1. Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(vii) (26)^{1/3 }

Answers (1)

Lets suppose y = (x)^{\frac{1}{3}} and let x = 27 and \Delta x = -1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{3}} - (x)^{\frac{1}{3}}
\Delta y = ({27 - 1})^{\frac{1}{3}} - (27)^{\frac{1}{3}}
\Delta y = ({26})^{\frac{1}{3}} - 3
({26})^{\frac{1}{3}} = \Delta y + 3
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{3 (x)^{\frac{2}{3}}}.(-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{3}} \ and \ \Delta x = -1)\\ dy = \frac{1}{3(27)^{\frac{2}{3}}}.(-1)\\ dy = \frac{1}{3\times 9}.(-1)\\dy = \frac{1}{27}.(-1) \\dy = -0.037
Now,
(27)^{\frac{1}{3}} = \Delta y +3\\ (27)^{\frac{1}{3}} = (-0.037) + 3\\ (27)^{\frac{1}{3}} = 2.963
Hence, (27)^{\frac{1}{3}} is approximately equal to 2.963



        

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