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Question
A ladder 13 m long is leaning against a vertical wall. The bottom of the ladder is dragged away from the wall along the ground at the rate of 2 cm/sec. How fast is the height on the wall decreasing when the foot of the ladder is 5 m away from the wall?
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Solution
Let:
x = distance from the wall to the foot of the ladder (in meters)
y = height of the ladder on the wall (in meters)
The length of the ladder is constant: L = 13 m
Apply the Pythagorean theorem
x2 + y2 = L2
Substituting L = 13:
x2 + y2 = 132 = 169
Differentiate with respect to time
Differentiating both sides of the equation with respect to time t:
`d/(dt) (x^2) + d/(dt)(y^2) = d/(dt) (169)`
This gives us:
`2x(dx)/(dt) + 2y(dy)/(dt) = 0`
Solve for `(dy)/(dt)`
Rearranging the equation:
`2x(dx)/(dt) + 2y(dy)/(dt) = 0`
⟹ `x(dx)/(dt) + y(dy)/(dt) = 0`
Thus: `(dy)/(dt) = -x/y (dx)/(dt)`
Substitute known values
We know: `(dx)/(dt)` = 2 m/sec (the rate at which the foot of the ladder is moving away from the wall)
When x = 5 m, we need to find y.
Using the Pythagorean theorem to find y when x = 5:
52 + y2 = 169
⟹ 25 + y2 = 169
⟹ y2 = 144
⟹ y = 12 m
Now substitute x = 5, y = 12, and `(dx)/(dt) = 2` into the equation for `(dy)/(dt)`
`(dy)/(dt) = -5/12 * 2 = -10/12 = -5/6` m/sec
The negative sign indicates that the height is decreasing.
Therefore, the height on the wall is decreasing at a rate of `5/6` m/sec
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