Question

A 10 foot ladder is leaning against a wall. Call x the distance from the top of the ladder to the ground, and call y the distance from the wall to the foot of the ladder. At the instant that the foot of the ladder is 6 feet away from the wall, the foot of the ladder is moving away from the wall at a rate of 4 ft/sec. At what rate is the top of the ladder falling down the wall at this time (in feet/sec)

Answer 1

$-5.3\:\text{ft/s}$

Step-by-step explanation:

We start by applying the Pythagorean theorem to the ladder, with its length L as the hypotenuse:

$L^2 = 100\:\text{ft}^2 = x^2 + y^2$

where x is the vertical distance from the top of the ladder to the ground and y is the horizontal distance from the bottom of the ladder to the wall. Taking the derivative of the above expression with respect to time, we get

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

Solving for dx/dt, we get

$\dfrac{dx}{dt} = -\left(\dfrac{y}{x}\right)\dfrac{dy}{dt} = -\left(\dfrac{\sqrt{L^2 - x^2}}{x}\right)\dfrac{dy}{dt}$

We know that

$\dfrac{dy}{dt} = 4\:\text{ft/s}$

when x = 6 ft. So the rate at which the top of the ladder is going down is

$\dfrac{dx}{dt} = -\left(\dfrac{\sqrt{100\:\text{ft}^2 - (6\:\text{ft})^2}}{6\:\text{ft}}\right)(4\:\text{ft/s})$

$\:\:\:\:\:\:\:= -5.3\:\text{ft/s}$

The negative sign means that the distance x is decreasing as y is increasing.