Question

Using differentials, find the approximate value of each of the following up to $3$ places of decimal.
(i) $\sqrt{25.3}$
(ii) $\sqrt{49.5}$
(iii) $\sqrt{0.6}$
(iv) $(0.009{)}^{\frac{1}{3}}$
(v) $(0.999{)}^{\frac{1}{10}}$
(vi) $(15{)}^{\frac{1}{4}}$
(vii) $(26{)}^{\frac{1}{3}}$
(viii) $(255{)}^{\frac{1}{4}}$
(ix) $(82{)}^{\frac{1}{4}}$
(x) $(401{)}^{\frac{1}{2}}$
(xi) $(0.0037{)}^{\frac{1}{2}}$

Answer 1

(i)

Consider $y=\sqrt{x}$. Let $x=25$ and $\Delta$ $x=0.3$.
Then,
$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{25.3}-\sqrt{25}=\sqrt{25.3}-5$
$\Rightarrow \sqrt{25.3}=\Delta y+5$
Now, $dy$ is approximately equal to $\Delta$y and is given by,
$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{2\sqrt{x}}\left(0.3\right)$ [as $y=\sqrt{x}$]
$=\frac{1}{2\sqrt{25}}\left(0.3\right)=0.03$
Hence, the approximate value of $\sqrt{25.3}$ is $5+.03=5.03$

(ii)

Consider $y=\sqrt{x}$.

Let $x=49$ and $\Delta x=0.5$.
Then,
$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{49.5}-\sqrt{49}=\sqrt{49.5}-7$
$\Rightarrow \sqrt{49.5}=7+\Delta y$
Now, dy is approximately equal to $\Delta$y and is given by,
$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{2\sqrt{x}}\left(0.5\right)$ [as $y=\sqrt{x}$]
$=\frac{1}{2\sqrt{49}}\left(0.5\right)=\left(0.5\right)=\frac{1}{14}\left(0.5\right)=0.035$
Hence, the approximate value of $\sqrt{49.5}$ is $7+.035=7.035$

(iii)

Consider $y=\sqrt{x}$.

Let $x=1$ and $\Delta x=-\mathrm{0.4.}$
Then,
$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{0.6}-1$
$\Rightarrow \sqrt{0.6}=1+\Delta y$
Now, dy is approximately equal to $\Delta$y and is given by,
$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{2\sqrt{x}}\left(\Delta x\right)$ [as $y=\sqrt{x}$]
$=\frac{1}{2\sqrt{1}}\left(2\right)=(-0.4)=-0.2$
Hence, the approximate value of $\sqrt{0.6}$ is $1-.02=.98$

(iv)

Consider $y=x^{\frac{1}{3}}$.

Let $x=.008,\Delta x=.001$
$\Delta y=(x+\Delta x{)}^{\frac{1}{3}}-x^{\frac{1}{3}}=(0.009{)}^{\frac{1}{3}}-0.2$
$\Rightarrow (0.009{)}^{\frac{1}{3}}=0.2+\Delta y$
Now, $dy$ is approximately equal to $\Delta$y and is given by,
$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{3(x{)}^{\frac{2}{3}}}\left(\Delta x\right)$ [as y=$x^{\frac{1}{3}}$]
$=\frac{1}{3\times 0.04}\left(0.001\right)=\frac{0.001}{0.12}=0.008$
Hence, the approximate value of $(0.009{)}^{\frac{1}{3}}$ is $0.2+0.008=0.208$.

(v)

Consider $y=x^{\frac{1}{10}}$. Let $x=1$ and $\Delta x=\mathrm{0.001.}$
Then,
$\Delta y=(x+\Delta x{)}^{\frac{1}{10}}-x^{\frac{1}{10}}=(0.009{)}^{\frac{1}{10}}-1$
$\Rightarrow (0.009{)}^{\frac{1}{10}}=1+\Delta y$
Now, $dy$ is approximately equal to $\Delta$y and is given by,
$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{3(x{)}^{\frac{9}{10}}}\left(\Delta x\right)$ [as $y=x^{\frac{1}{10}}$]
$=\frac{1}{10}(-0.001)=-0.0001$
Hence, the approximate value of $(0.999{)}^{\frac{1}{10}}$ is $1+(-0.0001)=\mathrm{0.9999.}$

(vi)

Consider $y=x^{\frac{1}{4}}$.

Let $x=16$ and $\Delta x=-1$.
Then,
$\Delta y={(x+\Delta x)}^{\frac{1}{4}}-x^{\frac{1}{4}}=(15{)}^{\frac{1}{4}}-(16{)}^{\frac{1}{4}}=(15{)}^{\frac{1}{4}}-2$
$\Rightarrow (15{)}^{\frac{1}{4}}=2+\Delta y$
Now, $dy$ is approximately equal to $\Delta$y and is given by,
$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{4(x{)}^{\frac{3}{4}}}\left(\Delta x\right)$ [as $y=x^{\frac{1}{4}}$]
$=\frac{1}{4(16{)}^{\frac{3}{4}}}(-1)=\frac{-1}{4\times 8}=\frac{-1}{32}=-0.03125$
Hence, the approximate value of $(15{)}^{\frac{1}{4}}$ is $2+(-0.03125)=1.96875$.

(vii)

Consider $y=x^{\frac{1}{3}}$.

Let $x=27$and $\Delta x=-1.$
Then,
$\Delta y={x+\Delta x}^{\frac{1}{3}}-x^{\frac{1}{3}}=(26{)}^{\frac{1}{3}}-(27{)}^{\frac{1}{3}}=(26{)}^{\frac{1}{3}}-3$
$\Rightarrow (26{)}^{\frac{1}{3}}-)=3+\Delta y$
Now, $dy$ is approximately equal to $\Delta y$ and is given by,
$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{3(x{)}^{\frac{2}{3}}}\left(\Delta x\right)$ [as $y=x^{\frac{1}{3}}$]
$=\frac{1}{3(27{)}^{2/3}}(-1)=-\frac{1}{27}=-.037$
Hence, the approximate value of $(26{)}^{\frac{1}{3}}$ is $3+(-0.0370)=\mathrm{2.9629.}$

(viii)

Consider $y=x^{\frac{1}{4}}$. Let $x=256$ and $\Delta x=-1$.
Then,
$\Delta y={(x+\Delta x)}^{\frac{1}{4}}-x^{\frac{1}{4}}=(255{)}^{\frac{1}{4}}-(256{)}^{\frac{1}{4}}=(255{)}^{\frac{1}{4}}-4$
$\Rightarrow (255{)}^{\frac{1}{4}}=4+\Delta y$
Now, $dy$ is approximately equal to $\Delta y$ and is given by,
$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{4(x{)}^{\frac{3}{4}}}\left(\Delta x\right)$ [as $y=x^{\frac{1}{4}}$]
$=\frac{1}{4(256{)}^{3/4}}(-1)=\frac{1}{4\left(64\right)}(-1)=-\frac{1}{256}=-.0039$
Hence, the approximate value of $(255{)}^{\frac{1}{4}}$ is $3+(-0.0039)=\mathrm{2.996.}$

(ix)

Consider $y=x^{\frac{1}{4}}$. Let $x=81$ and $\Delta x=1$.
Then,
$\Delta y={(x+\Delta x)}^{\frac{1}{4}}-x^{\frac{1}{4}}=(82{)}^{\frac{1}{4}}-(81{)}^{\frac{1}{4}}=(82{)}^{\frac{1}{4}}-3$
$\Rightarrow (82{)}^{\frac{1}{4}}=\Delta y+3$
Now, $dy$ is approximately equal to $\Delta y$ and is given by,
$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{4(x{)}^{\frac{3}{4}}}\left(\Delta x\right)$ [as $y=x^{\frac{1}{4}}$]
$=\frac{1}{4(81{)}^{3/4}}(-1)=\frac{1}{4\left(27\right)}(-1)=-\frac{1}{108}=-.009$
Hence, the approximate value of $(82{)}^{\frac{1}{4}}$ is $3+(-0.009)=\mathrm{2.99.}$

(x)

Consider $y=x^{\frac{1}{2}}$.

Let $x=400$ and $\Delta x=1$.
Then,
$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{401}-\sqrt{400}=\sqrt{401}-20$
$\Rightarrow \sqrt{401}=20+\Delta y$
Now, $dy$ is approximately equal to $\Delta y$ and is given by,
$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{2\sqrt{x}}\left(\Delta x\right)$ [as $y=x^{\frac{1}{2}}$]
$=\frac{1}{2\sqrt{400}}\left(1\right)=\frac{1}{40}=.025$
Hence, the approximate value of $\sqrt{401}$ is $20+.025=20.025$

(xi)

Consider $y=x^{\frac{1}{2}}$. Let $x=0.0036$ and $\Delta x=\mathrm{0.0001.}$
Then,
$\Delta y=(x+\Delta x{)}^{\frac{1}{2}}-(x{)}^{\frac{1}{2}}=(0.0037{)}^{\frac{1}{2}}-(.0036{)}^{\frac{1}{2}}=(0.0037{)}^{\frac{1}{2}}-0.06$
$\Rightarrow (0.0037{)}^{\frac{1}{2}}=0.06+\Delta y$
Now, $dy$ is approximately equal to $\Delta y$ and is given by,
$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{2\sqrt{x}}\left(\Delta x\right)$ [as $y=x^{\frac{1}{2}}$]
$=\frac{1}{2\sqrt{.0036}}\left(.0001\right)=\frac{1}{2\times .06}\left(.0001\right)=.00083$
Hence, the approximate value of $(0.0037{)}^{\frac{1}{2}}$ is $.06+.00083=.06083$