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A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.7 ft/s, how fast (in rad/s) is the angle (in radians) between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 6 ft from the wall.)

User Uhura
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To find the rate at which the angle between the ladder and the ground is changing, we can use trigonometry and related rates.
Let's denote the angle between the ladder and the ground as e (in radians). We want to find de/dt, the rate at which e is changing.
We are given that the ladder is 10 ft long and is sliding away from the wall at a rate of 0.7 ft/s. We want to find de/dt when the bottom of the ladder is 6 ft from the wall.
Using trigonometry, we know that the length of the ladder (10 ft) is the hypotenuse of a right triangle, and the distance of the bottom of the ladder from the wall (6 ft) is the adjacent side. The opposite side represents the height, which we can find using the Pythagorean theorem: height^2 + adjacent^2 = hypotenuse ^2
height^2 + 6^2 = 10^2
height^2 = 100 - 36
height^2 = 64
height = 8 ft
Now, we can differentiate the equation with respect to time:
2(height)(dheight/dt) + 0 = 0
2(8 ft)(dheight/dt) = 0
dheight/dt = 0
Since the height is not changing
(dheight/dt = 0), the angle e is not
changing either (de/dt = 0).
Therefore, the rate at which the angle is changing is 0 rad/s when the bottom of the ladder is 6 ft from the wall.
In conclusion, when the bottom of the ladder is 6 ft from the wall, the angle between the ladder and the ground is not changing.
User JB Hurteaux
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