Question

A ​25-foot ladder is leaning against a vertical wall​ (see figure) when Jack begins pulling the foot of the ladder away from the wall at a rate of 0.7ft/s. How fast is the top of the ladder sliding down the wall when the foot of the ladder is 24 ft from the​ wall?

Answer 1

To find the rate at which the top of the ladder is sliding down the wall, we can use related rates and similar triangles.

Step-by-step explanation:

To solve this problem, we can use similar triangles and the concept of related rates.

Let's define the height of the ladder as y and the distance from the wall as x. According to the problem, we have:

y = 25 - x

Now, we can differentiate both sides of this equation with respect to time t, remembering that y is also a function of x:

dy/dt = -dx/dt

We are given dx/dt (the rate at which the foot of the ladder is being pulled away from the wall) as 0.7 ft/s. So, we can substitute this value into the equation:

dy/dt = -0.7 ft/s

The negative sign indicates that y is decreasing (the top of the ladder is sliding down the wall). Finally, to find the rate at which the top of the ladder is sliding down the wall when x = 24 ft, we substitute x = 24 ft into the equation:

dy/dt = -0.7 ft/s

dy/dt = -0.7 ft/s

Answer 2

The top of the ladder is not sliding down the wall when the foot of the ladder is 24 ft from the wall.

Step-by-step explanation:

To find the rate at which the top of the ladder is sliding down the wall, we can use similar triangles. Let x be the distance from the top of the ladder to the wall. The length of the ladder is always 25 ft. The rate at which the foot of the ladder is being pulled away from the wall is 0.7 ft/s.

Using the Pythagorean theorem, we can write the equation:
x^2 + 24^2 = 25^2
Simplifying, we get:
x^2 = 25^2 - 24^2
x^2 = 625 - 576
x^2 = 49
x = 7

Now, we can take the derivative of both sides of the equation to find the rate at which x, the distance from the top of the ladder to the wall, is changing concerning time:

2x*dx/dt = 0
Solving for dx/dt, we get:
dx/dt = 0/2x = 0 ft/s

Therefore, the top of the ladder is not sliding down the wall when the foot of the ladder is 24 ft from the wall.