Final answer:
When the base of the ladder is 8 feet away from the wall, the top of the ladder is sliding down at approximately 2.67 feet per second. The closest rounded value to this answer is 2 feet per second.
Step-by-step explanation:
The student asks how quickly the top of a 10-foot ladder that leans against a wall is sliding down when the base is moving away horizontally at 2 feet per second and is 8 feet away from the wall. This is a problem involving related rates, which is a type of calculus problem. To solve it, we need to apply Pythagoras' theorem to the ladder, wall, and ground, considering them as a right-angled triangle.
Let x be the distance from the wall to the base of the ladder, and let y be the distance from the ground to the top of the ladder. Using Pythagoras' theorem:
x^2 + y^2 = 100 (since the ladder has a length of 10 feet, and 10 squared is 100).
Let's differentiate both sides of the equation with respect to time t:
2x(dx/dt) + 2y(dy/dt) = 0
Since we know that dx/dt is 2 ft/s (the speed at which the base of the ladder moves away from the wall), we can solve the equation for dy/dt when x is 8 feet:
16(dx/dt) + 2y(dy/dt) = 0
Since x is 8 feet, we can calculate y using Pythagoras' theorem:
y^2 = 100 - 64 = 36
y = 6 feet
Now plug x, dx/dt, and y back into the differentiated equation to solve for dy/dt:
16(2) + 2(6)(dy/dt) = 0
32 + 12(dy/dt) = 0
12(dy/dt) = -32
dy/dt = -32 / 12
dy/dt = -8/3 or approximately -2.67 ft/s
Therefore, the top of the ladder is sliding down the wall at an approximate rate of 2.67 feet per second when the base is 8 feet from the wall. This speed is not one of the options provided, so assuming that these choices are rounded, the closest answer would be Option B) 2 ft/s.