Posted by Michelle on Sunday, July 15, 2012 at 1:50am.
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
= 4/√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
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