Question

. a ladder 10 meters long is leaning against a wall. the base of the ladder is sliding away from the wall at a rate of 2 m/s. find the rate at which the top of the ladder is sliding down the wall when the base of the ladder is 6 meters away from the wall.

Answer 1

decending at 1.5 m/s

Explanation:

Let:

y = height of the ladder's top

Given:

`(dx)/(dt) =2\ m/s`

`y=√(10^2-x^2)`

`=(100-x^2)^{(1)/(2)`

`(dy)/(dx) =(1)/(2) (100-x^2)^{(1-(1)/(2) )}*(-2x^((2-1)))`

`=(1)/(2) (100-x^2)^{-(1)/(2) }*(-2x)`

`=-\frac{x}{(100-x^2)^{(1)/(2)} }`

`=-(x)/(√(100-x^2) )`

`(dy)/(dt) =(dy)/(dx) * (dx)/(dt)`

`=-(x)/(√(100-x^2) )*2`

`=-(2x)/(√(100-x^2) )`

when x = 6 m:

`(dy)/(dt) =-(2(6))/(√(100-(6)^2) )`

`=-1.5\ m/s`

decending at 1.5 m/s