# A 13ft ladder is placed against a verticle wall. Suppose the bottom of the ladder slides away from the wall at a constant rate of 2 feet per second. How fast is the top of the ladder sliding down the wall (negitive rate) when the bottom is 5ft from the wall? The ladder is sliding down the wall at a rate of _ ft/sec

## You have

x^2 + y^2 = 13^2

when x=5, y=12

2x dx/dt + 2y dy/dt = 0

Now plug in your numbers and solve for dy/dt.

## To find the rate at which the top of the ladder is sliding down the wall, we can use related rates. We need to find the rate at which the distance between the bottom of the ladder and the wall is changing, and use that information to determine the rate at which the top of the ladder is sliding down.

Let's use the Pythagorean theorem to relate the distances involved. Let's denote the distance between the bottom of the ladder and the wall as x, and the distance between the top of the ladder and the ground as y. We know that the ladder has a length of 13ft, so we have the equation: x^2 + y^2 = 13^2.

Differentiating both sides of the equation with respect to time (t), we get:

2x(dx/dt) + 2y(dy/dt) = 0.

We are given that the bottom of the ladder is sliding away from the wall at a constant rate of 2ft per second, so dx/dt = 2. We want to find dy/dt when the bottom is 5ft from the wall, so x = 5.

Plugging these values into the equation, we have:

2(5)(2) + 2y(dy/dt) = 0.

Simplifying, we get:

20 + 2y(dy/dt) = 0.

Now, let's solve for dy/dt:

2y(dy/dt) = -20,

dy/dt = -20 / 2y.

We can substitute the length of the ladder (y = 13) into the equation to get the final answer:

dy/dt = -20 / (2 * 13) = -20 / 26 = -0.769 ft/sec.

Therefore, the top of the ladder is sliding down the wall at a rate of approximately -0.769 ft/sec.