Question

A 5-meter ladder is leaning against the side of a house. The foot of the ladder is pulled away from the house at a rate of 0.4 m/sec. Determine how fast the top of the ladder is descending when the foot of the ladder is 3 meters from the house.

Answer 1

The top of the ladder is descending at 0.3 m/s.

Step-by-step explanation:

By Pythagoras theorem we know that

              Hypotenuse² = Base² + Perpendicular²

                   h² = b² + p²

We have for ladder

                        h = 5 m

                        b = 3 m

                        5² = 3² + p²

                        p = 4 m

                        $\frac{db}{dt}=0.4m/s\\\\\frac{dh}{dt}=0$

Differentiating h² = b² + p² with respect to time

                    $2h\times \frac{dh}{dt}=2b\times \frac{db}{dt}+2p\times \frac{dp}{dt}\\\\5\times 0=3\times 0.4+4\times \frac{dp}{dt}\\\\\frac{dp}{dt}=-0.3m/s$

The top of the ladder is descending at 0.3 m/s.