A ladder 5 m long rests against a vertical wall. If the bottom of the ladder slides away from
the wall at a rate of 1 m/s, how fast is the top of the ladder sliding down the wall when the
bottom of the ladder is 3 m from the wall?
Let's denote the distance from the bottom of the ladder to the wall as x and the distance from the top of the ladder to the wall as y.
From the Pythagorean theorem, we have:
x^2 + y^2 = 5^2
Differentiating both sides with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
=> x(dx/dt) = -y(dy/dt)
Given that dx/dt = 1 m/s and x = 3 m, we need to find dy/dt when x = 3 m.
From the Pythagorean theorem:
3^2 + y^2 = 5^2
=> y = 4 m
Now substitute the values of x, y, and dx/dt into the equation:
3(1) = -4(dy/dt)
=> dy/dt = -3/4 m/s
Therefore, the top of the ladder is sliding down the wall at a rate of 3/4 m/s.