Question

a ladder 5 meters long is leaning against a wall. if the foot of the ladder is being pulled away from the wall at a rate of 0.3m/min, how far is the foot of the ladder from the wall if it is moving as fast as the top of the ladder is dropping.

Answer 1

Answer:

The foot of the ladder is at 3.53 meters from the wall.

Step-by-step explanation:

Let the foot of the ladder is at x meters from the wall and it's height is y meters.

Rate of pulling away of foot = dx/dt

Rate of dropping of the top = - dy/dt  (-ve sign because y is decreasing)

From the given conditions :

dx/dt = -dy/dt

From the Pythagoras theorem we have

x² + y² = 5²

differentiating on both the sides,

2x. dx/dt + 2y. dy/dt = 0

=> 2x. dx/dt =  -2y. dy/dt

but dx/dt = -dy/dt,

hence

=> x = y

Again from the Pythagoras theorem

x² + y² = 5²

=> x² + x² = 5²

=> 2x² = 25

=> x = √12.5 = 3.53 m

Hence the foot of the ladder is at 3.53 meters from the wall.