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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.
Solution
Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x m away from the wall.
Then, by Pythagoras theorem, we have:
x2 + y2 = 169 ......................[Length of the ladder = 13 m]
⇒ `"y" = sqrt(169-"x"^2)`
Then, the rate of change of height (y) with respect to time (t) is given by,
`"dy"/"dx" = (-"x")/sqrt(169-"x"^2) "dx"/"dt"`
It is given that `"dx"/"dt" = 2 "cm"//"s".`
`"dy"/"dt" = (-2"x")/sqrt(169-"x"^2)`
Now, when x = 5 m, we have:
`"dy"/"dt" = (-2xx5)/sqrt(169-5^2) = (-10)/sqrt144 = -5/6`
Hence, the height of the ladder on the wall is decreasing at the rate of `5/6`cm/sec.
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