A ladder 10 meters long leans against a wall. The bottom of the ladder slides away from the wall at a rate of 2 meters per second. How fast is the top of the ladder sliding down the wall when the bottom is 6 meters from the wall?
Guide On Rating System
Vote
To solve this problem, we can use the Pythagorean theorem. The ladder, the wall, and the distance the bottom of the ladder slides from the wall form a right triangle.
Let's assume that the distance the top of the ladder slides down the wall is y meters per second.
According to the Pythagorean theorem, the distance from the wall to the top of the ladder is given by:
x^2 + y^2 = 10^2
Since the bottom of the ladder is sliding away from the wall at a rate of 2 meters per second, the horizontal distance from the wall to the bottom of the ladder, denoted by x, is changing over time. We are given that dx/dt = 2 m/s when x = 6 m.
Differentiating both sides of the equation with respect to time, we get:
2x(dx/dt) + 2yy' = 0
Substituting the given values, x = 6 m and dx/dt = 2 m/s, we have:
2(6)(2) + 2(6)y' = 0
12 + 12y' = 0
Dividing by 12, we get:
y' = -1
Thus, the top of the ladder is sliding down the wall at a rate of 1 meter per second when the bottom is 6 meters from the wall.
Got a question with my answer?
Cheapmath.com is completely free!
Asking question is free.
Getting answers is free.
+99