3)e³ˣ dx = (x/3)e³ˣ − (1/9)e³ˣ + C
How do you integrate eˣ sin x or eˣ cos x?HL: Use integration by parts twice. The original integral reappears, so collect it on one side and solve for it.
How do you integrate by substitution?HL: 1. Use the given u; find dx in terms of du. 2. Change the limits to u-values (or substitute back at the end). 3. Simplify fully before integrating. A substitution is given unless the integral has the form ∫ k g′(x) f(g(x)) dx.

5.17 Areas enclosed with the y-axis and volumes of revolution (AHL only)

Area between a curve and the y-axisHL: Area = ∫ₐᵇ |x| dy, with the curve written as x in terms of y and limits as y-values.
Volume of revolution about the x-axis and about the y-axisHL: About the x-axis: V = π ∫ₐᵇ y² dx (x-limits). About the y-axis: V = π ∫ₐᵇ x² dy (y-limits).
Find the area between y = x³ and the y-axis for 1 ≤ y ≤ 8HL: x = y^(1/3), so area = ∫₁⁸ y^(1/3) dy = [(3/4)y^(4/3)]₁⁸ = (3/4)(16 − 1) = 45/4.
The region under y = √x from x = 0 to x = 4 is rotated 2π about the x-axis. Find the volume.HL: V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π[x²/2]₀⁴ = 8π

5.18 First order differential equations (AHL only)

Euler’s method for dy/dx = f(x, y) with step hHL: xₙ₊₁ = xₙ + h yₙ₊₁ = yₙ + h f(xₙ, yₙ) Keep full calculator values; round only the final answer. A smaller h usually gives a better estimate.
One Euler step, h = 0.1, for dy/dx = xy with y(1) = 2HL: y(1.1) ≈ 2 + 0.1 × (1 × 2) = 2.2
How do you solve a separable differential equation?HL: 1. Rearrange to g(y) dy = f(x) dx. 2. Integrate both sides; add one constant. 3. Use the initial condition, then rearrange for y. Trap: ln|y| = f(x) + C gives y = Ae^(f(x)), not e^(f(x)) + C.
Homogeneous differential equation dy/dx = f(y/x): methodHL: Substitute y = vx, so dy/dx = v + x dv/dx. Separate the variables v and x, integrate, then replace v with y/x.
Integrating factor method for dy/dx + P(x)y = Q(x)HL: 1. Make the coefficient of dy/dx equal to 1. 2. I(x) = e^(∫P(x) dx), no constant needed. 3. d/dx(I y) = I Q; integrate, then divide by I.
Solve dy/dx + (2/x)y = 3xHL: I = e^(∫2/x dx) = x². d/dx(x²y) = 3x³, so x²y = (3/4)x⁴ + C. y = (3/4)x² + C/x²

5.19 The Maclaurin series (AHL only)

Maclaurin series of f(x)HL: f(x) = f(0) + x f′(0) + (x²/2!) f″(0) + (x³/3!) f‴(0) + …
Maclaurin series of eˣ, sin x and cos xHL: eˣ = 1 + x + x²/2! + x³/3! + … sin x = x − x³/3! + x⁵/5! − … cos x = 1 − x²/2! + x⁴/4! − …
Maclaurin series of ln(1 + x) and arctan xHL: ln(1 + x) = x − x²/2 + x³/3 − … arctan x = x − x³/3 + x⁵/5 − …
Maclaurin series of ln(1 + 2x) up to x³HL: Replace x with 2x: 2x − (2x)²/2 + (2x)³/3 = 2x − 2x² + (8/3)x³
Find the series of eˣ sin x up to x³HL: (1 + x + x²/2 + …)(x − x³/6 + …) = x + x² + x³/2 − x³/6 + … = x + x² + x³/3 + … (Do not miss cross terms such as (x²/2) × x.)
How do you find a Maclaurin series from a differential equation?HL: 1. Find y′(0) from the equation. 2. Differentiate the equation (implicitly) to get y″, y‴. 3. Substitute x = 0 each time. 4. Build the series from y(0), y′(0), y″(0), …