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  • Using differentials, find the approximate value of each of the following up to 3 places of decimal 26 57 ^ 1 / 3

1. Using differentials, find the approximate value of each of the following up to 3
places of decimal.
 (xii) (26.57) ^ {1/3}

Answers (1)

best_answer

Lets suppose y = (x)^{\frac{1}{3}} and let x = 27 and \Delta x = -0.43
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{3}} - (x)^{\frac{1}{3}}
\Delta y = ({27 - 0.43})^{\frac{1}{3}} - (27)^{\frac{1}{3}}
\Delta y = ({26.57})^{\frac{1}{3}} - 3
({26.57})^{\frac{1}{3}} = \Delta y + 3
Now, we cam say that \Delta y  is approximately equals to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{3 (x)^{\frac{2}{3}}}.(-0.43) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{3}} \ and \ \Delta x = -0.43)\\ dy = \frac{1}{3(27)^{\frac{2}{3}}}.(-0.43)\\ dy = \frac{1}{3\times 9}.(-0.43)\\dy = \frac{1}{27}.(-0.43) \\dy = -0.0159 = -0.016 (approx.)
Now,
(26.57)^{\frac{1}{3}} = \Delta y +3\\ (26.57)^{\frac{1}{3}} = (-0.016) + 3\\ (26.57)^{\frac{1}{3}} = 2.984
Hence, (0.0037)^{\frac{1}{2}} is approximately equal to 0.060  (because we need to take up to three decimal places)



        

Posted by

Gautam harsolia

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