✔ 最佳答案
(a) cosθ = (16 + x^2 - 25)/(2*4*x) = (x^2 - 9)/(8x) = x/8 - 9/(8x)(b) cosθ = (x^2 - 9)/(8x)sin^2θ = 1 - (x^2 - 9)^2/(64x^2)= [64x^2 - (x^2 - 9)^2]/(64x^2)sinθ = √[(x^2 + 8x - 9)(8x - x^2 + 9)]/(8x)= -√[(x + 9)(x - 1)(x - 9)(x + 1)]/(8x)dcosθ/dx= 1/8 + (9/8)(1/x^2)= (x^2 + 9)/(8x^2)(dx/dcosθ) = 8x^2/(x^2 + 9)dx/dt = (dx/dcosθ)(dcosθ/dθ)(dθ/dt)= [8x^2/(x^2 + 9)](-sinθ)(0.1)= [8x^2/(x^2 + 9)]{√[(x + 9)(x - 1)(x - 9)(x + 1)]/(8x)}(1/10)= {-x√[(x + 9)(x - 1)(9 - x)(x + 1)]}/[10(x^2 + 9)]Sub. x = 6dx/dt= -6√[(15)(5)(3)(7)]}/450= -√7/5 m/s(c) 0 <= φ <= πsinθ/5 = sinφ/4sinφ = (4/5)sinθcosφ(dφ/dt) = (4/5)(cosθ)(dθ/dt)Sub. θ = π/2 into cosθ = x/8 - 9/(8x)x/8 - 9/(8x) = 0x/8 = 9/(8x)x^2 = 3x = 3At that time, cosφ = 3/5So, (3/5)(dφ/dt) = (4/5)(0)(1/10)=> dφ/dt = 0(d) sin∠OPQ/x = sinθ/5sin∠OPQ = xsinθ/5(cos∠OPQ)(d∠OPQ/dt) = (1/5)[(xcosθ)(dθ/dt) + (sinθ)(dx/dt)]When P is vertically above Ocos∠OPQ = 4/5, cosθ = 0, sinθ = 1, dx/dt = {-3√[(12)(2)(6)(4)]}/180 = -1/5 m/sSo, d∠OPQ/dt = (1/5)(5/4)(0 - 1/5) = -1/20 and we conclude that the statement of the student is incorrect.
2011-02-28 20:53:13 補充：
應該那裡出了少少錯。本身就是√[(x + 9)(x - 1)(9 - x)(x + 1)]/(8x)
