A 10 foot ladder is placed against a vertical wall. Suppose the bottom of the ladder slides away from the wall at a constant rate of 4 feet per second. How fast is the top of the ladder sliding down the wall when the bottom is 8 feet from the wall
To solve this problem, we can use the Pythagorean theorem:
a^2 + b^2 = c^2
where:
a = distance between the bottom of the ladder and the wall
b = distance between the top of the ladder and the ground
c = length of the ladder
In this case, a = 8ft, b = unknown, and c = 10ft.
We can differentiate both sides of the equation with respect to time:
(d/dt)(a^2) + (d/dt)(b^2) = (d/dt)(c^2)
Since a = 8ft and it is sliding away from the wall at a constant rate of 4ft/s, we can write (d/dt)(a^2) = 2a * (d/dt)(a) = 2 * 8ft * (-4ft/s) = -64ft^2/s.
Since b and c are fixed lengths and do not change with time, (d/dt)(b^2) = 0 and (d/dt)(c^2) = 0.
Therefore, we have:
-64ft^2/s + 0 = 0
Now, we can solve for (d/dt)(b):
(d/dt)(b) = -64ft^2/s / (2a)
(d/dt)(b) = -64ft^2/s / (2 * 8ft)
(d/dt)(b) = -4ft/s
Therefore, the top of the ladder is sliding down the wall at a constant rate of 4ft/s.