A ladder 13 meters long rests on horizontal ground and leans against a vertical wall. The foot of the ladder is pulled away from the wall at the rate of 0.4 m/sec. How fast is the top sliding down the wall when the foot of the ladder is 5 m from the wall?
To solve this problem, we can use the concept of related rates. Let's denote the distance between the foot of the ladder and the wall as x (measured in meters), and the height of the ladder on the wall as y (also measured in meters).
We are given that dx/dt = 0.4 m/sec (the rate at which x is changing).
First, let's establish a relationship between x and y using the Pythagorean theorem:
x^2 + y^2 = 13^2 (since the ladder is 13 meters long)
Differentiating both sides of this equation implicitly with respect to time (t):
2x(dx/dt) + 2y(dy/dt) = 0
We're interested in finding dy/dt (the rate at which y is changing) when x = 5m. We know dx/dt = 0.4 m/sec.
Plugging in the known values into the equation, we get:
2(5)(0.4) + 2y(dy/dt) = 0
10(0.4) + 2y(dy/dt) = 0
4 + 2y(dy/dt) = 0
2y(dy/dt) = -4
dy/dt = -4 / (2y)
Now, we need to determine the value of y when x = 5m. To find that, we can use the Pythagorean relationship we established earlier:
5^2 + y^2 = 13^2
25 + y^2 = 169
y^2 = 169 - 25
y^2 = 144
y = √144
y = 12m
Plugging in the values of y and dy/dt into our equation:
dy/dt = -4 / (2 * 12)
dy/dt = -4/24
dy/dt = -1/6
Therefore, the top of the ladder is sliding down the wall at a rate of -1/6 m/sec when the foot of the ladder is 5m from the wall.
To find the rate at which the top of the ladder is sliding down the wall, we need to use related rates. Let's assign some variables to the given information:
- Let x represent the distance between the foot of the ladder and the wall.
- Let y represent the distance between the top of the ladder and the ground.
- Let θ represent the angle between the ladder and the ground.
From the problem statement, we know that the ladder is 13 meters long, so y + x = 13.
We are given that dx/dt (the rate at which x is changing) is 0.4 m/sec. We want to find dy/dt (the rate at which y is changing) when x = 5 m.
To find dy/dt, we can differentiate both sides of the equation y + x = 13 with respect to time (t). This gives us:
dy/dt + dx/dt = 0
Since dx/dt = 0.4 m/sec, we can substitute this value into the equation:
dy/dt + 0.4 = 0
Solving for dy/dt:
dy/dt = -0.4 m/sec
Therefore, the top of the ladder is sliding down the wall at a rate of -0.4 m/sec when the foot of the ladder is 5 m from the wall. The negative sign indicates that the top of the ladder is moving downward.
If the base of the ladder is x feet from the wall, and the ladder reaches up y feet, then we have
x^2 + y^2 = 13^2
taking derivatives wrt t,
x dx/dt + y dy/dt = 0
Now just plug in your numbers and solve for dy/dt
(note that at the moment in question, you conveniently have a 5-12-13 right triangle.)