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Question
A ladder 17 metre long is leaning against the wall. The base of the ladder is pulled away from the wall at a rate of 5 m/s. When the base of the ladder is 8 metres from the wall. How fast is the top of the ladder moving down the wall?
Solution
Let the height of the wall where the ladder touches are ‘y’ m.
The bottom of the ladder is at a distance of ‘x’ m from the wall.
Given x = 8, `("d"x)/"dt"` = 5
x2 + y2 = 172
Pythagoras Theorem
y2 = 172 – x2
= 289 – 64
= 225
∴ y = 15
Differentiating w.r.t. ‘t’
`2x ("d"x)/"dt" + 2y ("d"y)/"dt"` = 0 .....(÷ 2)
`x ("d"x)/"dt" + y ("d"y)/"dt"` = 0
`8(5) + 15 ("d"y)/"dt"` = 0
∴ `("d"y)/"dt" = 40/15`
= `- 8/5`
The top of the ladder is moving down the wall at the rate of `8/3` m/sec
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