Find the approximate value of (81.5)^{frac{1} | vedclass

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(A) Let $y = x^{\frac{1}{4}}$. Consider $x = 81$ and $\Delta x = 0.5$.Then,$\Delta y = (x + \Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} = (81.5)^{\frac{

Vedclass · Accepted Answer

$3.0046$

Vedclass · Answer

(A) Let $y = x^{\frac{1}{4}}$. Consider $x = 81$ and $\Delta x = 0.5$.Then,$\Delta y = (x + \Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} = (81.5)^{\frac{1}{4}} - (81)^{\frac{1}{4}} = (81.5)^{\frac{1}{4}} - 3$.This implies $(81.5)^{\frac{1}{4}} = 3 + \Delta y$.We know that $dy \approx \Delta y$,where $dy = \left( \frac{dy}{dx} ight) \Delta x$.Since $y = x^{\frac{1}{4}}$,we have $\frac{dy}{dx} = \frac{1}{4} x^{-\frac{3}{4}} = \frac{1}{4x^{\frac{3}{4}}}$.Substituting the values,$dy = \frac{1}{4(81)^{\frac{3}{4}}} 	imes 0.5 = \frac{0.5}{4 	imes (3^4)^{\frac{3}{4}}} = \frac{0.5}{4 	imes 3^3} = \frac{0.5}{4 	imes 27} = \frac{0.5}{108} \approx 0.0046$.Therefore,the approximate value is $3 + 0.0046 = 3.0046$.

Find the approximate value of $(81.5)^{\frac{1}{4}}$.

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