A ladder 20 ft long has one end on the ground and the other end in contact with a vertical wall. The lower end slips along the ground. If the lower end of the ladder is 16 t away from the wall, upper end is moving `lamda` time as fast as the lower end, then `lamda` is

Options

A. `1/3`
B. `2/3`
C. `4/3`
D. `5/3`

Answer

C

Explanation

Let OC be the wall and AB be the position of the ladder at any time `t` such that `OA=x` and `OB=y`.
Length of the ladder `AB=20ft`

Then `x^(2)+y^(2)=(20)^(2)`
`impliesx^(2)+y^(2)=400`……………i
`implies2x(dx)/(dt)+2y(dy)/(dt)=0`
`:.(dy)/(dt)=(-x)/y(dx)/(dt)=(-x)/(sqrt(400-x^(2))) . (dx)/(dt)` [From eq i]
At `x=16 ft, (dy)/(dt)`
`=(-16)/(sqrt(400-254)) . (dx)/(dt)=-4/3 (dx)/(dt)`
-ve sign indicates, then when `x` increases with time `y` decreases.
Hence the upper is moving `4/3` times as fast as the lower end.