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 concept of related rates. Related rates involve finding the rates at which two or more variables are changing with respect to each other.
a) To find how fast the ladder is moving away from the wall when the base of the ladder is 12 feet away, we need to relate the height of the ladder to the distance between the base of the ladder and the wall.
Let's denote the height of the ladder as h and the distance between the base of the ladder and the wall as x.
We know that the ladder is sliding down the wall at a constant rate of 2 feet per second, which means dh/dt = -2 (negative sign indicates the decrease in height).
We want to find dx/dt, the rate at which the distance between the base of the ladder and the wall is changing. To relate h and x, we can use the Pythagorean theorem:
h^2 + x^2 = 20^2
Differentiating both sides with respect to time (t) gives us:
2h (dh/dt) + 2x (dx/dt) = 0
Plugging in the values for dh/dt and x, we get:
2(12)(-2) + 2(12)(dx/dt) = 0
Simplifying the equation:
-48 + 24(dx/dt) = 0
24(dx/dt) = 48
dx/dt = 48/24 = 2 feet per second
Therefore, the ladder is moving away from the wall at a rate of 2 feet per second when the base of the ladder is 12 feet away.
b) To 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 away, we need to relate the height of the ladder to the angle.
Let's denote the angle between the ladder and the ground as θ.
Using trigonometry, we can write:
sin(θ) = h/20
Differentiating both sides with respect to time (t) gives us:
cos(θ) (dθ/dt) = (dh/dt)/20
Plugging in the values for dh/dt and h, we get:
cos(θ) (dθ/dt) = -2/20
Simplifying the equation:
cos(θ) (dθ/dt) = -1/10
We also know that the base of the ladder is 16 feet away, which means:
sin(θ) = x/16
Plugging in the value for x:
sin(θ) = 16/20 = 4/5
Using the Pythagorean identity, we can find cos(θ):
cos(θ) = √(1 - sin^2(θ)) = √(1 - (4/5)^2) = √(1 - 16/25) = √(9/25) = 3/5
Substituting cos(θ) and dθ/dt into the previous equation:
(3/5) (dθ/dt) = -1/10
Simplifying the equation:
(dθ/dt) = (-1/10) * (5/3) = -1/6 radians per second
Therefore, the angle between the ladder and the ground is changing at a rate of -1/6 radians per second when the base of the ladder is 16 feet from the wall.