Application of Derivatives (Rate of Change, Tangents, Maxima/Minima) — Practice

Question 1

Find the equations of the tangent and normal to the curve y = x² − 3x − 2 at the point (2, −4).

Accepted Answer

f(x) = x² − 3x − 2. f'(x) = 2x − 3. At x = 2: slope of tangent = f'(2) = 2(2) − 3 = 1. VERIFY POINT: f(2) = 4 − 6 − 2 = −4 ✓ (2, −4) is on the curve. TANGENT (slope 1, through (2,−4)): y − (−4) = 1(x − 2) → y + 4 = x − 2 → y = x − 6. NORMAL (slope = −1/1 = −1, through (2,−4)): y − (−4) = −1(x − 2) → y + 4 = −x + 2 → y = −x − 2.

Question 2

A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window that will admit maximum light through the whole opening.

Accepted Answer

Let width = 2r (so the semicircle has radius r) and height of rectangle = h. PERIMETER: The perimeter consists of: bottom (2r) + two sides (2h) + semicircular arc (πr) = 10. Constraint: 2r + 2h + πr = 10 → 2h = 10 − 2r − πr → h = (10 − 2r − πr)/2 = 5 − r − πr/2. AREA (objective, to maximise): A = 2r·h + πr²/2 (rectangle + semicircle). Substitute h: A(r) = 2r(5 − r − πr/2) + πr²/2 = 10r − 2r² − πr² + πr²/2 = 10r − 2r² − πr²/2. dA/dr = 10 − 4r − πr = 10 − r(4+π). Set dA/dr = 0: r = 10/(4+π). d²A/dr² = −(4+π) < 0 → MAXIMUM. h = 5 − r − πr/2 = 5 − r(1 + π/2) = 5 − [10/(4+π)] × (2+π)/2 = 5 − 5(2+π)/(4+π) = [5(4+π) − 5(2+π)]/(4+π) = 10/(4+π) = r. RESULT: Width = 2r = 20/(4+π) m, Height = h = 10/(4+π) m = r (height equals radius).

Question 3

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.

Accepted Answer

Let the cone have height h and base radius r, inscribed in a sphere of radius R. The cone's apex is at the top of the sphere, and the centre of the base circle is at distance (h−R) from the sphere's centre (or R−h if h < R... let's set it up carefully). SETUP: Place sphere centre at O. Let cone have apex at top of sphere, height h. The centre of the base is at distance (h−R) from O. The base radius r satisfies: r² + (h−R)² = R² (Pythagoras in sphere). So r² = R² − (h−R)² = R² − h² + 2hR − R² = 2hR − h². VOLUME: V = (1/3)πr²h = (1/3)π(2hR−h²)h = (π/3)(2Rh²−h³). OPTIMISE: dV/dh = (π/3)(4Rh − 3h²) = (πh/3)(4R − 3h). Set dV/dh = 0: h = 0 (trivial) or h = 4R/3. d²V/dh² = (π/3)(4R − 6h). At h = 4R/3: d²V/dh² = (π/3)(4R − 8R) = (π/3)(−4R) < 0. MAXIMUM at h = 4R/3. MAXIMUM VOLUME: r² = 2(4R/3)R − (4R/3)² = 8R²/3 − 16R²/9 = 24R²/9 − 16R²/9 = 8R²/9. V_max = (π/3)(8R²/9)(4R/3) = 32πR³/81. SPHERE VOLUME: V_sphere = (4/3)πR³ = 108πR³/81. RATIO: V_max/V_sphere = (32πR³/81)/(108πR³/81) = 32/108 = 8/27. The maximum cone volume = (8/27) × sphere volume. ✓

Practice — Application of Derivatives (Rate of Change, Tangents, Maxima/Minima)

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