Question

A ladder that is 15 meters long is leaning against a wall. The top of the ladder is being pulled up the wall at a rate of 3 meters per second. If x is the distance between the bottom of the ladder and the wall, find the rate of change of this distance at x = 10 meters. Assume that the top of the ladder maintains contact with the wall throughout.

A. –2.72 m/s
B. –3.354 m/s
C. –4.56 m/s
D. –5.75 m/s

Answer 1

The rate of change of the distance between the bottom of the ladder and the wall when the ladder is 10 meters away from the wall is ‑3.354 m/s, which is option B.

Step-by-step explanation:

The problem involves a 15-meter ladder leaning against a wall, where the top of the ladder moves up the wall at 3 meters per second. Using the Pythagorean theorem, we can relate the distance of the ladder's foot from the wall (x) to its length (L) and height (y) as L2 = x2 + y2. Differentiating both sides concerning time t, we get 2x(dx/dt) + 2y(dy/dt) = 0, where dy/dt is the speed at which the ladder slides up the wall, and dx/dt is the rate at which x changes.

Given that dy/dt is 3 m/s and x = 10 m, we first find y via the Pythagorean theorem: L2 = x2 + y2, which gives us y = 11.18 m. Then we plug the values into the differentiated formula: (2)(10)(dx/dt) + (2)(11.18)(3) = 0, solving for dx/dt. The rate of change of x is then found to be ‑3.354 m/s, so the correct answer is B.