Question Number 44291 by peter frank last updated on 25/Sep/18
$${withiout}\:{using}\:{calculator}\:{find}\:{approximate}\:{for} \\ $$$$\sqrt{\mathrm{9}.\mathrm{01}} \\ $$
Commented by maxmathsup by imad last updated on 25/Sep/18
$$\sqrt{\mathrm{9},\mathrm{01}}=\sqrt{\mathrm{9}+\mathrm{0},\mathrm{01}}=\sqrt{\mathrm{9}+\mathrm{10}^{−\mathrm{2}} }=\mathrm{3}\sqrt{\mathrm{1}+\frac{\mathrm{10}^{−\mathrm{2}} }{\mathrm{9}}}\:\:{but}\:{for}\:{x}\:\in\:{v}\left(\mathrm{0}\right) \\ $$$$\sqrt{\mathrm{1}+{x}}\:\sim\:\mathrm{1}+\frac{{x}}{\mathrm{2}}\:\Rightarrow\sqrt{\mathrm{1}+\frac{\mathrm{10}^{−\mathrm{2}} }{\mathrm{9}}}\:\sim\mathrm{1}+\frac{\mathrm{10}^{−\mathrm{2}} }{\mathrm{18}}\:\Rightarrow\sqrt{\mathrm{9},\mathrm{01}}\:\sim\mathrm{3}\left(\:\mathrm{1}+\frac{\mathrm{10}^{−\mathrm{2}} }{\mathrm{18}}\right)\:\Rightarrow \\ $$$$\sqrt{\mathrm{9},\mathrm{01}}\:\sim\mathrm{3}\:+\frac{\mathrm{10}^{−\mathrm{2}} }{\mathrm{6}}\:\:... \\ $$
Commented by Joel578 last updated on 26/Sep/18
$$\mathrm{Sir}\:\mathrm{please}\:\mathrm{explain}\:\:\sqrt{\mathrm{1}\:+\:{x}}\:\:\sim\:\mathrm{1}\:+\:\frac{{x}}{\mathrm{2}} \\ $$
Commented by Necxx last updated on 26/Sep/18
$${its}\:{linear}\:{approximation} \\ $$
Commented by abdo.msup.com last updated on 26/Sep/18
$${f}\left({x}\right)\:={f}\left(\mathrm{0}\right)\:+\frac{{x}}{\mathrm{1}!}{f}^{'} \left(\mathrm{0}\right)\:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\xi\left({x}\right)\:{with} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \xi\left({x}\right)=\mathrm{0}\:\Rightarrow{f}\left({x}\right)\:\sim{f}\left(\mathrm{0}\right)\:+{xf}^{'} \left({x}\right)\left({x}\rightarrow\mathrm{0}\right) \\ $$$${let}\:{take}\:{f}\left({x}\right)=\sqrt{\mathrm{1}+{x}}\:\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1}\:{and} \\ $$$${f}^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{1}}}\:\Rightarrow{f}^{'} \left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:\sim\mathrm{1}+\frac{{x}}{\mathrm{2}}\:. \\ $$
Commented by maxmathsup by imad last updated on 26/Sep/18
$${f}\left({x}\right)\sim{f}\left(\mathrm{0}\right)\:+{x}\:{f}^{'} \left(\mathrm{0}\right)\:\left({x}\rightarrow\mathrm{0}\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18
$${y}=\sqrt{{x}}\: \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}} \\ $$$${here}\:\:{x}=\mathrm{9}\:\:\:{x}+\bigtriangleup{x}=\mathrm{9}.\mathrm{01}\:\:\:\bigtriangleup{x}=\mathrm{0}.\mathrm{01} \\ $$$${y}=\sqrt{\mathrm{9}}\:=\mathrm{3}\:\:\:{y}+\bigtriangleup{y}={to}\:{find} \\ $$$$\frac{\bigtriangleup{y}}{\bigtriangleup{x}}\approx\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\bigtriangleup{y}}{\bigtriangleup{x}} \\ $$$$\bigtriangleup{y}=\frac{{dy}}{{dx}}\bigtriangleup{x} \\ $$$$\bigtriangleup{y}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{9}}}×\mathrm{0}.\mathrm{01}=\frac{\mathrm{0}.\mathrm{01}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{100}}\approx\mathrm{0}.\mathrm{1666}×\frac{\mathrm{1}}{\mathrm{100}}=\mathrm{0}.\mathrm{0016666} \\ $$$$\sqrt{\mathrm{9}.\mathrm{01}}\:=\mathrm{3}.\mathrm{0016666} \\ $$$$ \\ $$