A 25 foot ladder is placed against a vertical wall. Suppose the bottom of the ladder slides away from the wall at a constant rate of 3 feet per second. How fast is the top of the ladder sliding down the wall when the bottom is 7 feet from the wall
We have a right triangle formed by the ladder, the wall, and the ground.
Let x be the distance between the top of the ladder and the ground.
We can use the Pythagorean theorem to express the relationship between the distances:
x^2 + 7^2 = 25^2.
Simplifying this equation, we get:
x^2 = 625 - 49,
x^2 = 576,
x = 24.
Let y be the distance between the top of the ladder and the wall (height on the wall).
Thus, we have another right triangle formed by the ladder, the wall, and the height.
Using similar triangles, we can equate the ratios of corresponding sides:
7/24 = y/25.
Multiplying both sides by 25, we get:
y = (7/24)*25,
y = 7.29 feet.
We want to find how fast the top of the ladder (y) is sliding down the wall when the bottom is 7 feet from the wall.
Let's differentiate the equation with respect to time:
d(y)/dt = d((7/24)*25)/dt.
Taking constants outside the derivative, we get:
d(y)/dt = (7/24)*d(25)/dt.
Since the bottom is sliding away from the wall at a constant rate of 3 feet/second, the derivative of 25 with respect to time is 3:
d(y)/dt = (7/24)*3,
d(y)/dt = 21/8 feet/second.
Therefore, the top of the ladder is sliding down the wall at a rate of 21/8 feet/second when the bottom is 7 feet from the wall.