A 10 foot ladder is placed against a vertical all. Suppose the bottom of the ladder slides away from the wall at a constant rate of 4ft per second. How fast is the top of the ladder sliding down the wall when the bottom is 8 feet from the wall?
We can use related rates to solve this problem. Let's label the distance between the bottom of the ladder and the wall as x and the distance between the top of the ladder and the ground as y.
We are given that dx/dt = -4 ft/s (negative because x is getting smaller) and we need to find dy/dt.
Using the Pythagorean theorem, we know that x^2 + y^2 = 10^2.
Differentiating both sides of this equation with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0.
Substituting the given values, we have:
2(8)(-4) + 2y(dy/dt) = 0.
-64 + 2y(dy/dt) = 0.
Simplifying, we have:
2y(dy/dt) = 64.
dy/dt = 64/(2y).
To find y, we can use the Pythagorean theorem:
x^2 + y^2 = 10^2.
8^2 + y^2 = 10^2.
64 + y^2 = 100.
y^2 = 100 - 64.
y^2 = 36.
y = 6.
Substituting this value back into dy/dt = 64/(2y), we have:
dy/dt = 64/(2(6)).
dy/dt = 64/12.
dy/dt = 16/3 ft/s.
Therefore, when the bottom of the ladder is 8 feet from the wall, the top of the ladder is sliding down the wall at a rate of 16/3 ft/s.