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A Kite is 120 M High and 130 M of String is Out. If the Kite is Moving Away Horizontally at the Rate of 52 M/Sec, Find the Rate at Which the String is Being Paid Out.

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Question

A kite is 120 m high and 130 m of string is out. If the kite is moving away horizontally at the rate of 52 m/sec, find the rate at which the string is being paid out.

Sum
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Solution

\[\text { In the right triangle ABC },\]

\[\text { Here,} \]

\[A B^2 + B C^2 = A C^2 \]

\[ \Rightarrow x^2 + \left( 120 \right)^2 = y^2 \]

\[ \Rightarrow 2x\frac{dx}{dt} = 2y\frac{dy}{dt}\]

\[ \Rightarrow \frac{dy}{dt} = \frac{x}{y}\frac{dx}{dt}\]

\[ \Rightarrow \frac{dy}{dt} = \frac{50}{130} \times 52 \left[ \because x = \sqrt{\left( 130 \right)^2 - \left( 120 \right)^2} = 50 \right]\]

\[ \Rightarrow \frac{dy}{dt} = 20 m/\sec\]

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Chapter 12: Derivative as a Rate Measurer - Exercise 13.2 [Page 20]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 12 Derivative as a Rate Measurer
Exercise 13.2 | Q 25 | Page 20

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