Question

微积分，求解答

Answer 1

【Solution】
Let L be |AB|
L² = x² + y²
2LdL/dt = 2xdx/dt + 2ydy/dt
dL/dt = [xdx/dt + ydy/dt]/L [1]
sub x = 4,dx/dt = -15 km/hr,y = 3,dy/dt = 10 and L = 5 into equation [1],
we have dL/dt = [-4×15 + 3×10]/5 = -6 (km/hr)
C.let θ be the angle as shown in the picture,find the rate of chang of θ,in radians per hour,when x=4km and y=3km.
【Solution】
y = xtanθ
dy/dt = (dx/dt)tanθ + (xsec²θ)(dθ/dt)
So,
dθ/dt = [dy/dt - (dx/dt)tanθ]/xsec²θ
=[10 - (-15)×tanθ](cos²θ)/4
=[10 + 15×tanθ](cos²θ)/4
=[10 + 15×3/4](16/25)/4
=17/5 (rad/hr)