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Question
A 13 m long ladder is leaning against a wall, touching the wall at a certain height from the ground level. The bottom of the ladder is pulled away from the wall, along the ground, at the rate of 2 m/s. How fast is the height on the wall decreasing when the foot of the ladder is 5 m away from the wall?
Solution
`x^2 + y^2 = 169` ...(i)
The bottom of the ladder is beign pulled, so these distances are changing with time,
`2x dx/dt + 2y dy/dt =0` ...(ii)
`dy/dt = - x/y dx/dt` .....(iii)
at x=5m then from (i) y=12m
From equation(iii)
`dy/dt = -5/12xx2`
`dy/dt = - 5/6 m//s`
– sign indicates the height is decreasing.
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