A 20 foot ladder is sliding down a vertical wall at a constant rate of 2 feet per second.
a) How fast is the ladder moving away from the wall when the base of the ladder is 12 feet away from the wall?
b) Find the rate of change at which the angle between the ladder and the ground is changing when the base of the ladder is 16 feet from the wall.
To solve these problems, we can use the principles of trigonometry and related rates. Let's break down each question step by step.
a) To find the rate at which the ladder is moving away from the wall, we need to determine how the distance between the ladder and the wall is changing with respect to time. Let's denote this distance as "x" and the time as "t". We are given that dx/dt (the rate at which x is changing with respect to t) is 2 ft/s.
We can use the Pythagorean theorem to relate the distance x, the height h (which is fixed at 20 feet), and the length of the ladder (hypotenuse) L.
The Pythagorean theorem states: L^2 = x^2 + h^2
Differentiating both sides of the equation with respect to time (t), we get:
2L(dL/dt) = 2x(dx/dt)
Now we can substitute the given values:
2(20)(dL/dt) = 2(12)(2)
Simplifying the equation:
40(dL/dt) = 48
Finally, solving for (dL/dt), we find:
dL/dt = 48/40 = 6/5 ft/s
Therefore, the ladder is moving away from the wall at a rate of 6/5 ft/s when the base of the ladder is 12 feet away from the wall.
b) To find the rate of change of the angle between the ladder and the ground, we need to determine how the angle θ (theta) is changing with respect to time. Here, we are given that dθ/dt represents the rate of change of the angle.
Using trigonometry, we can relate θ to x and h:
tan(θ) = x/h
Differentiating both sides of the equation with respect to time (t), we get:
sec^2(θ)(dθ/dt) = (dx/dt)/h
Let's substitute the given values:
sec^2(θ)(dθ/dt) = (2/20)
Since sec^2(θ) is equal to 1 + tan^2(θ), we can rewrite the equation as:
(1 + tan^2(θ))(dθ/dt) = (2/20)
Simplifying the equation:
(dθ/dt) + tan^2(θ)(dθ/dt) = 1/10
The equation may look complicated, but we can use a trigonometric identity to simplify it:
sec^2(θ) = 1 + tan^2(θ)
Therefore, we have:
(dθ/dt) + sec^2(θ)(dθ/dt) = 1/10
Now we'll work with the given information. When the base of the ladder is 16 feet away from the wall, we can determine the corresponding value of θ using the right triangle formed by x, h, and L.
Using the Pythagorean theorem once again, we have:
L^2 = x^2 + h^2
L^2 = 16^2 + 20^2
L^2 = 256 + 400
L^2 = 656
L = sqrt(656) ≈ 25.61 ft
Now, we can use this value of L to determine the value of tan(θ):
tan(θ) = x/h = 16/20 = 4/5
Therefore, tan^2(θ) = (4/5)^2 = 16/25
Substituting these values into the equation:
(dθ/dt) + (16/25)(dθ/dt) = 1/10
Simplifying further:
(41/25)(dθ/dt) = 1/10
Solving for (dθ/dt):
dθ/dt = (1/10) / (41/25) = 5/410 rad/s
Hence, the rate of change at which the angle between the ladder and the ground is changing is 5/410 rad/s when the base of the ladder is 16 feet from the wall.