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two side of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06rad/s.find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3two side of a triangle have lengths 12m and 15m. the angle between them is increasing... 顯示更多 two side of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06rad/s.find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3 two side of a triangle have lengths 12m and 15m. the angle between them is increasing at a rate of 2degree/min. how fast is the length of the third side increasing when the angle between the sides of fixed length is 60degree
最佳解答:
two side of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06rad/s.find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3 let @ be angle between the sides of fixed length A = (1/2)(4)(5) sin@ = 10sin@ dA/dt = 10 cos@ d@/dt dA/dt = 10 cos(pi/3) (0.06) = 0.3 m^2/s two side of a triangle have lengths 12m and 15m. the angle between them is increasing at a rate of 2degree/min. how fast is the length of the third side increasing when the angle between the sides of fixed length is 60degree let @ be angle between the sides of fixed length let x be length of the third side x^2 = 144 + 225 - (2)(12)(15)cos@ x^2 = 369 - 360cos@ 2xdx/dt = 360sin@ d@/dt when @ = 60, x = sqrt(351) = 3sqrt(39) 2 [3sqrt(39)] dx/dt = 360sin60 (2) dx/dt = 10/sqrt(13) m/min 2007-10-03 12:36:40 補充： 第二條運算有少少錯, 快手得滯...when @ = 60, x = sqrt(189) = 3sqrt(21)2 [3sqrt(21)] dx/dt = 360sin60 (2)dx/dt = 60/sqrt(7) m/min
其他解答:
Area = a x b x sin c = 4 x 5 x sin c = 20 sin c d(Area) / dt = 20 d(sin c) / dt = 20 (cos c) x dc/dt = 20 (cos c) x 0.06 when the angle is pi/3, rate of change of area, d(Area) / dt = 20 (cos pi/3) x 0.06 = 0.6 m^2/s ----------------------------------------------------------------------- Let y be the length of the third side by cosine rule y^2 = 12^2 + 15^2 - 2 x 12 x 15 x cos c y^2 = 369 - 360 cos c y = [369 - 360 cos c]^(1/2) dy/dt = d[369 - 360 cos c]^(1/2) / dt dy/dt = (1/2)[369 - 360 cos c]^(-1/2) x (360 sin c) x dc/dt when the angle between the sides of fixed length is 60 degree, dy/dt = (1/2)[369 - 360 cos 60]^(-1/2) x (360 sin 60) x 2 dy/dt = 22.68 m/min 2007-10-03 12:37:37 補充： Area = (1/2) x a x b x sin c= (1/2) x 4 x 5 x sin c= 10 sin cd(Area) / dt = 10 d(sin c) / dt= 10 (cos c) x dc/dt= 10 (cos c) x 0.06when the angle is pi/3,rate of change of area, d(Area) / dt= 10 (cos pi/3) x 0.06= 0.3 m^2/s0D7DAC4ECAFB489C
related rate
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發問:two side of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06rad/s.find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3two side of a triangle have lengths 12m and 15m. the angle between them is increasing... 顯示更多 two side of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06rad/s.find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3 two side of a triangle have lengths 12m and 15m. the angle between them is increasing at a rate of 2degree/min. how fast is the length of the third side increasing when the angle between the sides of fixed length is 60degree
最佳解答:
two side of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06rad/s.find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3 let @ be angle between the sides of fixed length A = (1/2)(4)(5) sin@ = 10sin@ dA/dt = 10 cos@ d@/dt dA/dt = 10 cos(pi/3) (0.06) = 0.3 m^2/s two side of a triangle have lengths 12m and 15m. the angle between them is increasing at a rate of 2degree/min. how fast is the length of the third side increasing when the angle between the sides of fixed length is 60degree let @ be angle between the sides of fixed length let x be length of the third side x^2 = 144 + 225 - (2)(12)(15)cos@ x^2 = 369 - 360cos@ 2xdx/dt = 360sin@ d@/dt when @ = 60, x = sqrt(351) = 3sqrt(39) 2 [3sqrt(39)] dx/dt = 360sin60 (2) dx/dt = 10/sqrt(13) m/min 2007-10-03 12:36:40 補充： 第二條運算有少少錯, 快手得滯...when @ = 60, x = sqrt(189) = 3sqrt(21)2 [3sqrt(21)] dx/dt = 360sin60 (2)dx/dt = 60/sqrt(7) m/min
其他解答:
Area = a x b x sin c = 4 x 5 x sin c = 20 sin c d(Area) / dt = 20 d(sin c) / dt = 20 (cos c) x dc/dt = 20 (cos c) x 0.06 when the angle is pi/3, rate of change of area, d(Area) / dt = 20 (cos pi/3) x 0.06 = 0.6 m^2/s ----------------------------------------------------------------------- Let y be the length of the third side by cosine rule y^2 = 12^2 + 15^2 - 2 x 12 x 15 x cos c y^2 = 369 - 360 cos c y = [369 - 360 cos c]^(1/2) dy/dt = d[369 - 360 cos c]^(1/2) / dt dy/dt = (1/2)[369 - 360 cos c]^(-1/2) x (360 sin c) x dc/dt when the angle between the sides of fixed length is 60 degree, dy/dt = (1/2)[369 - 360 cos 60]^(-1/2) x (360 sin 60) x 2 dy/dt = 22.68 m/min 2007-10-03 12:37:37 補充： Area = (1/2) x a x b x sin c= (1/2) x 4 x 5 x sin c= 10 sin cd(Area) / dt = 10 d(sin c) / dt= 10 (cos c) x dc/dt= 10 (cos c) x 0.06when the angle is pi/3,rate of change of area, d(Area) / dt= 10 (cos pi/3) x 0.06= 0.3 m^2/s0D7DAC4ECAFB489C
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