Question

A ladder 10 metres long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at the rate of 1.2 metres per second, find how fast the top of the ladder is sliding down the wall, when the bottom is 6 metres away from the wall.

Answer 1

Let AB be the ladder, where AB = 10 meters. Let at time t seconds, the end A of the ladder be x metres from the wall and the end B be y metres from the ground.

Since, OAB is a right angled triangle, by Pythagoras theorem.

x² + y² = 10²

i.e. y² = 100 – x²

Differentiating w.r.t. t, we get

`2ydy/dt = 0 - 2xdx/dt`

∴ `dy/dt = x/y.dx/dt`              ...(1)

Now, `dx/dt = (12"metres")/sec` is the rate at wh the bottom of the ladder s pulled horizontally and `dy/dt` is the rate which the top of ladder B is sliding.

When x = 6, y² = 100 – 36 = 64

∴ y = 8

∴ (1) gives, `dy/dt = -(6)/(8)(1.2)`

= `(6)/(8) xx (12)/(10)`

= `-(9)/(10)`

= – 0.9

Hence, the top of the ladder is sliding down the wall, at the rate of `(0.9"metre")/sec`.