A 26 foot ladder is placed against a wall. If the top of the ladder is sliding down the wall at 2 feet per second, at what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall?
Let's use the Pythagorean theorem to solve this problem.
According to the Pythagorean theorem, in a right triangle, the square of the hypotenuse (the longest side, which is the ladder in this case) is equal to the sum of the squares of the other two sides.
In this case, let's call the distance between the ladder's bottom and the wall as x. This means that the height (or distance covered by the ladder along the wall) would be 10 - x.
Now we can apply the Pythagorean theorem:
x^2 + (10 - x)^2 = 26^2
Expanding and simplifying, we get:
x^2 + 100 - 20x + x^2 = 676
2x^2 - 20x + 100 - 676 = 0
2x^2 - 20x - 576 = 0
Dividing the equation by 2:
x^2 - 10x - 288 = 0
Using the quadratic formula to solve for x, we get:
x = (-(-10) ± √((-10)^2 - 4(1)(288))) / (2(1))
x = (10 ± √(100 + 1152)) / 2
x = (10 ± √(1252)) / 2
x = (10 ± √(4 * 313)) / 2
x = (10 ± 2√(313)) / 2
x = 5 ± √(313)
Since we are given that the bottom of the ladder is 10 feet away from the wall, x = 5 + √(313).
To find the rate at which the bottom of the ladder is moving away from the wall, let's differentiate the equation with respect to time (t):
x = 5 + √(313)
Taking the derivative:
(dx/dt) = d/dt (5 + √(313))
(dx/dt) = 0 + 0.5 * (313)^(-0.5) * (d/dt(313))
Since the derivative of a constant is 0, we can ignore the first term:
(dx/dt) = 0.5 * (313)^(-0.5) * 0
(dx/dt) = 0
Therefore, the rate at which the bottom of the ladder is moving away from the wall when the bottom of the ladder is 10 feet away from the wall is 0 feet per second.