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Tamil Nadu Board of Secondary EducationHSC Science Class 12

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, - Mathematics

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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?

Numerical
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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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Chapter 7: Applications of Differential Calculus - Exercise 7.1 [Page 8]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 7 Applications of Differential Calculus
Exercise 7.1 | Q 9. (ii) | Page 8

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