Question

Using differentials, find the approximate value of each of the following up to $3$ decimal places.
i) $\sqrt{25.3}$
ii) $(0.999{)}^{\frac{1}{10}}$
ii) ${15}^{\frac{1}{4}}$

Answer 1

i)

$\sqrt{25.3}$

Let $y=\sqrt{x}$

Let $x=25,\Delta x=0.3$

$\frac{dy}{dx}=\frac{d\sqrt{x}}{dx}=\frac{1}{2\sqrt{x}}$

$\Delta y=\frac{dy}{dx}\Delta x$

$=\frac{1}{2\sqrt{x}}\left(0.3\right)$

$=\frac{1}{2\sqrt{25}}\left(0.3\right)$

$=0.03$

$\Delta y=f(x+\Delta x)-f\left(x\right)$

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

$0.03=\sqrt{25+0.3}-\sqrt{25}$

$\sqrt{25.3}=5.03$

ii)

$(0.999{)}^{\frac{1}{10}}$

Let $y=x^{\frac{1}{10}}$

where, $x=1,\Delta x=-0.001$

$y=x^{\frac{1}{10}}$

$\frac{dy}{dx}=\frac{d\left(x^{\frac{1}{10}}\right)}{dx}=\frac{1}{10(x{)}^{\frac{9}{10}}}$

$\Delta y=\frac{dy}{dx}\Delta x$

$\Delta y=\frac{1}{10(x{)}^{\frac{9}{10}}}\Delta x$

$\Delta y=\frac{1}{10(1{)}^{\frac{9}{10}}}\times (-0.001)$

$\Delta y=-0.0001$

$\Delta y=f(x+\Delta x)-f\left(x\right)$

$-0.0001=(1+(-0.001){)}^{\frac{1}{10}}-(1{)}^{\frac{1}{10}}$

$(0.999{)}^{\frac{1}{10}}=0.9999$

iii)

${15}^{\frac{1}{4}}$

Let $y=x^{\frac{1}{4}}$

where $x=16,\Delta x=-1$

$\frac{dy}{dx}=\frac{d\left(x^{\frac{1}{4}}\right)}{dx}=\frac{1}{4}{x}^{\frac{-3}{4}}$

$\Delta y=\frac{dy}{dx}\Delta x$

$\Delta y=\frac{1}{4}\left(16{)}^{\frac{-3}{4}}\right(-1)$

$\Delta y=\frac{-1}{32}$

$\Delta y=-0.03125$

$y=f(x+\Delta x)-f\left(x\right)$

$-0.03125=(16+(-1){)}^{\frac{1}{4}}-(16{)}^{\frac{1}{4}}$

$(15{)}^{\frac{1}{4}}=1.96875$