Question

The top of a ladder slides down a vertical wall at a rate of 2 feet per second. At the moment when the bottom of the ladder is 4 feet from the wall, it slides away from the wall at a rate of 3 feet per second. How long is the ladder

Answer 1

The length of the ladder is 7.21 feet

Explanation:

Let

y = Height of the wall

x = Distance between the wall and the ladder (bottom)

L = Length of the ladder

So, the given parameters are:

`(dy)/(dt) = -2ft/s`

When
`x = 4; (dx)/(dt) = 3ft/s`

The ladder, the wall and the ground forms a right-angled triangle where the hypotenuse is the length of the ladder; So:

`L^2 = x^2 + y^2`

Differentiate with respect to time

`2L(dL)/(dt) = 2x(dx)/(dt) + 2y(dy)/(dt)`

The length of the ladder does not change with time. So:

`2L*0 = 2x(dx)/(dt) + 2y(dy)/(dt)`

`0 = 2x(dx)/(dt) + 2y(dy)/(dt)`

Rewrite as:

`2x(dx)/(dt) =- 2y(dy)/(dt)`

Divide both sides by 2

`x(dx)/(dt) =- y(dy)/(dt)`

Recall that:

`(dy)/(dt) = -2ft/s` and
`x = 4; (dx)/(dt) = 3ft/s`

So:

`4 * 3 = -y * -2`

`12 = 2y`

`6 = y`

`y = 6`

Substitute
`y = 6` and
`x =4` in:
`L^2 = x^2 + y^2`

`L^2 =4^2 + 6^2`

`L^2 =52`

`L = \sqrt{52`

`L = 7.21ft`