Find the approximate value of (32.15)^{frac{1 | vedclass

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Question

(A) Let $y = x^{\frac{1}{5}}$.Let $x = 32$ and $\Delta x = 0.15$.Then,$\Delta y = (x + \Delta x)^{\frac{1}{5}} - x^{\frac{1}{5}} = (32.15)^{\frac{1}{5

Vedclass · Accepted Answer

$2.00187$

Vedclass · Answer

(A) Let $y = x^{\frac{1}{5}}$.Let $x = 32$ and $\Delta x = 0.15$.Then,$\Delta y = (x + \Delta x)^{\frac{1}{5}} - x^{\frac{1}{5}} = (32.15)^{\frac{1}{5}} - (32)^{\frac{1}{5}} = (32.15)^{\frac{1}{5}} - 2$.So,$(32.15)^{\frac{1}{5}} = 2 + \Delta y$.Now,$dy$ is approximately equal to $\Delta y$,given by $dy = \left(\frac{dy}{dx}\right) \Delta x$.Since $y = x^{\frac{1}{5}}$,we have $\frac{dy}{dx} = \frac{1}{5} x^{-\frac{4}{5}} = \frac{1}{5(x^{\frac{1}{5}})^4}$.Substituting $x = 32$ and $\Delta x = 0.15$:$dy = \frac{1}{5(32^{\frac{1}{5}})^4} \cdot (0.15) = \frac{1}{5(2)^4} \cdot (0.15) = \frac{0.15}{5 \cdot 16} = \frac{0.15}{80} = 0.001875$.Rounding to five decimal places,$dy \approx 0.001875$.Thus,$(32.15)^{\frac{1}{5}} \approx 2 + 0.001875 = 2.001875$.

Find the approximate value of $(32.15)^{\frac{1}{5}}$.

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