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The length of a diagonal of a cube increases at a rate of 0.1 cm/s-1. Find the rate of change of the volume of the cube when the diagonal is 30cm long.The length of a rectangle increases at a rate of 2cm/s-1 while the width decreases at a rate of 1cm/s-1. Find the rate of change of the area when the leugth and... 顯示更多 The length of a diagonal of a cube increases at a rate of 0.1 cm/s-1. Find the rate of change of the volume of the cube when the diagonal is 30cm long. The length of a rectangle increases at a rate of 2cm/s-1 while the width decreases at a rate of 1cm/s-1. Find the rate of change of the area when the leugth and width of the rectangle are 12cm and 8cm respectively. A pulley is set up on the edge of a dock, 6cm above the water level. A rope passed through the pulley and is attached to a boat at water level. If the rope is pulled at a rate of 2ms-1. Find the speed at which the boat approaches the dock when if is 8m from the dock. The area of a circle increases at a rate 12兀cm^2 s-1. Find the rate of change of the radius and cicumference when radius is 3cm. 更新: A particle moves along a straight line so that its displacement s m, from the orgin O, after t seconds is given by s=4t+2t^2-t^3, where T >=0. A. Find the displacement of the particle from O when its velocity is zero. B. Find the displacement of the particle when its velocity is 5cm/s-1. 更新 2: A rocket R rises vertically at 250m/s-1 fromA. A video camera locating at C which is 800m fromA is to keep the rocket in sight all the time. Let ∠RCA = θ. Find the rate of change of θ with respect to time when RA =600m. 更新 3: 回答者: Karin & Victor 第4題錯左= = 有人識另2題呀??

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The length of a diagonal of a cube increases at a rate of 0.1 cm/s-1. Find the rate of change of the volume of the cube when the diagonal is 30cm long Diagonal = 30cm ==> Length = 30/sqrt(3) cm Diagonal increases at 0.1 cm/s ==> Length increases at 0.1/sqrt(3) cm/s Volume V at time t = (30/sqrt(3) + 0.1/sqrt(3))^3 dV/dt = 3[(30/sqrt(3) + 0.1t/sqrt(3))^2](0.1/sqrt(3)) At t = 0, dV/dt = 3[(30/sqrt(3))^2](0.1/sqrt(3)) = 30sqrt(3) = 51.96 Therefore volume increase at a rate 51.96 cm^3/s The length of a rectangle increases at a rate of 2cm/s-1 while the width decreases at a rate of 1cm/s-1. Find the rate of change of the area when the leugth and width of the rectangle are 12cm and 8cm respectively. Area A = (12+2t)(8-t) = 96+4t-t^2 dA/dt = 0 + 4 - 2t At t = 0, dA/dt = 4 Therefore, area increases at a rate 4 cm^2/s A pulley is set up on the edge of a dock, 6cm above the water level. A rope passed through the pulley and is attached to a boat at water level. If the rope is pulled at a rate of 2ms-1. Find the speed at which the boat approaches the dock when if is 8m from the dock. Length of rope at start = sqrt(6^2+8^2) = 10m Length of rope at time t = 10 -2t Distance D of boat from the dock at time t = sqrt[(10-2t)^2 - 6^2] dD/dt ={(1/2)/sqrt[(10-2t)^2 - 6^2]}{2(10-2t)(-2)} At t = 0, dD/dt = -40/16 = -2.5 Therefore, the boat approaches dock at 2.5m/s The area of a circle increases at a rate 12兀cm^2 s-1. Find the rate of change of the radius and cicumference when radius is 3cm. At time t, radius = R; Area = [R^2](pi) = 9(pi) + 12(pi)t R = sqrt(9+12t) dR/dt = (1/2)(12)/sqrt(9+12t) At time t = 0, radius increase at rate 6cm/s circumference increase at a rate 12(pi) cm/s 2008-04-29 09:11:24 補充: Correction to Q4. R = sqrt(9+12t) dR/dt = (1/2)(12)/sqrt(9+12t) At time t = 0, radius increase at rate 2cm/s Circumference = 2(pi)R circumference increase at a rate d(2(pi)R)/dt = 2(pi)dR/dt AT time t = 0, circumference increase at a rate 4(pi) cm/s 2008-04-29 09:20:45 補充: Q5A ds/dt = 4+4t -3t^2 velocity = 0 when (2-t)(2+3t)=0 t = 2 or t = -2/3 (rejected) The displacement of particle when its velocity is zero = 4(2) + 2(2)^2 - (2)^3 = 8 m 2008-04-29 09:21:09 補充: Q5B velocity = 5 when 4+4t-3t^2 = 5 3t^2-4t+1=0 (3t-1)(t-1)=0 t = 1/3 or t = 1 When t = 1/3, displacement = 4(1/3) + 2(1/3)^2 - (1/3)^3 = 41/27 m When t = 1, displacement = 4(1) + (1)^2 - (1)^3 = 4m

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