First slide

Differentials; errors and approximations

Question

$Use differential to approximate \sqrt{16.6}$

Moderate

A

4.4
B

1.3712
C

4.075
D

4.175

Solution

$\begin{array}{l} Take y = \sqrt{x}, let x = 16 and Δ x = 0.6 \\ Δ y = \sqrt{(x + Δ x)} - \sqrt{x} \end{array}$
$\begin{array}{l} = \sqrt{16.6} - \sqrt{16} \\ = \sqrt{16.6} - 4 \\ or \sqrt{16.6} = Δ y + 4 \\ Now dy is approximately equal to Δ y and is given by \end{array}$
$\begin{array}{l} d y = (\frac{d y}{d x}) Δ x \\ = \frac{1}{2 \sqrt{x}} (0.6) \\ d y = \frac{1}{(2 \sqrt{16})} (0.6) \\ - \frac{1}{8} (0.6) \\ = 0.075 \end{array}$
$Thus, the approximate value of \sqrt{16.6} = 4 + 0.075 = 4.075$

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