A ladder 10 metres long is leaning versus a vertical wall. If the bottom the the ladder is traction horizontally far from the wall surface at the rate of 1.2 metres per second, discover how quick the top of the ladder is sliding down the wall, as soon as the bottom is 6 metres far from the wall.

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Let ab be the ladder, where abdominal = 10 meters. Let in ~ time t seconds, the finish A the the ladder be x metres from the wall and the end B be y metres from the ground.

Since, OAB is a right angled triangle, through Pythagoras theoremx2 + y2 = 102i.e. Y2 = 100 – x2Differentiating w.r.t. T, we get

`2ydy/dt = 0 - 2xdx/dt`

∴ `dy/dt = x/y.dx/dt` ...(1)

Now, `dx/dy = (12"metres")/sec` is the price at wh the bottom that the ladder s traction horizontally and `dy/dx` is the price which the optimal of ladder B is sliding.

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which the height of ladder B is sliding.When x = 6, y2 = 100 – 36 = 64∴ y = 8

∴ (1) gives, `dy/dx = -(6)/(8)(1.2)`

= `(6)/(8) xx (12)/(10)`

= `-(9)/(10)`

= – 0.9Hence, the top of the ladder is sliding down the wall, at the rate of `(0.9"metre")/sec`.


Concept: Derivatives together a price Measure
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Chapter 2: Applications the Derivatives - exercise 2.1
Q 15Q 14Q 16
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Balbharati Mathematics and Statistics 2 (Arts and Science) 12th typical HSC Maharashtra State Board
Chapter 2 Applications that DerivativesExercise 2.1 | Q 15 | web page 72
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