Using differentials, find the approximate value of $\sqrt{49.5} .$

Solution

consider y = $\sqrt{x}$ , Let x = 49 and $∆ x$ = 0.5.

Then,

$∆ y = \sqrt{x + ∆ x} - \sqrt{x} = \sqrt{49.5} - \sqrt{49} = \sqrt{49.5} - 7 \Rightarrow \sqrt{49.5} = 7 + ∆ y$

Now, dy is approximately equal to $∆ y$ and is given by,

$dy = (\frac{dy}{dx}) ∆ x = \frac{1}{2 \sqrt{x}} (0 . 5) . . . . . . . [∵ y = \sqrt{x}] = \frac{1}{2 \sqrt{49}} (0 . 5) = \frac{1}{14} (0 . 5) = 0.035 Hence the approximate value of \sqrt{49.5} is 7 + 0.035 = 7.035 .$

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Using differentials, find the approximate value of $\sqrt{49.5} .$

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Relations And Functions

Using differentials, find the approximate value of 49.5.

Some More Questions From Relations and Functions Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Using differentials, find the approximate value of $\sqrt{49.5} .$