Find the approximate value of (26.57)^{frac{1 | vedclass

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Question

(A) Let $y = x^{\frac{1}{3}}$.Consider $x = 27$ and $\Delta x = -0.43$,so that $x + \Delta x = 26.57$.Then,$\Delta y = (x + \Delta x)^{\frac{1}{3}} -

Vedclass · Accepted Answer

$2.984$

Vedclass · Answer

(A) Let $y = x^{\frac{1}{3}}$.Consider $x = 27$ and $\Delta x = -0.43$,so that $x + \Delta x = 26.57$.Then,$\Delta y = (x + \Delta x)^{\frac{1}{3}} - x^{\frac{1}{3}} = (26.57)^{\frac{1}{3}} - (27)^{\frac{1}{3}} = (26.57)^{\frac{1}{3}} - 3$.This implies $(26.57)^{\frac{1}{3}} = 3 + \Delta y$.We use the approximation $dy \approx \Delta y$,where $dy = \left(\frac{dy}{dx}\right) \Delta x$.Since $y = x^{\frac{1}{3}}$,we have $\frac{dy}{dx} = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3(x^{\frac{1}{3}})^2}$.At $x = 27$,$\frac{dy}{dx} = \frac{1}{3(3)^2} = \frac{1}{27}$.Thus,$dy = \left(\frac{1}{27}\right) (-0.43) = -\frac{0.43}{27} \approx -0.0159$.Therefore,$(26.57)^{\frac{1}{3}} \approx 3 - 0.0159 = 2.9841$.The approximate value is $2.984$.

Find the approximate value of $(26.57)^{\frac{1}{3}}$.

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