#### Please solve RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 25 maths textbook solution

Answer: $\frac{d y}{d t}=20m/sec$

Hint: Here, we use the Pythagoras theorem $P R^{2}=P Q^{2}+Q R^{2}$

Given: A kite is 120 m high and 130 m string is out.

Solution:

\begin{aligned} &Q R=x \\ &P R=y \end{aligned}

Then from figure by applying Pythagoras theorem,

\begin{aligned} &P R^{2}=P Q^{2}+Q R^{2} \\\\ &y^{2}=(120)^{2}+x^{2} \end{aligned}.......(i)

Differentiating the above equation with respect to time,

\begin{aligned} &\frac{d y^{2}}{d t}=\frac{d\left((120)^{2}+x^{2}\right)}{d t} \\\\ &2 y \frac{d y}{d t}=0+2 x \frac{d x}{d t} \end{aligned}

So , $\frac{d x}{d t}=52m/sec$

$2 y \frac{d y}{d t}=2 x \times 52$    ......(ii)

\begin{aligned} &y^{2}=x^{2}+(120)^{2} \\\\ &130^{2}=x^{2}+14400 \end{aligned}

Therefore, $x=50$ m

So, let use equation (i) and (ii)

\begin{aligned} &2 y \frac{d y}{d t}=2 x \times 52 \\\\ &2(130) \frac{d y}{d t}=2(50) \times 52 \\\\ &\frac{d y}{d t}=\frac{2(50) \times 52}{2 \times 130}=20m/sec \end{aligned}