Question:

A ladder is leaning against a wall, with its base 5 meters away from the wall. The ladder is initially 10 meters long and its bottom end is sliding away from the wall at a constant rate of 2 meters per second. At what rate is the top of the ladder sliding down the wall when the bottom of the ladder is 7 meters away from the wall?

Solution:

Let's denote the distance between the bottom of the ladder and the wall as x (in meters) and the distance between the top of the ladder and the ground as y (in meters). We are asked to find the rate at which y is changing with respect to x when x = 7m.

Using the Pythagorean theorem, we have:

x^2 + y^2 = 10^2

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

2x(dx/dt) + 2y(dy/dt) = 0 --(1)

We are given that dx/dt = 2 m/s, and we want to find dy/dt when x = 7m. However, we need to find the value of y first.

Substituting x = 7 into the equation x^2 + y^2 = 10^2, we can solve for y:

7^2 + y^2 = 10^2 49 + y^2 = 100 y^2 = 51 y = √51

Now we can substitute x = 7m and y = √51 into equation (1) and solve for dy/dt:

2(7)(2) + 2(√51)(dy/dt) = 0 28 + 2√51(dy/dt) = 0 2√51(dy/dt) = -28 dy/dt = -28 / (2√51) dy/dt = -14 / √51

Therefore, when the bottom of the ladder is 7 meters away from the wall, the top of the ladder is sliding down the wall at a rate of -14 / √51 meters per second.