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

To find the rate at which the area of the equilateral triangle is decreasing, we can differentiate the area equation with respect to time and then substitute the known values to solve for the rate.

Let's denote the side length of the triangle as s(t), where t represents time. From the problem, we know that s(t) = 17 - 2.3t, where t is measured in seconds.

Now, let's differentiate the area equation A = (sqrt(3)/4)(s^2) with respect to time (t):

dA/dt = (sqrt(3)/4)(2s)(ds/dt)

Since s = 17 - 2.3t, we can substitute that into the equation:

dA/dt = (sqrt(3)/4)(2(17 - 2.3t))(ds/dt)

Now, let's find ds/dt, which represents the rate at which the side length is changing:

ds/dt = -2.3 cm/s

Substitute this value into the equation:

dA/dt = (sqrt(3)/4)(2(17 - 2.3t))(-2.3)

Now, let's evaluate this expression at the instant when the equilateral triangle disappears. To find that instant, we need to solve for t when s(t) equals zero:

0 = 17 - 2.3t

2.3t = 17

t = 17/2.3

t ≈ 7.39 seconds

Substitute this value of t into the equation:

dA/dt = (sqrt(3)/4)(2(17 - 2.3(7.39)))(-2.3)

Now, we can calculate the rate at which the area of the triangle is decreasing at the instant it disappears by evaluating this expression.