Question

A ladder

13 m

long leans against a wall. The foot of the ladder is pulled along the ground away from the wall, at the rate of

1.5 m / \sec

. How fast is the angle

θ

between the ladder and the ground is changing when the foot of the ladder is

12 m

away from the wall.

Accepted Answer

Given, length of ladder $= 13 m$

Let the bottom of the ladder at distance $x m$ from the wall and its top be at a height of $y m$ from the ground.

Let $∠ A C B = θ$ ,

We need to find the rate of change of an angle $' θ'$

In $△ A B C$

$\tan θ = \frac{y}{x}$

Also, $x^{2} + y^{2} = (13)^{2}$ [ Pythagorean theorem]

$\Rightarrow x^{2} (1 + \tan^{2} θ) = 169$

$\Rightarrow \sec^{2} θ = \frac{169}{x^{2}}$

Differentiating both sides with respect to $' t'$ , we get

$\Rightarrow 2 \sec^{2} θ \tan θ \frac{d θ}{d t} = 169 (- \frac{2}{x^{3}}) \frac{d x}{d t}$

$\Rightarrow \frac{d θ}{d t} = \frac{- 338 \times 1.5}{2 \sec^{2} θ \times \tan θ \times x^{3}} [\frac{d x}{d t} = 1.5 m / s]$

Since, $x = 12 m$

$∴ \frac{d θ}{d t} = \frac{- 338 \times 1.5}{(12)^{3} \times 2 \sec^{2} θ \times \tan θ} . . . . . . (i)$

$∵ y = \sqrt{169 - x^{2}} = \sqrt{169 - 144} = 5 m$

So, $\sec θ = \frac{13}{12}$ and $\tan θ = \frac{5}{12}$

From equation $(i)$ we have,

$\frac{d θ}{d t} = \frac{- 338 \times 1.5}{(12)^{3} \times 2 \times {(\frac{13}{12})}^{2} \times \frac{5}{12}}$

$= - \frac{338 \times 1.5}{10 \times 169}$

$= - 0.3$ rad / sec

Thus, the angle $θ$ is decreasing at the rate of $0.3$ rad/sec.

R D Sharma Solutions for Chapter: Derivative as a Rate Measurer, Exercise 1: EXERCISE

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Derivative as a Rate Measurer, Exercise 1: EXERCISE

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