A 15 ft ladder is placed against a vertical wall. the bottom of the ladder is sliding away from the wall at a rate of 2 ft/sec. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 9 ft from the wall?

rfw

To find how fast the top of the ladder is sliding down the wall, we can use related rates. Let's assign variables to the distances involved:

Let x be the distance between the bottom of the ladder and the wall (changing),
And let y be the distance between the top of the ladder and the ground (also changing).

We are given that dx/dt = -2 ft/sec, as the bottom of the ladder is sliding away from the wall. We want to find dy/dt when x = 9 ft.

Now, we know that the ladder forms a right triangle with the wall, and the ladder is always 15 ft long. We can express this relationship using the Pythagorean theorem:

x^2 + y^2 = 15^2

To differentiate both sides of this equation with respect to time t:

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

Now we can plug in the given values and solve for dy/dt:

2(9)(-2) + 2y(dy/dt) = 0
-36 + 2y(dy/dt) = 0
2y(dy/dt) = 36
dy/dt = 36 / (2y)

Now, to find the value of y when x = 9 ft, we can use the Pythagorean theorem:

9^2 + y^2 = 15^2
81 + y^2 = 225
y^2 = 225 - 81
y^2 = 144
y = 12 ft

Finally, we can substitute the value of y = 12 ft into the equation for dy/dt:

dy/dt = 36 / (2 * 12)
dy/dt = 36 / 24
dy/dt = 3/2 ft/sec

Therefore, the top of the ladder is sliding down the wall at a rate of 3/2 ft/sec when the bottom of the ladder is 9 ft from the wall.