A ladder 8 meters long is leaning against a wall. If the top of the ladder is slipping down at the rate of 8 meters per second, how fast is the bottom moving away from the wall when it is 7 meters from the wall?
Please give your answer in , enter it without the units.
When the ladder base is 7m from the wall, the top is √(64-49) = √15 from the ground.
When the base is x meters from the wall,
x^2 + h^2 = 64
2x dx/dt + 2h dh/dt = 0
2(7)dx/dt + 2√15 (-8) = 0
dx/dt = 16√15/14 = 4.43m/s
To solve this problem, we can use the concept of related rates. Let's denote the distance between the bottom of the ladder and the wall as x (in meters) and the distance between the top of the ladder and the ground as y (in meters).
We know that the ladder is 8 meters long, so we have the equation:
x^2 + y^2 = 8^2
Differentiating both sides of this equation with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
We are given that dy/dt = -8 (since the top of the ladder is slipping down at a rate of 8 meters per second). We need to find dx/dt when x = 7.
Substituting the values into the equation, we have:
2(7)(dx/dt) + 2y(-8) = 0
Simplifying the equation, we get:
14(dx/dt) - 16y = 0
To solve for dx/dt, we need to find the value of y when x = 7. To do this, we can substitute x = 7 into the equation x^2 + y^2 = 8^2 and solve for y.
7^2 + y^2 = 8^2
49 + y^2 = 64
y^2 = 64 - 49
y^2 = 15
y = √15
Now we can substitute the values of y and x into the equation 14(dx/dt) - 16y = 0 to find dx/dt:
14(dx/dt) - 16√15 = 0
14(dx/dt) = 16√15
dx/dt = 16√15 / 14
Simplifying further, we get:
dx/dt = 8√15 / 7
Therefore, the speed at which the bottom of the ladder is moving away from the wall when it is 7 meters from the wall is 8√15 / 7, without the units.
To solve this problem, we can use related rates. Let's define the variables:
h = height of the ladder
x = distance from the wall to the bottom of the ladder
We are given that the top of the ladder is slipping down at a rate of 8 meters per second, so dh/dt = -8 m/s since the height is decreasing. We need to find dx/dt, the rate at which the bottom is moving away from the wall when it is 7 meters from the wall.
Using the Pythagorean theorem, we have the equation: x^2 + h^2 = 8^2 (since the ladder is 8 meters long).
Differentiating both sides of the equation with respect to time t, we get:
2x(dx/dt) + 2h(dh/dt) = 0
We want to find dx/dt when x = 7 and dh/dt = -8. Plugging these values into the equation above, we have:
2(7)(dx/dt) + 2h(-8) = 0
14(dx/dt) - 16h = 0
We need to find h when x = 7. Using the Pythagorean theorem again:
7^2 + h^2 = 8^2
49 + h^2 = 64
h^2 = 15
h = √15
Plugging h = √15 into the equation above, we get:
14(dx/dt) - 16√15 = 0
Now we can solve for dx/dt:
14(dx/dt) = 16√15
dx/dt = (16√15)/14
Simplifying the expression, we get:
dx/dt = 8√15/7
Therefore, the bottom of the ladder is moving away from the wall at a rate of 8√15/7 meters per second when it is 7 meters from the wall.