? 發問於 科學及數學數學 · 1 十年 前

related rate

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

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  • 1 十年 前
    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

  • J
    Lv 5
    1 十年 前

    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/s

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