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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, at what rate, the area of the triangle formed by the ladder, wall, and the floor, is changing?
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`
Area of triangle formed by the ladder, wall and the floor is A = `1/2` xy
Differentiating w.r.t. ‘t’
`"dA"/"dt" = 1/2[x ("d"y)/"dt" + y ("d"x)/"dt"]`
= `1/2[8(- 8/3) + 15(5)]`
= `1/2[(225 - 64)/3]`
= `161/6`
= 26.83
∴ Area of the triangle is increasing at the rate of 26.83 m2/sec.
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