Calculus Math Problem

Question

A ladder 13 meters long rests on horizontal ground and leans against a vertical wall. The foot of the ladder is pulled away from the wall at the rate of 0.5 m/sec. How fast is the top sliding down the wall when the foot of the ladder is 5 m from the wall?

Answer 1

Let x = distance of the bottom of the ladder from the wall

y = distance from the top of the ladder to the ground

Then, x^2 + y^2 = 169

Given: dx/dt = 0.5

Find: dy/dt when x = 5

Solution: Since x^2 + y^2 = 169, 2x(dx/dt) + 2y(dy/dt) = 0

dy/dt = -x(dx/dt) / y

= -5(0.5) / y = -2.5 / y

Since x^2 + y^2 = 169 and x = 5, y = 12.

So, dy/dt = -2.5 / 12

The ladder is sliding down the wall at the rate of 5/24 m / sec.