Understand · explore · practise
28 original mixed questions on differentiation, numerical methods, integration, differential equations and 3D vectors, with independent hints and worked solutions.
Before you startPure 2 differentiation, numerical methods, integration and vectors.
01 / Check what the method establishes
Attempt a question on paper, then choose a hint or worked solution when useful. There is no timer. Keep exact expressions until rounding is requested, state domains, and distinguish a candidate from a verified answer.
Choose → calculate → check → interpret
The interval model shows why a change of sign establishes existence but not necessarily uniqueness.
a = −2, f(a) = −6; b = −0.5, f(b) = 0.375.
The opposite signs and continuity guarantee at least one root inside this interval.
Opposite signs do not prove uniqueness. Equal signs alone do not prove that an interval contains no roots. An endpoint value of zero is already a root, rather than a strict sign change.
02 / Combine differentiation rules
Write the rule before simplifying, then check signs on either side of a stationary point.
For y = x²e⁻²ˣ, find and classify both stationary points.
Factor y′ before solving.
y′ = 2xe⁻²ˣ(1 − x). Hence x = 0 or 1. Its sign is negative, positive, negative across these points: (0,0) is a minimum and (1,e⁻²) a maximum.
For x > 0, find the maximum value of (ln x)/x.
Use the quotient rule; e is the only stationary input.
y′ = (1 − ln x)/x². It is positive for 0 < x < e and negative for x > e, so the maximum is 1/e at x = e.
03 / Inverse and parametric derivatives
State where the derivative formula is valid. Divide by dx/dt only when it is nonzero.
f(x) = x³ + x. Find the derivative of f⁻¹ at input 10.
First solve f(x) = 10.
f(2) = 10 and f′(2) = 13. Since f′ = 3x² + 1 > 0 everywhere, the inverse exists and (f⁻¹)′(10) = 1/13.
x = t² + 1, y = t³ − t. Find the tangent at t = 2 and state where the quotient formula fails.
dy/dx = (3t² − 1)/(2t).
At t = 2, (x,y) = (5,6) and the slope is 11/4. Tangent: y − 6 = (11/4)(x − 5). The quotient is undefined at t = 0 because dx/dt = 0; that point requires separate analysis.
04 / Implicit slopes and concavity
A zero numerator gives a horizontal tangent only when the derivative is defined.
Find the normal to x² + xy + y² = 7 at (1,2).
Differentiate xy with the product rule.
2x + y + (x + 2y)y′ = 0, so y′ = −(2x + y)/(x + 2y). At (1,2) the tangent slope is −4/5; the normal is y − 2 = (5/4)(x − 1).
Find where y = x²e⁻ˣ has y″ < 0.
Factor out the positive exponential.
y′ = e⁻ˣ(2x − x²), and y″ = e⁻ˣ(x² − 4x + 2). The quadratic is negative between its roots, so 2 − √2 < x < 2 + √2.
05 / Related rates and reliable rounding
A rate needs units. A rounded root needs a bracket inside the appropriate rounding interval.
A sphere has radius 3 cm when its volume increases at 18π cm³ s⁻¹. Find dr/dt.
V = 4πr³/3, so dV/dt = 4πr² dr/dt.
18π = 36π dr/dt, giving dr/dt = 1/2 cm s⁻¹.
Show that the positive root of x³ − 2 = 0 is 1.2599 to 4 decimal places.
Use the rounding boundaries 1.25985 and 1.25995, not just consecutive 4-decimal numbers.
f(1.25985) ≈ −0.000338334953 and f(1.25995) ≈ 0.000137869450. Continuity gives a root between them; f′ = 3x² > 0 there gives uniqueness. The whole open bracket rounds to 1.2599.
06 / Iteration and Newton updates
Keep unrounded working values, and distinguish solving a rearrangement from proving convergence.
For f(x) = x³ − 2, start at x₀ = 1 and find x₁ and x₂ exactly.
xₙ₊₁ = xₙ − (xₙ³ − 2)/(3xₙ²).
x₁ = 4/3. Then x₂ = 4/3 − (10/27)/(16/3) = 91/72. These are approximations, not exact roots.
Consider xₙ₊₁ = √(3 + xₙ), x₀ = 1. Find the next two values and the possible nonnegative limit.
Square the limit relation but retain the nonnegative square-root condition.
x₁ = 2, x₂ = √5. A limit L satisfies L² − L − 3 = 0 and L ≥ 0, so L = (1 + √13)/2. Near L, g′(L) = 1/(2L) < 1, supporting local convergence; the negative quadratic root is not a fixed point of this square-root iteration.
07 / Check failure and retain offsets
Numerical algorithms have assumptions; periodic models may have a nonzero mean level.
Why can Newton’s method not start at x₀ = 1 for f(x) = x³ − 3x + 1?
Evaluate f and f′ separately.
f(1) = −1, so it is not a root; f′(1) = 0. The Newton correction divides by zero. Choose a different starting value after inspecting a bracket or graph.
A level is h(t) = 10 + 3cos t − 4sin t. Express it as 10 + Rcos(t + α), and find its full-period range.
Match Rcosα = 3 and Rsinα = 4.
R = 5, α = arctan(4/3) in the first quadrant. Thus h = 10 + 5cos(t + α), with range 5 ≤ h ≤ 15. Dropping the offset 10 changes the model.
08 / Recognise standard and reverse-chain forms
A constant factor from the inner derivative must be included.
Find ∫[4e²ˣ − 3sin(3x)] dx.
Integrate each term and divide by its inner derivative.
The result is 2e²ˣ + cos(3x) + C. Differentiating recovers 4e²ˣ − 3sin(3x).
Find ∫2x/(x² + 5) dx for real x.
The numerator is the derivative of the positive denominator.
The integral is ln(x² + 5) + C. The argument is positive for every real x, so absolute-value notation is unnecessary here.
09 / Substitution and integration by parts
Change bounds when changing the variable; keep the boundary term in integration by parts.
Evaluate ∫₀¹ x√(1 + x²) dx exactly.
Use u = 1 + x²; the new bounds are 1 and 2.
The integral is ½∫₁²u¹ᐟ² du = [u³ᐟ²/3]₁² = (2√2 − 1)/3.
Evaluate ∫₁ᵉ ln x dx.
Take u = ln x and dv = dx.
An antiderivative is x ln x − x. At e it is 0; at 1 it is −1. The integral is 1.
10 / Partial fractions and signed area
A definite integral counts area below the axis negatively.
Evaluate ∫₀¹ 1/[(x + 1)(x + 2)] dx.
Decompose as 1/(x + 1) − 1/(x + 2).
The antiderivative is ln(x + 1) − ln(x + 2). Evaluation gives ln(2/3) − ln(1/2) = ln(4/3).
For y = x² − 1 on 0 ≤ x ≤ 2, find the signed integral and total area with the x axis.
Split at x = 1, the zero inside the interval.
F(x) = x³/3 − x. On [0,1] the integral is −2/3; on [1,2] it is 4/3. The net integral is 2/3, while the total area is 2.
11 / Compare exact and approximate areas
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Find the total area between y = x and y = x² for 0 ≤ x ≤ 2.
The curves meet at 0 and 1. Between 0 and 1, x is above x²; after 1 the order reverses.
Area = ∫₀¹(x − x²)dx + ∫₁²(x² − x)dx = 1/6 + 5/6 = 1.
Estimate ∫₀²x² dx with two equal strips. Compare with the exact value and find percentage overestimate relative to the exact value.
Width h = 1; ordinates 0, 1, 4.
T = ½(0 + 2×1 + 4) = 3; exact = 8/3. Excess = 1/3, and percentage error = (1/3)/(8/3) × 100 = 12.5%. The upward-curving graph lies below each chord.
12 / Initial conditions and limiting models
Check your solution in both the differential equation and the initial condition.
Solve dy/dx = 2x(y + 1), y(0) = 2.
Integrate dy/(y + 1) = 2x dx on the solution branch.
ln|y + 1| = x² + C. The initial value gives y + 1 = 3ex², hence y = 3ex² − 1. The omitted equilibrium y = −1 does not meet y(0) = 2.
A tank model is dV/dt = 12 − 0.3V, V(0) = 10, with V in litres and t in minutes. Find V(t) and its limiting value.
Equilibrium is V = 40. Solve for the difference from equilibrium.
V = 40 − 30e⁻⁰·³ᵗ. Its derivative is 9e⁻⁰·³ᵗ, equal to 12 − 0.3V, and V(0) = 10. It approaches 40 litres from below for t ≥ 0, assuming the physical tank can accommodate that volume.
13 / Geometric rates and exponential decay
State the physical time interval as well as the formula.
A radius obeys dr/dt = −2/r², with r(0) = 3. Find r(t) and when the model first reaches zero.
Multiply by r² and integrate, while r > 0.
r³ = 27 − 6t, so r = (27 − 6t)¹ᐟ³ for 0 ≤ t < 4.5. The limiting radius is zero at 4.5 time units; the original rate law is undefined at r = 0 and cannot be continued through it.
A quantity decays exponentially from 80 to 20 in 6 hours. Find its decay constant k in Q = 80e⁻ᵏᵗ, and its half-life.
The factor 1/4 is two halvings.
e⁻⁶ᵏ = 1/4 gives k = ln4/6 = ln2/3 per hour. The half-life is ln2/k = 3 hours.
14 / Geometry and directed angles
Use displacement for separation and signed components for positive-axis angles.
A = (−1,2,3), B = (3,−2,1). Find AB, its length and a unit vector from A to B.
End minus start, then divide by its magnitude.
AB = (4,−4,−2); its length is 6. A unit vector is (2/3,−2/3,−1/3).
For v = (2,−1,−2), find the angle with the positive z axis and the acute angle with the xy plane.
The magnitude is 3; the perpendicular-to-plane component has magnitude 2.
Positive-z angle = arccos(−2/3) ≈ 131.8°. Plane angle = arcsin(2/3) ≈ 41.8°. Keep the negative sign for the directed positive-axis question.
15 / Check all conditions before concluding
Reattempt any question that needed a hint. Then return to the corresponding lesson for another independent example.
A 4 kg particle is in equilibrium. Force (8,−4,12) N is removed. Find the new acceleration.
The remaining forces sum to the negative of the removed force.
The new resultant is (−8,4,−12) N, so acceleration is (−2,1,−3) m s⁻². Equilibrium described zero acceleration before removal; it did not require zero velocity.
Find every horizontal tangent on x² + xy + y² = 3, checking the derivative denominator.
Use y′ = −(2x + y)/(x + 2y), then substitute y = −2x into the curve.
Substitution gives 3x² = 3, hence (x,y) = (1,−2) or (−1,2). The denominators are −3 and 3, both nonzero, so these are valid horizontal tangents: y = −2 and y = 2.
Section 1 of 15 · Check what the method establishes