A ladder is initially 5 meters long and is leaning against a wall. The bottom of the ladder is being pulled away from the wall at a rate of 2 meters per second. How fast is the angle between the ladder and the ground changing when the bottom is 4 meters from the wall?
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To solve this problem, we can start by drawing a diagram:
Let's call the distance from the bottom of the ladder to the wall x and the angle between the ladder and the ground θ.
Since we're given that the bottom of the ladder is being pulled away from the wall at a rate of 2 meters per second, we know that dx/dt = 2.
We want to find dθ/dt, the rate at which the angle between the ladder and the ground is changing.
Using trigonometry, we can relate x and θ. In a right triangle formed by the ladder, the wall, and the ground, the sine of θ is equal to the opposite side (x) divided by the hypotenuse (the length of the ladder). So, sin(θ) = x/5.
To find dθ/dt, we can take the derivative of both sides of this equation with respect to time:
d/dt(sin(θ)) = d/dt(x/5)
cos(θ) * dθ/dt = (1/5) * dx/dt
We know that dx/dt = 2, so we can substitute that into the equation:
cos(θ) * dθ/dt = (1/5) * 2
cos(θ) * dθ/dt = 2/5
cos(θ) = √(1 - sin^2(θ)) = √(1 - (x/5)^2)
Now we can substitute that into the equation:
√(1 - (x/5)^2) * dθ/dt = 2/5
To find dθ/dt when x = 4, we can substitute that into the equation:
√(1 - (4/5)^2) * dθ/dt = 2/5
Simplifying, we have:
√(1 - 16/25) * dθ/dt = 2/5
√(9/25) * dθ/dt = 2/5
(3/5) * dθ/dt = 2/5
Now we can solve for dθ/dt:
dθ/dt = (2/5) * (5/3)
dθ/dt = 2/3
Therefore, the angle between the ladder and the ground is changing at a rate of 2/3 radians per second when the bottom of the ladder is 4 meters from the wall.
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