3 Calculus

14)a) L = (AD + 1)/cosθ = [√(AC² - 1) + 1]/cosθ = [√(1/sin²θ - 1) + 1]/cosθ= [√(cos²θ/sin²θ) + 1]/cosθ = (cosθ/sinθ + 1)/cosθ = secθ + cscθb) L = secθ + cscθ ⇒ dL/dθ = tanθsecθ - cotθcscθc) When L is minimum, dL/dθ = 0.tanθsecθ - cotθcscθ = 0 ⇒ tanθsecθ = cotθcscθ⇒ (tanθsecθ)/(cotθcscθ) = 1 ⇒ tan³θ = 1 ⇒ θ = π/4d²L/dθ² = tanθ(secθtanθ) + secθ(sec²θ) - cotθ(- cscθcotθ) - cscθ(- csc²θ)= tan²θsecθ + sec³θ + cot²θcscθ + csc³θWhen θ = π/4, tanθ, secθ, cscθ and csc³θ all > 0, d²L/dθ² > 0.∴ L is minimum when θ = π/4.L = sec(π/4) +csc(π/4) = √2 + √2 = 2√2 whichis minimum. 15)a) V = [(½x/π)²π](½ × 24 - x) = ¼x²(12 - x)/πb) V = ¼x²(12 - x)/π = ¼(12x² - x³)/πdV/dx = ¼(24x - 3x²)/πWhen V is maximum, dV/dx = 0.¼(24x -3x²)/π = 0 ⇒ x(8 - x) =0 ⇒ x = 0(rejected) or 8.d²L/dθ² = ¼(24 - 6x)/πWhen x = 8, d²L/dθ² = ¼(24 - 6 × 8)/π = - 6/π < 0.∴ V is maximum when x = 8.Baseradius = √(½ × 8/π) = 2/√π cmHeight = ½ × 24 - 8 = 4 cm 16)a) Velocity = (t⁴ - 2t² + 3)/t = (t³ - 2t + 3/t) ms⁻¹b) Let v be the velocity.v = t³ - 2t+ 3/tdv/dt = 3t² - 2 -3/t²When the acceleration = 0, dv/dt = 0.3t² - 2 -3/t² = 0 ⇒ 3t⁴ - 2t² - 3 = 0t² = {2 ± √[2² - 4 × 3 × (- 3)]}/(2 × 3) = (2 ± √40)/6 = (1 ± √10)/3t = ± √[(1 + √10)/3] or ± √[(1 - √10)/3]∵ √[(1 - √10)/3] is not real∴ t = ± √[(1 + √10)/3]∵ - √[(1 + √10)/3] < 0∴ t = √[(1 + √10)/3]i.e. The particle has zero acceleration when t = √[(1 + √10)/3]. I HOPE THIS CAN HELP YOU!~PLEASE FEEL FREE AGAIN TO ASK IF YOU STILL HAVE ANYPROBLEM!~ ^_^