Name: Derivates and Rate of Change
Uploaded: 2019-07-19
Duration: 12 min 59 s
Description: In this video, we look at finding the derivates and talk about the rate of change. It would help you understand the concepts, and you will be able to solve the questions in exams.

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.

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.

Important Questions on Derivative as a Rate Measurer

Learn and Prepare for any exam you want !

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.

Important Questions on Derivative as a Rate Measurer

Learn and Prepare for any exam you want !

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.