Question

A 20-foot ladder is leaning against the wall. If the base of the ladder is sliding away from the wall at the rate of 3 feet per second, find the rate at which the top of the ladder is sliding down when the top of the ladder is 8 feet from the ground.

Answer 1

6.87 ft/s is the rate at which the top of ladder slides down.

Step-by-step explanation:

Given:

Length of the ladder is,
`L=20\ ft`

Let the top of ladder be at height of 'h' and the bottom of the ladder be at a distance of 'b' from the wall.

Now, from triangle ABC,

AB² + BC² = AC²

`h^2+b^2=L^2\\h^2+b^2=20^2\\h^2+b^2=400----1`

Differentiating the above equation with respect to time, 't'. This gives,

`(d)/(dt)(h^2+b^2)=(d)/(dt)(400)\\\\(d)/(dt)(h^2)+(d)/(dt)(b^2)=0\\\\2h(dh)/(dt)+2b(db)/(dt)=0\\\\h(dh)/(dt)+b(db)/(dt)=0--------2`

In the above equation the term
`(dh)/(dt)` is the rate at which top of ladder slides down and
`(db)/(dt)` is the rate at which bottom of ladder slides away.

Now, as per question,
`h=8\ ft, (db)/(dt)=3\ ft/s`

Plug in
`h=8` in equation (1) and solve for
`b`. This gives,

`8^2+b^2=400\\64+b^2=400\\b^2=400-64\\b^2=336\\b=√(336)=18.33\ ft`

Now, plug in all the given values in equation (2) and solve for
`(dh)/(dt)`

`8* (dh)/(dt)+18.33* 3=0\\8* (dh)/(dt)+54.99=0\\8* (dh)/(dt)=-54.99\\ (dh)/(dt)=-(54.99)/(8)=-6.87\ ft/s`

Therefore, the rate at which the top of ladder slide down is 6.87 ft/s. The negative sign implies that the height is reducing with time which is true because it is sliding down.