Find the approximate value of (401)^{frac{1}{ | vedclass

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Question

(A) Let $y = x^{\frac{1}{2}}$.We choose $x = 400$ and $\Delta x = 1$ such that $x + \Delta x = 401$.Then,$\Delta y = \sqrt{x + \Delta x} - \sqrt{x} =

Vedclass · Accepted Answer

$20.025$

Vedclass · Answer

(A) Let $y = x^{\frac{1}{2}}$.We choose $x = 400$ and $\Delta x = 1$ such that $x + \Delta x = 401$.Then,$\Delta y = \sqrt{x + \Delta x} - \sqrt{x} = \sqrt{401} - \sqrt{400} = \sqrt{401} - 20$.This implies $\sqrt{401} = 20 + \Delta y$.Using the approximation $dy \approx \Delta y$,we have $dy = \left(\frac{dy}{dx}ight) \Delta x$.Since $y = x^{\frac{1}{2}}$,we have $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.Substituting the values,$dy = \frac{1}{2\sqrt{400}} 	imes 1 = \frac{1}{2 	imes 20} = \frac{1}{40} = 0.025$.Therefore,the approximate value of $\sqrt{401} = 20 + 0.025 = 20.025$.

Find the approximate value of $(401)^{\frac{1}{2}}$.

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