Using differentials, find the approximate value of each of the following up to 3 places of decimal. $(iii)\;\sqrt{0.6}$

Question

This is third part of multipart q1

Answer 1

Toolbox:

• Let $y=f(x)$
• $\Delta x$ denote a small increment in $x$
• $\Delta y=f(x+\Delta x)-f(x)$
• $dy=\big(\large\frac{dy}{dx}\big)\Delta x$

Step 1:

Let $y=\sqrt{0.6}$

$\Rightarrow y=\sqrt x$

$x=0.64$

$\Delta x=-0.04$

$\Delta y=\sqrt{x+\Delta x}-\sqrt x$

$\quad\;\;=\sqrt{0.64+(-0.04)}-\sqrt{0.64}$

$\quad\;\;=\sqrt{0.6}-0.8$

$\sqrt{0.6}=\Delta y+0.8$

$\sqrt{0.6}=\Delta y+0.8$------(1)

Step 2:

$dy$ is the approximate value of $\Delta y$

$dy=\large\frac{dy}{dx}$$\times \Delta x$

$\large\frac{dy}{dx}=\frac{1}{2\sqrt x}$[Differentiating with respect to $x$]

$dy=\large\frac{1}{2\sqrt x}$$.(-0.04)$

$\quad=\large\frac{-0.04}{2\sqrt{0.64}}$

$\quad=\large\frac{-0.04}{2\times 0.8}$

$\quad=-0.025$

Step 3:

Substitute the value of $dy$ in equation (1)

$\sqrt{0.60}=0.8-0.025$

$\qquad\;\;=0.775$