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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?
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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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