Practice Maths

★ Topic Review — Applications of Differential Calculus — Solutions

← Back to Applications of Differential Calculus›Maths Methods

Topic coverage: Instantaneous rates of change • Tangent and normal lines • Displacement, velocity and acceleration • Stationary points • First derivative test • Curve sketching • Optimisation

Fluency

  1. Fluency

    Find the equation of the tangent to y = x³ − 2x² + 1 at x = 2.

    f′(x) = 3x² − 4x. f′(2) = 12 − 8 = 4. f(2) = 8 − 8 + 1 = 1. Point: (2, 1).

    Tangent: y − 1 = 4(x − 2) → y = 4x − 7.

  2. Fluency

    Find the equation of the normal to y = x² + 3x at x = 1.

    f′(x) = 2x + 3. f′(1) = 5. f(1) = 1 + 3 = 4. Point: (1, 4).

    Normal gradient = −1⁄5.

    Normal: y − 4 = −1⁄5(x − 1) → y = −x⁄5 + 21⁄5.

  3. Fluency

    A particle has displacement x(t) = t³ − 3t² + 4 (metres, t in seconds). Find the velocity and acceleration at t = 2, and determine the direction of motion.

    v(t) = x′(t) = 3t² − 6t. v(2) = 12 − 12 = 0 m/s.

    a(t) = v′(t) = 6t − 6. a(2) = 12 − 6 = 6 m/s².

    At t = 2, the particle is momentarily at rest (v = 0). Since a > 0, velocity is about to become positive — the particle is changing direction to the positive direction.

  4. Fluency

    Find and classify all stationary points of f(x) = x³ − 6x² + 9x + 1 using the first derivative test.

    f′(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3).

    f′ = 0 at x = 1 and x = 3.

    f(1) = 1 − 6 + 9 + 1 = 5. f(3) = 27 − 54 + 27 + 1 = 1.

    Sign: f′(0) = 9 > 0 (+); f′(2) = 3(−1) < 0 (−); f′(4) = 9 > 0 (+).

    x = 1: +→− → local maximum at (1, 5).

    x = 3: −→+ → local minimum at (3, 1).

  5. Fluency

    Determine the end behaviour of y = 3x4 − x² + 5 and state whether the function has a global minimum.

    Leading term 3x4: even degree, positive coefficient.

    As x → +∞, y → +∞. As x → −∞, y → +∞.

    Because the function rises to +∞ in both directions and is continuous, it has a global minimum (at its lowest stationary point).

Understanding

  1. Understanding

    Find the point(s) on y = x² + x where the tangent is parallel to the line y = 5x − 3.

    f′(x) = 2x + 1. Set equal to gradient of given line: 2x + 1 = 5 → x = 2.

    f(2) = 4 + 2 = 6. Point: (2, 6).

    Tangent equation: y − 6 = 5(x − 2) → y = 5x − 4.

  2. Understanding

    A particle has displacement x = 2t³ − 12t² + 18t (cm, t ≥ 0 s).

    1. (a) Find when the particle is at rest.
    2. (b) Find the total distance travelled from t = 0 to t = 4.

    (a) v(t) = 6t² − 24t + 18 = 6(t² − 4t + 3) = 6(t − 1)(t − 3) = 0 at t = 1 s and t = 3 s.

    (b) x(0) = 0; x(1) = 2 − 12 + 18 = 8; x(3) = 54 − 108 + 54 = 0; x(4) = 128 − 192 + 72 = 8.
    Distances: 0→1: 8 cm; 1→3: |0−8| = 8 cm; 3→4: |8−0| = 8 cm.
    Total distance = 24 cm.

  3. Understanding

    For f(x) = 2x³ + 3x² − 12x − 7:

    1. (a) Find all stationary points and classify them.
    2. (b) State the intervals where f is increasing and decreasing.

    (a) f′(x) = 6x² + 6x − 12 = 6(x² + x − 2) = 6(x + 2)(x − 1). f′ = 0 at x = −2 and x = 1.
    f(−2) = −16 + 12 + 24 − 7 = 13. f(1) = 2 + 3 − 12 − 7 = −14.
    Sign: f′(−3) = 6(1)(−4) < 0 (−); f′(0) = −12 < 0... wait: f′(0) = 6(−2+2)(0−1)... recalculate: f′(0) = 6(2)(−1) = −12 < 0 (−); f′(2) = 6(4)(1) = 24 > 0 (+).
    x = −2: − (to the left)... Test x = −3: 6(1)(−4) = −24 < 0. Hmm, this would be − on the left of −2. Test x = −1: 6(1)(−2) = −12 < 0. That can’t be right for a local max. Re-check: f′(x) = 6(x+2)(x−1). Test x = −3: 6(−3+2)(−3−1) = 6(−1)(−4) = 24 > 0 (+). Test x = 0: 6(2)(−1) = −12 < 0 (−). Test x = 2: 6(4)(1) = 24 > 0 (+).
    x = −2: +→− → local maximum at (−2, 13).
    x = 1: −→+ → local minimum at (1, −14).

    (b) Increasing on (−∞, −2) ∪ (1, ∞); decreasing on (−2, 1).

  4. Understanding

    Use the 5-step method to sketch y = x³ − 3x, labelling all intercepts and stationary points.

    Step 1 — y-intercept: f(0) = 0. Point: (0, 0).

    Step 2 — x-intercepts: x³ − 3x = x(x² − 3) = x(x − √3)(x + √3) = 0 → x = 0, ±√3 (≈ ±1.73).

    Step 3 — Stationary points: f′(x) = 3x² − 3 = 3(x − 1)(x + 1) = 0 → x = ±1.
    f(1) = 1 − 3 = −2. f(−1) = −1 + 3 = 2.
    Sign: f′(−2) = 9 > 0; f′(0) = −3 < 0; f′(2) = 9 > 0.
    x = −1: +→− → local max (−1, 2). x = 1: −→+ → local min (1, −2).

    Step 4 — End behaviour: Odd cubic, positive leading term: y → −∞ as x → −∞; y → +∞ as x → +∞.

    Step 5: Classic S-shaped cubic crossing x-axis at −√3, 0, √3; local max at (−1, 2), local min at (1, −2).

  5. Understanding

    A 6 m length of wire is bent to form three sides of a rectangle (the fourth side lies along a wall). Find the dimensions that maximise the area of the rectangle.

    Let x = the two equal sides (width). Then the third side (length) = 6 − 2x. Domain: 0 < x < 3.

    Area A = x(6 − 2x) = 6x − 2x².

    A′(x) = 6 − 4x = 0 → x = 1.5 m.

    A′(1) = 2 > 0; A′(2) = −2 < 0. Maximum confirmed.

    Length = 6 − 2(1.5) = 3 m. Width = 1.5 m. Maximum area = 1.5 × 3 = 4.5 m².

Problem Solving

  1. Problem Solving

    The curve y = ax³ + bx passes through (1, −2) and has a tangent gradient of 1 at that point. Find a and b, and find the equation of the tangent at (1, −2).

    f(x) = ax³ + bx, f′(x) = 3ax² + b.

    f(1) = −2: a + b = −2  …(1)

    f′(1) = 1: 3a + b = 1  …(2)

    (2) − (1): 2a = 3 → a = 3⁄2. Then b = −2 − 3⁄2 = −7⁄2.

    Tangent at (1, −2): y − (−2) = 1(x − 1) → y = x − 3.

  2. Problem Solving

    A particle moves along a line with displacement x = t³ − 6t² + 12t − 5 (m, t ≥ 0 s).

    1. (a) Show that the particle never reverses direction by showing v(t) ≥ 0 for all t ≥ 0.
    2. (b) Find the acceleration when v is at its minimum, and interpret the result.

    (a) v(t) = 3t² − 12t + 12 = 3(t² − 4t + 4) = 3(t − 2)². Since (t−2)² ≥ 0 always, v(t) ≥ 0 for all t. The particle never moves in the negative direction, so it never reverses. □

    (b) Minimum of v: v(t) = 3(t−2)² has its minimum value of 0 at t = 2 (and is a stationary point of inflection on x vs t). a(t) = v′(t) = 6t − 12. a(2) = 12 − 12 = 0 m/s². When velocity is at its minimum (zero), the acceleration is also zero — the particle is momentarily at rest and there is no net force causing it to reverse.

  3. Problem Solving

    A farmer wants to build a rectangular pen divided into 3 equal sections by internal fences parallel to the width. The total fencing available is 240 m. Find the overall dimensions that maximise the total enclosed area.

    Let x = width of each section (m), y = length of the whole pen (m).

    Fencing: 4 widths + 2 lengths (4 internal + external width fences, 2 length fences): 4x + 2y = 240 → y = (240 − 4x)⁄2 = 120 − 2x.

    Area A = x · y = x(120 − 2x) = 120x − 2x². Domain: 0 < x < 60.

    A′(x) = 120 − 4x = 0 → x = 30 m. y = 120 − 60 = 60 m.

    A′(20) = 40 > 0; A′(40) = −40 < 0. Maximum confirmed.

    Dimensions: 30 m × 60 m. Maximum area = 1800 m².

  4. Problem Solving

    The profit from selling x items is P(x) = −2x³ + 18x² − 30x for x ≥ 0. Find the production level that maximises profit, and verify it is a maximum using the first derivative test.

    P′(x) = −6x² + 36x − 30 = −6(x² − 6x + 5) = −6(x − 1)(x − 5).

    P′ = 0 at x = 1 and x = 5.

    P(1) = −2 + 18 − 30 = −14 (a loss).

    P(5) = −250 + 450 − 150 = 50.

    Sign: P′(0) = −30 < 0 (−); P′(3) = −6(2)(−2) = 24 > 0 (+); P′(6) = −6(5)(1) = −30 < 0 (−).
    x = 1: −→+ (local min). x = 5: +→− (local max).

    Maximum profit = 50 (thousand dollars) at x = 5 items.

  5. Problem Solving

    A ladder of length L = 5 m leans against a vertical wall. The base of the ladder slides away from the wall at 0.5 m/s.

    Using Pythagoras: if the base is x metres from the wall, the height up the wall is h = √(25 − x²).
    1. (a) When the base is 3 m from the wall, what is the height of the top of the ladder?
    2. (b) Find h′(x) = dh⁄dx.
    3. (c) Using dx⁄dt = 0.5, find the rate at which the height is decreasing when x = 3 m (i.e. find dh⁄dt).

    (a) h = √(25 − 9) = √16 = 4 m.

    (b) h = (25 − x²)1⁄2. h′(x) = 1⁄2(25 − x²)−1⁄2 · (−2x) = −x ⁄ √(25 − x²).

    (c) dh⁄dt = (dh⁄dx) · (dx⁄dt) = [−3 ⁄ √(25−9)] × 0.5 = (−3⁄4) × 0.5 = −0.375 m/s. The top of the ladder is sliding down at 0.375 m/s.