A pole 12 meters long is leaning against a vertical wall of a building. if the foot of the pole is being pulled away from the wall at the rate of 2 meter per minute, how fast is the top of the pole sliding down the wall at the instant when the foot of the pole is 5 meters from the wall?
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To solve this problem, we can use related rates and trigonometry. Let's define the following variables:
- x: the distance between the foot of the pole and the wall (changing over time)
- y: the height of the pole where it touches the wall (changing over time)
- h: the length of the pole (fixed at 12 meters)
We want to find how fast the top of the pole (y) is sliding down the wall when the foot of the pole (x) is 5 meters from the wall.
Initially, we have a right triangle formed by the pole, the wall, and the ground, where the hypotenuse is the pole.It's important to note that this right triangle will remain similar as the pole moves.
Using the Pythagorean Theorem, we can relate x, y, and h:
x^2 + y^2 = h^2
Differentiating with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since dx/dt = -2 (the foot of the pole is being pulled away from the wall at a rate of 2 m/min), and we need to find dy/dt when x = 5, we can substitute these values into the equation:
2(5)(-2) + 2y(dy/dt) = 0
Simplifying, we get:
-20 + 2y(dy/dt) = 0
To find dy/dt, we can solve for it:
2y(dy/dt) = 20
dy/dt = 20 / (2y)
Now, we need to find y when x = 5. To do this, we use the equation x^2 + y^2 = h^2:
(5)^2 + y^2 = (12)^2
25 + y^2 = 144
y^2 = 144 - 25
y^2 = 119
y ≈ √119
Substituting this value into the expression for dy/dt, we have:
dy/dt = 20 / (2√119)
dy/dt ≈ 10 / √119
Therefore, the top of the pole is sliding down the wall at a rate of approximately 10 / √119 meters per minute, when the foot of the pole is 5 meters from the wall.
To find the rate at which the top of the pole is sliding down the wall, we can use related rates.
Let's label the distance between the foot of the pole and the wall as x (in meters) and the height of the pole as y (in meters).
Given:
dx/dt = 2 m/min (the rate at which the foot of the pole is being pulled away from the wall)
We need to find dy/dt (the rate at which the top of the pole is sliding down the wall).
We also know the length of the pole (12 meters).
To solve this problem, we can use the Pythagorean theorem. According to the theorem, the length of the pole is equal to the square root of the sum of the squares of its height and the distance between the foot of the pole and the wall:
x^2 + y^2 = 12^2
Differentiating both sides of the equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
We know dx/dt = 2, and we want to find dy/dt when x = 5.
Substituting the values into the equation, we have:
2(5)(2) + 2y(dy/dt) = 0
10 + 2y(dy/dt) = 0
2y(dy/dt) = -10
dy/dt = -10 / (2y)
Now, we can determine the value of y using the Pythagorean theorem:
x^2 + y^2 = 12^2
5^2 + y^2 = 144
25 + y^2 = 144
y^2 = 144 - 25
y^2 = 119
Taking the square root of both sides, we have y = √119 (approximately 10.92) meters.
Substituting this value back into our equation for dy/dt, we get:
dy/dt = -10 / (2 * 10.92)
dy/dt ≈ -0.457 meters per minute
Therefore, at the instant when the foot of the pole is 5 meters from the wall, the top of the pole is sliding down the wall at a rate of approximately 0.457 meters per minute.