A 20 ft ladder is leaning against the side of a house. If the bottom of the ladder is moving away from the house at the rate of 2 feet per second, how fast is the top of the ladder moving down the wall when the bottom of the ladder is 12 feet from the house
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To solve this problem, we can use related rates, which involve finding the rate at which one quantity is changing with respect to another.
Let's start by labeling the important variables in this problem:
Let x represent the distance between the bottom of the ladder and the house (which is changing).
Let y represent the height of the ladder on the wall (which is changing).
Let z represent the length of the ladder (which is constant at 20 ft).
We are given that the rate at which the bottom of the ladder is moving away from the house is 2 ft/s (dx/dt = 2 ft/s).
We need to find the rate at which the top of the ladder is moving down the wall when x = 12 ft (dy/dt when x = 12 ft).
Now, we can relate these variables using the Pythagorean theorem:
x^2 + y^2 = z^2
Differentiating both sides of this equation with respect to time (t), we get:
(2x)(dx/dt) + (2y)(dy/dt) = 0
Since we're interested in finding dy/dt when x = 12 ft, we plug in these values into the equation:
(2*12)(dx/dt) + (2y)(dy/dt) = 0
Simplifying the equation further:
24(2) + 2y(dy/dt) = 0
48 + 2y(dy/dt) = 0
Now, we can solve for dy/dt:
2y(dy/dt) = -48
(dy/dt) = -48 / (2y)
To substitute the value of y, we can use the Pythagorean theorem with x = 12 ft:
(12)^2 + y^2 = (20)^2
144 + y^2 = 400
y^2 = 400 - 144
y^2 = 256
y = 16 ft
Substituting this value into the equation:
(dy/dt) = -48 / (2*16)
(dy/dt) = -48 / 32
(dy/dt) = -3/2 ft/s
Therefore, when the bottom of the ladder is 12 feet from the house, the top of the ladder is moving down the wall at a rate of -3/2 ft/s. The negative sign indicates that the top of the ladder is moving downward.