What ft/s the latter is sliding down the wall

Question

 A 17 foot ladder is leaning against a wall. If the top slips down the wall at a rate of 4 ft/s, how fast will the foot be moving away from the wall when the top is 12 feet above the ground?

Answer 1

Equation needed: Pythagorean Theorem (x²+y²=c²). We can take x as the length of the bottom of the latter to the wall in feet, y as the height of the floor to the top of the ladder that is touching the wall, and c as the length of the latter itself.

We are given the following values:

c = 17 feet

dy/dt = -4ft/sec (it is negative as the latter is sliding down the wall)

y = 12 feet

dc/dt = 0 feet/sec (as the ladder does not expand nor contract)

We are asked to find dx/dt. When y is 12 feet, we need to first find the value of x. By using Pythagorean Theorem, we get: x = sqrt(145) which is approximately 12.04 feet from the wall.

To find dx/dt, we take the derivative of the Pythagorean Theorem equation: 2x(dx/dt) + 2y(dy/dt) = 2c(dc/dt)

Solving for dx/dt we get:

dx/dt = [c(dc/dt) - y(dy/dt)] / x.

Plugging in the known values we get:

dx/dt = [48*sqrt(145)] / 145 which is approximately 3.99 feet/sec. <- (Answer)