Applied calculus help

This is part of my homework and I dont understand how to do it can anyone help me?
The top of a 45 ft ladder is sliding down a wall at a rate of 15 feet per second. How fast is the base of the ladder sliding away from the wall at the instant when the top of the ladder is 27 feet from the ground?
ft/sec

Let y = ladder top's distance from ground. Let x = ladder bottom's distance from wall's base.
Let t = time in seconds.
x^2 + y^2 = 45^2
d/dt(x^2 + y^2) = d/dt(45^2)
2x(dx/dt) + 2y(dy/dt) = 0
When y = 27, x = (45^2 - 27^2)^(1/2) = 36.
2(36)dx/dt + 2(27)15 = 0
72dx/dt + 810 = 0
72dx/dt = -810
dx/dt = -11.25
The minus sign indicates movement away from the wall's base.

This is a related rates problem, which means you need to set up equations to represent things you are looking for. First, lets use the pythagorean theorem to represent the sides of the triangle formed by the ground (lets call this b for base), the height (h), and the ladder (45):

b² + h² = 45²

What you are ultimately looking for is the rate of change of b. So, lets solve the above equation for b, and then take its derivative.

b² + h² = 45²
b = √(45² - h²) = (45² - h²)^(1/2)
db/dt = (1/2)(45² - h²)^(-1/2)(0 - 2h(dh/dt))

Now just plug in your value for h ("top of the ladder is 27 feet from the ground") and your value for dh/dt ("sliding down a wall at a rate of 15 feet per second", this will be negative):

db/dt = (1/2)(45² - h²)^(-1/2)(0 - 2h(dh/dt))
db/dt = (1/2)(45² - (27)²)^(-1/2)(0 - 2(27)(-15))
db/dt = (1/2)(1296)^(-1/2)(-810)
db/dt = -11.25

So, when the ladder is 27 feet from the ground, the base of the ladder is sliding away from the wall at 11.25 ft/sec! (The negative means it's moving away from the ladder)

Hope this helps!