That's not the way to do it. As it is, you are analysing it at the condition when the lenght of the string is already 500 ft.

You must analyze it in the conditions before the length of the string is 500 ft.

Here it is assumed that the string is in a straight line, like the hypotenuse of a right triangle. [In reality, it is not--it sags a lot.]

Your right triangle should be

----x at the top, horizontal leg.

----y=300 as the vertical leg

----z as the hypotenuse

So,

z = sqrt(x^2 +300^2)

z = sqrt(x^2 +90,000)

You are to find dz/dt when z = 500 ft.

z = sqrt(x^2 +90,000)

Differentiate both sides with respect to time t,

dz/dt = (1/2)(x^2 +90,000)^(-1/2) *2x *dx/dt

dz/dt = x (dx/dt) / sqrt(x^2 +90,000) --------------(i)

When z = 500ft, what is dz/dt?

You are solving for that.

When z = 500ft, what is dx/dt?

dx/dt is always 25 ft/sec anytime.

When z = 500ft, what is x?

z = sqrt(x^2 +90,000)

500 = sqrt(x^2 +90,000)

Square both sides,

250,000 = x^2 +90,000

x^2 = 250,000 -90,000

x^2 = 160,000

x = 400 ft

[Your 3-4-5 triangle.]

Hence, at z=500ft,

dz/dt = x (dx/dt) / sqrt(x^2 +90,000) --------------(i)

dz/dt = 400(25) / sqrt(400^2 +90,000)

dz/dt = 400(25) / sqrt(250,000)

dz/dt = 400(25) /500

dz/dt = 20 ft/sec

Therefore, she must let the string out at the rate of 20 feet per second at that time. -----------------answer.