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
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

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

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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