Physics
posted by Anonymous .
An equilateral triangle initially has side length equal to 17 cm. Each vertex begins moving in a straight line towards the midpoint of the opposite side at a constant rate of 2.3 cm/s, continuously forming progressively smaller equilateral triangles until it disappears. At what rate is the area of the triangle decreasing at the instant it vanishes? The area A of an equilateral triangle with side length s is
A= (sqrt(3)/4)(s^2)

dA/ds = sqrt 3 (1/2) s
dA/dt = .5 sqrt 3 sds/dt
but
altitude h = (s/2) sqrt 3
dh/dt = .5 sqrt 3 ds/dt
so
ds/dt = [2/sqrt 3] dh/dt
dA/dt=.5 sqrt 3*[2/sqrt 3]s dh/dt
= s dh/dt
evaluate when s = 17 and dh/dt = 2.3 
whoa, at the instant it vanished?
dA/dt = s dh/dt
as s > 0 and dh/dt is constant
dA/dt > 0 
wow got to right the question and solution down
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