#### Please solve RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 17 maths textbook solution

Answer: $\frac{d y}{d t}=-\frac{1}{\sqrt{5}} m / s e c$  and $x=3 \sqrt{2} m$

Hint: Here we use the theorem of Pythagoras

Given: The top of ladder is 6 meter long is resting against a vertical wall on a level pavement, when the ladder begins to side outward.

Solution: Let $AC$  be the position of the ladder initially then $AC=6$ m

$DE$ be the position of the ladder after being pulled at the rate of 0.5 m/sec.

So $\frac{d x}{d t}=0.5 m/sec$ …….. (i)

$\Delta ABC$ , let’s apply Pythagoras theorem

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

$y^{2}+x^{2}=h^{2}$       ……… (ii)

Here h = 6 and x=4

\begin{aligned} &y^{2}=6^{2}-4^{2} \\\\ &y^{2}=36-16 \\\\ &y^{2}=20=4 \times 5=2 \sqrt{5} \end{aligned}

Differentiate equation (ii)

\begin{aligned} &y^{2}+x^{2}=h^{2} \\\\ &2 x \frac{d x}{d t}=-2 y \frac{d y}{d t} \end{aligned}

Let’s put value of $x = 4$ and $y = 2\sqrt{5}$   and $\frac{d x}{d t}=0.5$  ........equation (iii)

\begin{aligned} &2 \times 4 \times 0.5=-2 \times 2 \sqrt{5} \times \frac{d y}{d t} \\\\ &\Rightarrow \frac{d y}{d t}=-\frac{1}{\sqrt{5}} \mathrm{~m} / \mathrm{sec} \end{aligned}

From equation (iii) we get,

$2 x \frac{d x}{d t}=-2 y \frac{d y}{d t}$     as we know from question $\left[\frac{d x}{d t}=\frac{d y}{d t}\right]$

$x=-y$

Substituting $x=-y$  in $x^{2}+y^{2}=36$

We get

\begin{aligned} &x^{2}+x^{2}=36 \\\\ &2 x^{2}=36 \\\\ &x^{2}=18 \\\\ &x=3 \sqrt{2} \end{aligned}