# A 24 ft. ladder is leaning against a house while the base is pulled away at a constant rate of 1 ft/s. At what rate is the top of the ladder sliding down the side of the house when the base is:

Using Pythagoras theorem, The ladder and house makes up a right triangle: \(x^2+y^2=24^2\) where x is the horizontal distance the ladder is from the wall and y is the vertical distance the ladder is up the wall. 24 ft. is hypotenuse of the triangle. Firstly, We need to calculate the derivative of this equation that gives us:- \(2x \cfrac { dx }{ dt } + 2y \cfrac { dy }{ dt } =0\)

Here, The Derivative of hypotenuse is 0 as the length of the ladder is not changing.

Solving the problem further gives us \(2y \cfrac { dy }{ dt } = -2x \cfrac { dx }{ dt }\) or \(y \cfrac { dy }{ dt } = -x \cfrac { dx }{ dt }\)

We know the rate of moving the ladder is 1 ft/s so we can say \(\cfrac { dx }{ dt } = 1\) Hence, \(y \cfrac { dy }{ dt } = -x\) or \(\cfrac { dy }{ dt } = \cfrac { -x }{ y }\)

We need to calculate \(\cfrac { dy }{ dt }\) as it gives us the rate of descent of the ladder. Below cases gives us the value of x with different values. We solve for y and substiture x and y values in the above equation to find the rate of descent. We will solve of y from \(x^2+y^2=24^2\)