An 18 foot ladder is leaning against a building. if the bottom of the ladder is sliding along the pavement directly away from the building at 3 feet/second, how fast is the top of the ladder moving down when the foot of the ladder is 2 feet from the wall?
We can use the Pythagorean theorem to solve this problem. Let x be the distance between the foot of the ladder and the wall, and let y be the height of the ladder.
We have the equation:
x^2 + y^2 = 18^2
Differentiating with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
We are given that dx/dt = 3 ft/s, and we want to find dy/dt when x = 2 ft.
Plugging in the values, we have:
2(2)(3) + 2y(dy/dt) = 0
12 + 2y(dy/dt) = 0
2y(dy/dt) = -12
dy/dt = -6/y
To find y when x = 2, we can substitute into the Pythagorean theorem equation:
2^2 + y^2 = 18^2
4 + y^2 = 324
y^2 = 320
y = √320 = 4√10
Now, we can find dy/dt when y = 4√10:
dy/dt = -6/(4√10) = -3/(2√10) = -3√10/20 = -3√10/20 ft/s
Therefore, the top of the ladder is moving down at a rate of -3√10/20 ft/s when the foot of the ladder is 2 feet from the wall.