Question

A 13 m long ladder is leaning against a wall, touching the wall at a certain height from the ground level. The bottom of the ladder is pulled away from the wall, along the ground, at the rate of 2 m/s. How fast is the height on the wall decreasing when the foot of the ladder is 5 m away from the wall?

Answer 1

`x^2 + y^2 = 169`         ...(i)

The bottom of the ladder is beign pulled, so these distances are changing with time,
`2x  dx/dt + 2y  dy/dt =0`    ...(ii)

`dy/dt = - x/y dx/dt`     .....(iii)

at  x=5m then from (i) y=12m 
From equation(iii)
`dy/dt = -5/12xx2`

`dy/dt = - 5/6 m//s`

– sign indicates the height is decreasing.