Question

A ladder 10 feet long rests against a vertical wall. Initially, the top of the ladder is 8 feet above the ground. If the top of the ladder is sliding down the wall at a rate of 2 feet per second, how fast is the bottom of the ladder sliding away from the wall 1 second after the ladder starts sliding? Answer the question in a complete sentence and include appropriate units

Answer 1

Step-by-step explanation:

Let vertical height of ladder from ground be y and

horizontal distance of the base of the ladder from the wall be x respectively.

Length of the ladder = l (constant) = 10 ft

Using Pythagoras theorem:

${l}^{2} = {y}^{2} + {x}^{2}$

Differentiate both sides w.r.t time

$0 = 2y \frac{dy}{dt} + 2x \frac{dx}{dt}$

$y \frac{dy}{dt} + x \frac{dx}{dt} = 0$

We know that (After 1 sec, y = 6 ft and x = 8 ft ; dy/dt = 2 ft/sec)

$6\times 2 + 8\frac{dx}{dt} = 0$

$\frac{dx}{dt} = - 1.5 ft \: per \: sec$

( Ignore - ive sign)

Therefore, bottom of the ladder is sliding away from the wall at a speed of 1.5 ft/sec one second after the ladder starts sliding.