posted by Michelle .
a 4m ladder is leaning against a wall when its base starts to slide away. By the time the base is 2m from the wall, the base is moving at the rate of 1m/sec.
a) the speed at which the top of the ladder is sliding down the wall
b) the rate at which the area of the triangle formed by the ladder, wall and ground is changing.
c) the rate at which the angle theta between the ladder and the ground is changing.
with x=distance of base from wall, y the height,
x^2 + y^2 = 16
2x dx/dt + 2y dy/dt = 0
when x=2, dx/dt=1
y = √12, so
2(2) + 2√12 dy/dt = 0
dy/dt = -4/2√12 = -1/√3
a = xy/2
2da/dt = y dx/dt + x dy/dt
= √12*1 - 2/√3
so da/dt = 2/√3
tanθ = y/x
sec^2θ dθ/dt = (xy' - yx')/x^2
when x=2, tanθ = √12/2 = √3, so θ=π/6
sec^2θ = 4/3, so
4/3 dθ/dt = (2(-1/√3)-(√12)(1))/4
dθ/dt = -√3 / 2
as always, double-check my algebra