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# Using differentials, find the approximate value of each of the following up to 3 places of decimal. $(x)\;(401)^{\Large\frac{1}{2}}$

This is tenth part of multipart q1

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 $y=x^{\Large\frac{1}{2}}$
Also let $x=400$
$\Delta x=1$
So that $x+\Delta x=401$
$\Delta y=(x+\Delta x)^{\Large\frac{1}{2}}-x^{\Large\frac{1}{2}}$
$\quad=(400+1)^{\Large\frac{1}{2}}-(400)^{\Large\frac{1}{2}}$
$\quad=(401)^{\Large\frac{1}{2}}-(400)^{\Large\frac{1}{2}}$
$\quad=(401)^{\Large\frac{1}{2}}-20$
$\Delta y=(401)^{\Large\frac{1}{2}}-20$
$(401)^{\Large\frac{1}{2}}=\Delta y+20$
$\large\frac{dy}{dx}=\frac{1}{2}$$x^{-\Large\frac{1}{2}} \quad=\large\frac{1}{2x^{\Large\frac{1}{2}}} Step 2: Now \Delta y is approximately equal to dy dy=\big(\large\frac{dy}{dx}\big)$$\Delta x$
$\quad=\large\frac{1}{2x^{\Large\frac{1}{2}}}$$\Delta x \quad=\large\frac{1}{2(400)^{\Large\frac{1}{2}}}$$\times 1$
$\quad=\large\frac{1}{2\times 20}$$\times 1$
$\quad=\large\frac{1}{40}$
$\quad=0.025$
Step 3:
Approximate value of $\Delta y=dy=0.025$
Hence from (1) we have
$\sqrt{401}=20+0.025$
$\qquad\;\;=20.025$