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中五 maths 幾條問題 ...@oo
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(1) (i) The volume of the cup V = (1/3)πr^2h = 30 h = 90/πr^2 Slant height s = √(h^2 + r^2) Steel sheet area A = πrs = πr√(h^2 + r^2) = πr√[(90/πr^2)^2 + r^2] A^2 = E = πr^2[8100/π^2r^4 + r^2] = 8100/πr^2 + πr^4 dE/dr = -16200/πr^3 + 4πr^3 = 0 => 4π^2r^6 = 16200 => r = 2.726 m d^E/dr^2 = 48600/πr^4 + 12πr^2 > 0 => min r = 2.726 m =>h = 90/(π*2.726^2) = 3.855 m (ii) The minimum amount of steel = π(2.726)√(3.855^2 + 2.726^2) = 40.44 m^2 (2)(a) y = 5x^2 dy/dx = lim h -> 0 [5(x + h)^2 - 5x^2]/h = lim h -> 0 [5(x^2 + 2hx + h^2) - 5x^2]/h = lim h -> 0 [5x^2 + 10hx + 5h^2 - 5x^2]/h = lim h -> 0 [10hx + 5h^2]/h = lim h -> 0 [10x + 5h] = 10x (b) y = 2t^2 - 4t t = √(3x^2+1) dy/dx = (dy/dt)(dt/dx) = (4t - 4)(1/2)(3x^2 + 1)^(-1/2) (6x) = 4(t - 1)3x/√(3x^2+1) = 12x[√(3x^2+1) - 1]/√(3x^2+1) = 12x - 12x/√(3x^2+1) 2009-12-15 22:49:35 補充： 收到.以後會小心不答他的問題.
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中五 maths 幾條問題 ...@oo
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1. A cylindrical open cup container with circular base is required to contain 30cubic meters of water. Let the height of the cylinder be h meters and the radiusof the base be r meters as shown in Figure B2 balow. Findi) the height and the radius of the cylinder so that the amount of sheet steelused is... 顯示更多 1. A cylindrical open cup container with circular base is required to contain 30 cubic meters of water. Let the height of the cylinder be h meters and the radius of the base be r meters as shown in Figure B2 balow. Find i) the height and the radius of the cylinder so that the amount of sheet steel used is minimum. ii) the minimum amount of sheet steel used. 2. a) Differentiate y=5x^2 with respect to x from the first principle. b) If y=2t^2-4t and t=開方3x^2+1, find dy/dx in terms of x.最佳解答:
(1) (i) The volume of the cup V = (1/3)πr^2h = 30 h = 90/πr^2 Slant height s = √(h^2 + r^2) Steel sheet area A = πrs = πr√(h^2 + r^2) = πr√[(90/πr^2)^2 + r^2] A^2 = E = πr^2[8100/π^2r^4 + r^2] = 8100/πr^2 + πr^4 dE/dr = -16200/πr^3 + 4πr^3 = 0 => 4π^2r^6 = 16200 => r = 2.726 m d^E/dr^2 = 48600/πr^4 + 12πr^2 > 0 => min r = 2.726 m =>h = 90/(π*2.726^2) = 3.855 m (ii) The minimum amount of steel = π(2.726)√(3.855^2 + 2.726^2) = 40.44 m^2 (2)(a) y = 5x^2 dy/dx = lim h -> 0 [5(x + h)^2 - 5x^2]/h = lim h -> 0 [5(x^2 + 2hx + h^2) - 5x^2]/h = lim h -> 0 [5x^2 + 10hx + 5h^2 - 5x^2]/h = lim h -> 0 [10hx + 5h^2]/h = lim h -> 0 [10x + 5h] = 10x (b) y = 2t^2 - 4t t = √(3x^2+1) dy/dx = (dy/dt)(dt/dx) = (4t - 4)(1/2)(3x^2 + 1)^(-1/2) (6x) = 4(t - 1)3x/√(3x^2+1) = 12x[√(3x^2+1) - 1]/√(3x^2+1) = 12x - 12x/√(3x^2+1) 2009-12-15 22:49:35 補充： 收到.以後會小心不答他的問題.
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