Question

A ladder 10 meters long is leaning against a vertical wall with its other end on the ground. The top end of the ladder is sliding down the wall. When the top end is 6 meters from the ground, it is sliding down at 2 m/sec. How fast is the bottom moving away from the wall at this instant?

Answer 1

To solve the problem, we apply the Pythagorean theorem and differentiate with respect to time to find that the bottom of the ladder is moving away from the wall at 1.5 m/sec when the top is 6 meters above the ground.

Step-by-step explanation:

The problem posed involves applying the principles of related rates, which is a concept in differential calculus. The student is asked to find out how fast the bottom of a 10-meter ladder is moving away from the wall if the top is sliding down at 2 m/sec and is currently 6 meters above the ground. To solve this problem, we can use the Pythagorean theorem to relate the height of the ladder against the wall and the distance from the wall to the base of the ladder.

Steps to Solve the Problem:

1. Let y be the vertical distance from the ground to the top of the ladder. As given, dy/dt = -2 m/sec (negative because y is decreasing).
2. Let x be the horizontal distance from the wall to the bottom of the ladder. We need to find dx/dt when y = 6 m.
3. Using the Pythagorean theorem,
   we have x² + y² = 10².
4. Differentiating both sides with respect to time t gives us 2x(dx/dt) + 2y(dy/dt) = 0.
5. Now plug in y = 6 and dy/dt = -2 into the differentiated equation.
6. Solve for dx/dt, which is the rate at which the bottom is moving away from the wall.

To calculate dx/dt, we get x = √(10² - 6²) = √(100 - 36) = √64 = 8 meters. Plugging into the differentiated equation:

2(8)(dx/dt) + 2(6)(-2) = 0
16(dx/dt) - 24 = 0
16(dx/dt) = 24
dx/dt = 1.5 m/sec

This means the bottom of the ladder is moving away from the wall at 1.5 m/sec when the top is 6 meters above the ground.