A 25 foot ladder is leaning on a 9 foot fence. The base of the ladder is being pulled away from the fence at the rate of 10 feet/minute. How fast is the top of the ladder approaching the ground when the base is 9 from the fence? [Note: The ladder is protruding over the top of the fence.]

Before I start typing a rather messy solution, do you know if the correct answer is

-10.1788 ft/min ?

Not correct.

A 25 foot ladder is leaning on a 9 foot fence. The base of the ladder is being pulled away from the fence at the rate of 10 feet/minute. How fast is the top of the ladder approaching the ground when the base is 9 from the fence? [Note: The ladder is protruding over the top of the fence.]

To find the rate at which the top of the ladder is approaching the ground, we need to use related rates. Let's assign variables to the given information:

Let x be the distance between the base of the ladder and the fence.
Let y be the distance from the top of the ladder to the ground.

We are given that the rate at which the base of the ladder is being pulled away from the fence is -10 ft/min (negative because it is moving away from the fence).

We need to find the rate at which the top of the ladder is approaching the ground, which is dy/dt.

Now, we can use the Pythagorean theorem to relate x and y:

x^2 + y^2 = 25^2

Differentiating this equation implicitly with respect to time t, we get:

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

Since we are interested in finding dy/dt when x = 9 ft, we substitute the values into the equation:

2(9)(-10) + 2y dy/dt = 0

Simplifying this equation:

-180 + 2y dy/dt = 0

2y dy/dt = 180

dy/dt = 180 / (2y)

Now, we need to find the value of y when x = 9 ft. Using the Pythagorean theorem, we substitute the values into the equation:

9^2 + y^2 = 25^2

81 + y^2 = 625

y^2 = 544

y = √544 = 23.32 ft

Substituting y = 23.32 into the equation dy/dt = 180 / (2y):

dy/dt = 180 / (2 * 23.32)

dy/dt ≈ 3.87 ft/min

Therefore, when the base of the ladder is 9 ft from the fence, the top of the ladder is approaching the ground at a rate of approximately 3.87 ft/min.