Question

a 25 feet ladder leans against a vertical wall. the foot of the ladder is pulled away from the wall horizontally at the rate of 6 ft/sec. what is the rate, in feet/sec, at which the top of the ladder is sliding down the wall when the top of the ladder is 15 feet above the ground?

Answer 1

The rate at which the top of the ladder is sliding down the wall when it is 15 feet above the ground is approximately -6.8 ft/sec.

Step-by-step explanation:

To find the rate at which the top of the ladder is sliding down the wall, we can use related rates and trigonometry. Let the horizontal distance between the foot of the ladder and the wall be x, and let y be the height of the top of the ladder above the ground. Using the Pythagorean theorem, we have:

x^2 + y^2 = 25^2

Differentiating both sides with respect to time, we have:

2x(dx/dt) + 2y(dy/dt) = 0

Solving for dy/dt, we get:

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

Substituting the given values, x = 15 ft and dx/dt = 6 ft/sec, we can find y using the Pythagorean theorem:

15^2 + y^2 = 25^2

y^2 = 400 - 225 = 175

y = sqrt(175) ft ≈ 13.23 ft

Now we can substitute these values into the equation for dy/dt:

dy/dt = -15(6)/13.23 = -90/13.23 ft/sec ≈ -6.8 ft/sec

Therefore, the rate at which the top of the ladder is sliding down the wall when it is 15 feet above the ground is approximately -6.8 ft/sec.