Question

A 50 ft ladder is placed against a building. The top of the ladder is sliding down the building at the rate of 2 ft per minute. Find the rate at Which the BASE of the ladder is slipping away from the building at the instant that the base is 30 ft from the building

Answer 1

At the moment when the ladder's base is 30 ft from the building, it's slipping away from the building at a rate of
`\( (3)/(2) \)` feet per minute.

Let's denote:

- \(x\) as the distance between the base of the ladder and the building.

- \(y\) as the height of the ladder on the building.

Given the ladder's length, \(y = 50\) ft, and it's sliding down the building at a rate of \(2\) ft per minute.

We're interested in finding the rate at which the base of the ladder (\(x\)) is slipping away from the building when \(x = 30\) ft.

Using the Pythagorean theorem for the relation between \(x\) and \(y\):

`\[x^2 + y^2 = 50^2\]`

`\[y = √(50^2 - x^2)\]`

Differentiating both sides of the equation with respect to time (t):

`\[(dy)/(dt) = (d)/(dt)\left(√(50^2 - x^2)\right)\]`

Now, differentiate with the chain rule:

`\[(dy)/(dt) = (d)/(dx)\left(√(50^2 - x^2)\right) \cdot (dx)/(dt)\]`

When \(x = 30\):

`\[y = √(50^2 - 30^2) = 40\text{ ft}\]`

Differentiating
`\(√(50^2 - x^2)\)`:

`\[(dy)/(dt) = (-x)/(√(50^2 - x^2)) \cdot (dx)/(dt)\]`

At
`\(x = 30\):`

`\[(dy)/(dt) = (-30)/(40) \cdot 2 = -(3)/(4) \cdot 2 = -(3)/(2)\text{ ft/min}\]`

Therefore, when the base is 30 ft from the building, it is slipping away from the building at a rate of
`\((3)/(2)\)` ft/min.