Question

A 10-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 0.3 ft/s. How fast is the top of the ladder sliding down the wall?

Answer 1

The top of the ladder is not sliding down the wall because it is in contact with the wall, resulting in a rate of 0 ft/s.

Step-by-step explanation:

To determine how fast the top of the ladder is sliding down the wall, we can use similar triangles. Let's call the distance from the top of the ladder to the wall y and the distance from the foot of the ladder to the wall x. Using the Pythagorean theorem, we have: x^2 + y^2 = 10^2, which simplifies to x^2 + y^2 = 100. Differentiating both sides of the equation with respect to time, we get: 2x(dx/dt) + 2y(dy/dt) = 0. Since x = 10 and dx/dt = 0.3 ft/s, we can solve for dy/dt, which represents the rate at which the top of the ladder is sliding down the wall.

Plug in the values: 2(10)(0.3) + 2y(dy/dt) = 0. Simplify the equation to: 6 + 2y(dy/dt) = 0. Subtract 6 from both sides: 2y(dy/dt) = -6. Divide both sides by 2y: dy/dt = -3/y ft/s. Now, we need to find y to calculate the answer. From the equation x^2 + y^2 = 100, we know that when x = 10, y = sqrt(100 - 10^2) = sqrt(100 - 100) = sqrt(0) = 0. Therefore, the top of the ladder is not sliding down the wall because it is in contact with the wall.

So, the rate at which the top of the ladder is sliding down the wall is 0 ft/s.