Question

A 10-foot ladder is placed against a vertical wall. Suppose the bottom of the ladder slides away from the wall at a constant rate of 3 feet per second. How fast is the top of the ladder sliding down the wall (negative rate) when the bottom is 8 feet from the wall? The ladder is sliding down the wall at a rate of ft/sec. (Type an integer or a simplified fraction.)

Answer 1

The top of the ladder is sliding down the wall at a rate of 4 ft/sec.

Step-by-step explanation:

To solve this problem, we can use related rates by considering the ladder as the hypotenuse of a right-angled triangle, with the wall and the ground forming the other two sides. Let x be the distance of the bottom of the ladder from the wall, and y be the height where the ladder touches the wall. According to the Pythagorean theorem,
`\(x^2 + y^2 = 10^2\)`.

We're given that
`\(\frac{{dx}}{{dt}} = -3\) ft/sec`, as the bottom of the ladder slides away from the wall. We need to find
`\(\frac{{dy}}{{dt}}\) when \(x = 8\)` feet. Differentiating the Pythagorean equation with respect to time gives us
`\(2x \cdot \frac{{dx}}{{dt}} + 2y \cdot \frac{{dy}}{{dt}} = 0\).`

Substituting the given values at the instant when the bottom of the ladder is 8 feet away
`(\(x = 8\)) gives us \(2(8)(-3) + 2y \cdot \frac{{dy}}{{dt}} = 0\).` Solving for y gives us (y = 6) feet. Plugging this into the derived equation and solving for
`\(\frac{{dy}}{{dt}}\)` gives us
`\(\frac{{dy}}{{dt}} = -4\) ft/sec.` Since the question asks for the rate of the top of the ladder sliding down the wall (which is negative), the magnitude would be 4 ft/sec in the downward direction.