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  2. Using differentials, find the approximate value of each of the following up to 3 places of decimal square root of 49.5
# NCERT

1. Using differentials, find the approximate value of each of the following up to 3 places of decimal. 

(ii)\sqrt { 49.5 }

Answers (1)
G Gautam harsolia

Lets suppose y = \sqrt x and let x = 49 and \Delta x = 0.5
Then,
\Delta y = \sqrt{x+\Delta x} - \sqrt x
\Delta y = \sqrt{49+0.5} - \sqrt 49
\Delta y = \sqrt{49.5} - 7
\sqrt{49.5} = \Delta y +7
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2\sqrt x}.(0.5) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = \sqrt x \ and \ \Delta x = 0.5)\\ dy = \frac{1}{2\sqrt 49}.(0.5)\\ dy = \frac{1}{14}.(0.5)\\ dy = 0.035
Now,
\sqrt{49.5} = \Delta y +7\\ \sqrt {49.5} = 0.035 + 7\\ \sqrt{49.5} = 7.035
Hence, \sqrt{49.5} is approximately equal to 7.035



        

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