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) ${(0.009)^{1/3}}$

(v) ${(0.999)^{1/10}}$

(vi) ${(15)^{114}}$

(vii) ${(26)^{1/3}}$

(viii) ${(255)^{1/4}}$

(ix) ${(82)^{1/4}}$

(x) ${(401)^{1/2}}$

(xi) ${(0.0037)^{1/2}}$

(xii) ${(26.57)^{1/3}}$

(xiii) ${(81.5)^{1/4}}$

(xiv) ${(3.968)^{3/2}}$

(xv) ${(32.15)^{1/5}}$

(i) Let $y = \sqrt x ,x = 25,\Delta x = 0.3$

$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{{2\sqrt x }}$, and
$\Delta y = \left( {\cfrac{{dy}}{{dx}}} \right)\Delta x = \left( {\cfrac{1}{{2\sqrt x }}} \right)\Delta x = \cfrac{1}{{2 \times 5}} \times 0.3 = 0.03$

Also $\Delta y = \sqrt {x + \Delta x} - \sqrt x$
$\Rightarrow 0.03 = \sqrt {253} - \sqrt {25} \Rightarrow \sqrt {253} = 0.03 + 5 = 5.030$

Therefore the approximate value of :$\sqrt {25.3} =5.030$

(ii)

For Solution Refer Practice Questions, Q. No. 35

(iii) Let $y = \sqrt x ,$
$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{{2\sqrt x }}$ and $\Delta y = \left( {\cfrac{{dy}}{{dx}}} \right)\Delta x$

$\Rightarrow \Delta y = \cfrac{1}{{2\sqrt {064} }} \times \Delta r = \cfrac{1}{{2(0.8)}} \times \left( { - 0.04} \right) = \cfrac{{ - 0. \cdot 04}}{{16}} = - 0.025$

Also, $\Delta y = \sqrt {x + \Delta x} - \sqrt x$
$- 0.025 = \sqrt {0.6} - \sqrt {0.64}$ or $\sqrt {0.6} = 0.8 - 0.025 = 0.775$

Therefore the approximate value of :$\sqrt {0.6} =0.775$

(iv) Let $y = {x^{1/3}},x = 0.008,\Delta x = 0.001$

$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{3}{x^{ - 2/3}}$ and $\Delta y = \left( {\cfrac{{dy}}{{dx}}} \right)\Delta x$

$= \cfrac{1}{3}{(0.008)^{ - 2/3}}(0.001) = \cfrac{1}{{3{{(0.2)}^{3 \times (2/3)}}}} \times 0.001$
$= \cfrac{1}{{3(0.04)}} \times 0.001 = \cfrac{{0.001}}{{0.12}} = \cfrac{1}{{120}} = 0.008$

Also, $\Delta y = \sqrt[3]{{x + \Delta x}} - \sqrt[3]{x}$
$\Rightarrow 0.008 = \sqrt[3]{{0009}} - \sqrt[3]{{0008}}$
$\Rightarrow \sqrt[3]{{0009}} = 0.008 + 0.2 = 0.208$

Therefore the approximate value of :${(0.009)^{1/3}}=0.208$

(v) Let $y = {x^{1/10}},x = 1,\Delta x = - 0.001$
$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{{10}}{x^{ - 9/10}}$ and $\Delta y = \left( {\cfrac{{dy}}{{dx}}} \right)\Delta x$

$\Rightarrow \Delta y = \cfrac{1}{{10{x^{9/10}}}} \times ( - 0.001) = \cfrac{{ - 0.001}}{{10{{(1)}^{9/10}}}} = \cfrac{{ - 0.001}}{{10}} = - 0.0001$

Also, $\Delta y = {(x + \Delta x)^{1/10}} - {(x)^{1/10}}$
$\Rightarrow - 0.0001 = {(0.999)^{1/10}} - {(1)^{1/10}}$

$\Rightarrow {(0.999)^{1/10}} = 1 - 0.0001 = 0.999$

Therefore the approximate value of :${(0.999)^{1/10}} = 0.999$

(vi) Let $y = {x^{1/4}},x = 16,\Delta x = - 1$
$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{4}{x^{ - 3/4}}$ and $\Delta y = \left( {\cfrac{{dy}}{{dx}}} \right)\Delta x$

$\Rightarrow \Delta y = \left( {\cfrac{1}{4}{x^{ - 3/4}}} \right)( - 1) = \left( {\cfrac{{ - 1}}{{4{{(16)}^{3/4}}}}} \right) = \cfrac{{ - 1}}{{4 \times {2^3}}} = \cfrac{{ - 1}}{{32}} = - 0.03125$

Also $\Delta y = {(x + \Delta x)^{1/4}} - {(x)^{1/4}} \Rightarrow - 0.03125 = {(15)^{1/4}} - {(16)^{1/4}}$
$\; \Rightarrow {\left( {15} \right)^{1/4}} = - 0.03125 + 2 = 1.96875 \cong 1.969$

Therefore the approximate value of :${(15)^{114}}=1.969$

(vii) Let $y = {x^{1/3}},x = 27,\Delta x = - 1$
$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{3}{x^{ - 2/3}}$ and $\Delta y = \left( {\cfrac{{dy}}{{dx}}} \right)\Delta x$

$\Rightarrow \Delta y = \left( {\cfrac{1}{3}{x^{ - 2/3}}} \right)( - 1) = \left( {\cfrac{{ - 1}}{{3{{(27)}^{2/3}}}}} \right) = \cfrac{{ - 1}}{{3 \times 9}} = \cfrac{{ - 1}}{{27}} = - 0.037$

AIso, $\Delta y = {(x + \Delta x)^{1/3}} - {(x)^{1/3}}$
$\Rightarrow - 0.0370 = {(26)^{1/3}} - {(27)^{1/3}}$
$\Rightarrow \;{\left( {26} \right)^{1/3}} = 3 - 0.037 = 2.963$

Therefore the approximate value of :${(26)^{1/3}}=2.963$

(viii) Let $y = {x^{1/4}},x = 256,\Delta x = - 1$

$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{4}{x^{ - 3/4}}$ and $\Delta y = \left( {\cfrac{{dy}}{{dx}}} \right)\Delta x$

$\Rightarrow \Delta y = \left( {\cfrac{1}{4}{x^{ - 3/4}}} \right)( - 1) = \left( {\cfrac{{ - 1}}{{4{{(256)}^{3/4}}}}} \right) = \cfrac{{ - 1}}{{4 \times 64}} = \cfrac{{ - 1}}{{256}}$
Also, $\Delta y = {(x + \Delta x)^{1/4}} - {(x)^{1/4}}$

$\Rightarrow \cfrac{{ - 1}}{{256}} = {(255)^{1/4}} - {(256)^{1/4}}$
$\Rightarrow {(255)^{1/4}} = 4 - \cfrac{1}{{256}} = \cfrac{{1023}}{{256}} = 3.996$

Therefore the approximate value of :${(255)^{1/4}}=3.996$

(ix) Let $y = {x^{1/4}}$,
$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{{4{x^{3/4}}}}$ and $\Delta y = \left( {\cfrac{{dy}}{{dx}}} \right)\Delta x$

$\Rightarrow \Delta y = \left( {\cfrac{1}{{4{{(81)}^{3/4}}}}} \right)(1) = \cfrac{1}{{4(27)}} = \cfrac{1}{{108}}$

Also, $\Delta y = {(x + \Delta x)^{1/4}} - {(x)^{1/4}}$

$\Rightarrow \cfrac{1}{{108}} = {(82)^{1/4}} - {(81)^{1/4}}$
$\Rightarrow {(82)^{1/4}} = 3 + \cfrac{1}{{108}} = 3.009$

Therefore the approximate value of :${(82)^{1/4}}=3.009$

(x) Let $y = {x^{1/2}},\,x = 400,\Delta x = 1$
$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{{2{x^{1/2}}}}$ and $\Delta y = \left( {\cfrac{{dy}}{{dx}}} \right)\Delta x$

$\Rightarrow \Delta y = \left( {\cfrac{1}{{2{{(400)}^{1/2}}}}} \right)(1) = \cfrac{1}{{2(20)}} = \cfrac{1}{{40}}$

Also, $\Delta y = {(x + \Delta x)^{1/2}} - {(x)^{1/2}}$
$\Rightarrow \cfrac{1}{{40}} = {(401)^{1/2}} - {(400)^{1/2}}$

$\Rightarrow {(401)^{1/2}} = 20 + \cfrac{1}{{40}} = 20.025$.

Therefore the approximate value of :${(401)^{1/2}}=20.025$.

(xi) Let $y = {x^{1/2}},x = 0.0036,\Delta x = 0.0001$
$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{{2\sqrt x }}$ and $\Delta y = \left( {\cfrac{{dy}}{{dx}}} \right)\Delta x$

$\Rightarrow \Delta y = \cfrac{1}{{2{{(0.0036)}^{1/2}}}} \times (0.0001) = \cfrac{{0.0001}}{{2(0.06)}} = \cfrac{1}{{1200}}$

Also,$\Delta y = {(x + \Delta x)^{1/2}} - {(x)^{1/2}} \Rightarrow \cfrac{1}{{1200}} = {(0.0037)^{1/2}} - {(0.0036)^{1/2}}$
$\Rightarrow {(0.0037)^{1/2}} = {(0.0036)^{1/2}} + \cfrac{1}{{1200}} = 0.06 + \cfrac{1}{{1200}} = 0.06083 \cong 0.061$

Therefore the approximate value of :${(0.0037)^{1/2}}=0.061$

(xii) Let $y = {x^{1/3}},x = 27,\Delta x = - 0.43$

$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{{3{x^{2/3}}}}$ and $\Delta y = \left( {\cfrac{{dy}}{{dx}}} \right)\Delta x$
$\Rightarrow \Delta y = \left( {\cfrac{1}{{3{{(27)}^{2/3}}}}} \right)( - 0.43) = \cfrac{{ - 0.43}}{{3(9)}} = \cfrac{{ - 0.43}}{{27}} = - 0.015926$

Also, $\Delta y = {(x + \Delta x)^{1/3}} - {(x)^{1/3}}$
$\Rightarrow$ $- 0.015926 = {(26.57)^{1/3}} - {(27)^{1/3}}$
$\Rightarrow {(26.57)^{1/3}} = - 0.015926 + 3 = 2.984$

Therefore the approximate value of :${(26.57)^{1/3}}=2.984$

(xiii) Let $y = {x^{1/4}},x = 81,\Delta x = 0.5$
$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{{4{x^{3/4}}}}$ and $\Delta y = \left( {\cfrac{{dy}}{{dx}}} \right)\Delta x$

$\Rightarrow \Delta y = \left( {\cfrac{1}{{4{{(81)}^{3/4}}}}} \right)(0.5) = \cfrac{1}{{4 \times {3^3}}} \times 0.5 = \cfrac{{0.5}}{{108}} = \cfrac{5}{{1080}}$

Also, $\Delta y = {(x + \Delta x)^{1/4}} - {(x)^{1/4}}$
$\Rightarrow \cfrac{5}{{1080}} = {(81.5)^{1/4}} - {(81)^{1/4}}$
$\Rightarrow {(81.5)^{1/4}} = 3 + \cfrac{5}{{1080}} = 3.0046 \cong 3.005$

Therefore the approximate value of :${(81.5)^{1/4}}= 3.005$

(xiv)

For Solution : Refer Practice Questions., Q.No. 33

Therefore the approximate value of :${(3.968)^{3/2}}= 7.904$

(xv) Let $y = {x^{1/5}},x = 32,\Delta x = 0.15$
$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{{5{x^{4/5}}}}$ and $\Delta y = \left( {\cfrac{{dy}}{{dx}}} \right)\Delta x$

$\Rightarrow \Delta y = \cfrac{1}{{5{{(32)}^{4/5}}}} \times (0.15) = \cfrac{1}{{5(16)}} \times 0.15 = \cfrac{{15}}{{8000}}$

Also, $\Delta y = {\left( {x + \Delta x} \right)^{1/5}} - {\left( x \right)^{1/5}} \Rightarrow \cfrac{{15}}{{8000}} = {\left( {32.15} \right)^{1/5}} - {\left( {32} \right)^{1/5}}$
$\Rightarrow \& {\left( {32.15} \right)^{1/5}} = \cfrac{{15}}{{8000}} + 2 = 2.00187$

Therefore the approximate value of : ${(32.15)^{1/5}}=2.00187$