Question

A 12ft ladder is leaning up against a wall when it starts to slip down. When the bottom portion is 8ft from the wall, it's moving outward at 3ft/sec. How fast is the top falling down at that time?

a) 3 ft/s
b) 6 ft/s
c) 4 ft/s
d) 2 ft/s

Answer 1

The top of the ladder is falling at a rate of 2 ft/s at that time.(Option d)

Step-by-step explanation:

The problem involves related rates, connecting the rates of change of different variables. As the ladder slides down, the distance between the base of the ladder and the wall changes, causing the top of the ladder to fall. We're given that the bottom of the ladder is 8ft from the wall and moving outward at 3ft/s. To find how fast the top is falling, we'll use the Pythagorean theorem to relate the ladder's length, the distance of the bottom from the wall, and their rates of change.

Let x represent the distance from the wall to the base of the ladder, and y represent the height of the ladder. According to the Pythagorean theorem,
`\(x^2 + y^2 = \text{length of the ladder}^2\)`. We differentiate this equation with respect to time to relate the rates:
`\(2x (dx)/(dt) + 2y (dy)/(dt) = 2\text{length of the ladder} \frac{d\text{length of the ladder}}{dt}\)`.

Substituting the given values, when
`\(x = 8\) ft` and
`\((dx)/(dt) = 3\) ft/s,` and
`\(\text{length of the ladder} = 12\) ft`, we can solve for
`\((dy)/(dt)\)`. Solving the equation, we get
`\((dy)/(dt) = (-3x)/(y) = (-3 * 8)/(y)\)`. Given that y = 10 ft, substituting this value yields
`\((dy)/(dt) = -2\) ft/s`. The negative sign indicates that the top of the ladder is falling, and its magnitude is 2 ft/s. Therefore, the top of the ladder is falling at a rate of 2 ft/s at that instant.

The correct calculation results in
`\((dy)/(dt) = -2\) ft/s`.(Option d)