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explain solution RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 12 maths

Answers (1)


Answer: -0.3 rad/sec

Hint: Here we use the Pythagoras theorem,

A B^{2}+B C^{2}=A C^{2}

Given: A ladder 13 m leans against at a wall and the foot of the ladder is pulled along the ground away from the wall,

at the rate of 1.5 m/sec.

Solution: Suppose AC be the position of the ladder initially, then h = AC=13 m, DE  be the position of the ladder after being pulled at the rate of 1.5 m/sec.

Let the bottom of the ladder be at distance of x m from the wall and it’s top be at a height y m from the ground

Therefore, \frac{d x}{d t}=1.5 m/sec   .......(i)                                                                                                                                                


By Pythagoras theorem,

Considering, \Delta A B C, A B^{2}+B C^{2}=A C^{2}

y^{2}+x^{2}=h^{2}            .......(ii)

\tan \theta=\frac{A B}{B C}=\frac{y}{x}

\Rightarrow x \tan \theta=y            ......(iii)


Put the value of y in equation (ii) , we get

\begin{aligned} &x^{2}+(x \tan \theta)^{2}=(13)^{2} \\\\ &x^{2}+x^{2} \tan ^{2} \theta=169 \\\\ &x^{2}\left(1+\tan ^{2} \theta\right)=169 \\\\ &\sec ^{2} \theta=\frac{169}{x^{2}} \end{aligned}......(iv)


Differentiate the above equation w.r.t ‘t’

\begin{aligned} &\frac{d}{d t}\left(\sec ^{2} \theta\right)=\frac{d}{d t}\left(\frac{169}{x^{2}}\right) \\\\ &\Rightarrow 2 \sec \theta \sec \theta \tan \theta \frac{d \theta}{d t}=169 \frac{d}{d t}\left(x^{-2}\right) \end{aligned}

                                                    \left[\because \frac{d}{d x}\left\{f(x)^{n}\right\}=n f(x)^{n-1} \frac{d}{d x} f(x)\right]

\Rightarrow 2 \sec ^{2} \theta \tan \theta \frac{d \theta}{d t}=169(-2) x^{-2-1} \frac{d x}{d t} \quad\left[\frac{d}{d x} x^{n}=n x^{n-1}\right]

\begin{aligned} &\Rightarrow 2 \sec ^{2} \theta \tan \theta \frac{d \theta}{d t}=-338 x^{-3} \frac{d x}{d t} \\\\ &\Rightarrow \frac{d \theta}{d t}=\frac{-338}{2 x^{3} \sec ^{2} \theta \tan \theta} \frac{d x}{d t} \end{aligned}

\begin{aligned} &\Rightarrow \frac{d \theta}{d t}=\frac{-338}{2 x^{3} \sec ^{2} \theta \tan \theta}(1.5) \\\\ &\Rightarrow \frac{d \theta}{d t}=\frac{-169 \times 1.5}{x^{3} \sec ^{2} \theta \tan \theta} \end{aligned}........(v)


Now the foot of the ladder is pulled along the ground away from the wall,

So, x=12m

\Rightarrow y=\sqrt{h^{2}-x^{2}}              .......from equation (ii)

\Rightarrow y=\sqrt{169-144} \quad\left[\because h=13 \Rightarrow h^{2}=169 \& x=12 \Rightarrow x^{2}=144\right]

\Rightarrow y=\sqrt{25}=5 m     .........(iv)


Thus equation (v)

\Rightarrow \frac{d \theta}{d t}=\frac{-169 \times 1.5}{(12)^{3}\left(\frac{169}{144}\right)\left(\frac{5}{12}\right)}\left[\begin{array}{l} \because \sec ^{2} \theta=\frac{169}{x^{2}}=\frac{169}{144} \\ \tan \theta=\frac{y}{x}=\frac{5}{12} \end{array}\right]

\begin{aligned} &\Rightarrow \frac{d \theta}{d t}=\frac{-169 \times 1.5}{(12)^{3} \times \frac{169}{\left(12^{2}\right)} \times 5}=\frac{-1.5}{5} \\\\ &=-0.3 \mathrm{rad} / \mathrm{sec} \end{aligned}


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