1. Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(x) ( 401 ) ^{1/2 }

Answers (1)

Let's suppose y = (x)^{\frac{1}{2}} and let x = 400 and \Delta x = 1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{2}} - (x)^{\frac{1}{2}}
\Delta y = ({400 + 1})^{\frac{1}{2}} - (400)^{\frac{1}{2}}
\Delta y = ({401})^{\frac{1}{2}} - 20
({401})^{\frac{1}{2}} = \Delta y + 20
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2 (x)^{\frac{1}{2}}}.(1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{2}} \ and \ \Delta x = 1)\\ dy = \frac{1}{2(400)^{\frac{1}{2}}}.(1)\\ dy = \frac{1}{2\times 20}.(1)\\dy = \frac{1}{40}.(1) \\dy = 0.025
Now,
(401)^{\frac{1}{2}} = \Delta y +20\\ (401)^{\frac{1}{2}} = (0.025) + 20\\ (401)^{\frac{1}{2}} = 20.025
Hence, (401)^{\frac{1}{2}} is approximately equal to 20.025



        

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