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  • Using differentials, find the approximate value of each of the following up to 3 places of decimal. square root of 25.3

1. Using differentials, find the approximate value of each of the following up to 3
places of decimal. (i)  \sqrt {25.3 }

Answers (1)

best_answer

Let's suppose y = \sqrt x and let x = 25 and \Delta x = 0.3
Then,
\Delta y = \sqrt{x+\Delta x} - \sqrt x
\Delta y = \sqrt{25+0.3} - \sqrt 25
\Delta y = \sqrt{25.3} - 5
\sqrt{25.3} = \Delta y +5
Now, we can say that \Delta y  is approximate equals to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2\sqrt x}.(0.3) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = \sqrt x \ and \ \Delta x = 0.3)\\ dy = \frac{1}{2\sqrt 25}.(0.3)\\ dy = \frac{1}{10}.(0.3)\\ dy = 0.03
Now,
\sqrt{25.3} = \Delta y +5\\ \sqrt {25.3} = 0.03 + 5\\ \sqrt{25.3} = 5.03
Hence, \sqrt{25.3} is approximately equal to 5.03 



        

Posted by

Gautam harsolia

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