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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?
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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, `dx/dt` = 5
x2 + y2 = 172
Pythagoras Theorem
y2 = 172 – x2
= 289 – 64
= 225
∴ y = 15
Differentiating w.r.t. ‘t’
`2x (dx)/dt + 2y (dy)/dt` = 0 .....(÷ 2)
`x (dx)/dt + y (dy)/dt` = 0
`8(5) + 15 (dy)/dt` = 0
∴ `(dy)/dt = 40/15`
= `- 8/3`
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 (dy)/dt + y (dx)/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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