application of differentiation

1. Let PQ = x and a be the angle between PQ and the vertical, then
x^2 = (3 + 3 cos a)^2 + (9 - 3 sin a)^2
Differentiate both sides w.r.t. t,
2 x (dx/dt) = 2 (3 + 3 cos a) (-3 sin a) (da/dt) + 2 (9 - 3 sin a) (-3 cos a) (da/dt)
ie. x (dx/dt) = - 9 sin a (1 + cos a) (da/dt) - 9 cos a (3 - sin a) (da/dt)

When a = 0, x = sqrt (6^2 + 9^2) = 3 sqrt (13),
sin a = 0, cos a = 1, and as a is decreasing, so da/dt = -1.5, so
3 sqrt(13) (dx/dt) = -9 (3) (-1.5).
therefore, dx/dt = 3.74 cm/s

2. Let angle DAP = a, angle CBP = b, then
4 sqrt(3) = 3 tan a + 3 tan b
Differentiate both sides w.r.t. t,
(da/dt) (sec a)^2 + (db/dt) (sec b)^2 = 0

When a = pi/6, b = pi/3, sec(pi/6) = 2/sqrt(3), sec(pi/3) = 2, da/dt = 3, so
3 (4/3) + (db/dt) (4) = 0
therefore, db/dt = -1 rad/min

1. 設@為P與垂直方向的角度

L = PQ = sqrt[(9 - 3sin@)^2 + (3 + 3cos@)^2]

= sqrt[(27 - 54sin@ + 9 + 18cos@ + 9(sin^2@ + cos^2@)]

= sqrt[45 - 54sin@ + 18cos@]

= 3sqrt(5 - 6sin@ + 2cos@)

dL/d@ = 3 [1/2sqrt(5 - 6sin@ + 2cos@)][-6cos@ - 2sin@]

= -3(3cos@ + sin@) / sqrt(5 - 6sin@ + 2cos@)

而d@/dt = 1.5 rads^-1

dL/dt = dL/d@‧d@/dt

= -3(3cos@ + sin@) / sqrt(5 - 6sin@ + 2cos@) ‧ 1.5

= -4.5(3cos@ + sin@) / sqrt(5 - 6sin@ + 2cos@)

當P在最高點時，@ = 0

繩子長度變率, dL/dt at @ = 0

= -4.5(3cos0 + sin0) / sqrt(5 - 6sin0 + 2cos0)

= -4.5(3) / sqrt(5 - 0 + 2)

= -27 / [2sqrt(7)] ms^-1 (負值為變短)


2. 設DP = x, 角DAP = @

tan@ = x/3

而CP = 4sqrt3 - x

設角CBP = &

tan& = (4sqrt3 - x) / 3 = (4sqrt3) / 3 - tan@ ... (1)

而d@/dt = 3 radmins^-1

將(1)對t微分

sec^2& d&/dt = -sec^2@ d@/dt

(tan^2& + 1) d&/dt = -(tan^2@ + 1) d@/dt ... (2)

當@ = pi/6, x = 3tan(pi/6) = 3sqrt3

tan& = (4sqrt3) / 3 - tan(pi/6) = sqrt3

代入(2):

{[((sqrt3)]^2 + 1} d&dt = -((sqrt3)^2 + 1)(3)

d&/dt = -(4)(3) / (4) = -3 radmins^-1