1. Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(iv) ( 0.009 ) ^{1/3 }

Answers (1)

Lets suppose y = (x)^{\frac{1}{3}} and let x = 0.008 and \Delta x = 0.001
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{3}} - (x)^{\frac{1}{3}}
\Delta y = ({0.008+ 0.001})^{\frac{1}{3}} - (0.008)^{\frac{1}{3}}
\Delta y = ({0.009})^{\frac{1}{3}} - 0.2
({0.009})^{\frac{1}{3}} = \Delta y + 0.2
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}}}.(0.001) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{3}} \ and \ \Delta x = 0.001)\\ dy = \frac{1}{3(0.008)^{\frac{2}{3}}}.(0.001)\\ dy = \frac{1}{0.12}.(0.001)\\ dy = 0.008
Now,
(0.009)^{\frac{1}{3}} = \Delta y +0.2\\ (0.009)^{\frac{1}{3}} = (0.008) + 0.2\\ (0.009)^{\frac{1}{3}} = 0.208
Hence, (0.009)^{\frac{1}{3}} is approximately equal to 0.208



        

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