Question Number 190076 by Spillover last updated on 26/Mar/23
Answered by som(math1967) last updated on 27/Mar/23
$${a}.\:\:\:{f}^{\boldsymbol{{l}}} \left({x}\right)=\:\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\:\frac{\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{{x}}+\boldsymbol{{h}}\right)}−\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{sinx}}}}{\boldsymbol{{h}}} \\ $$$$\:=\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{sinx}}−\boldsymbol{{sin}}\left(\boldsymbol{{x}}+\boldsymbol{{h}}\right)}{\boldsymbol{{h}}\left\{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{{x}}+\boldsymbol{{h}}\right)\right\}\left(\mathrm{1}+\boldsymbol{{sinx}}\right)} \\ $$$$=\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{−\mathrm{2}{sin}\left(\frac{{h}}{\mathrm{2}}\right){cos}\left({x}+\frac{{h}}{\mathrm{2}}\right)}{{h}\left(\mathrm{1}+\boldsymbol{{sinx}}\right)\left\{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{{x}}+\boldsymbol{{h}}\right)\right\}} \\ $$$$=−\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {{lim}}\frac{\boldsymbol{{sin}}\frac{\boldsymbol{{h}}}{\mathrm{2}}\boldsymbol{{cos}}\left(\boldsymbol{{x}}+\frac{\boldsymbol{{h}}}{\mathrm{2}}\right)}{\frac{\boldsymbol{{h}}}{\mathrm{2}}\left(\mathrm{1}+\boldsymbol{{sinx}}\right)\left\{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{{x}}+\boldsymbol{{h}}\right)\right\}} \\ $$$$=−\frac{\boldsymbol{{cosx}}}{\left(\mathrm{1}+\boldsymbol{{sinx}}\right)^{\mathrm{2}} } \\ $$$$\:\therefore\boldsymbol{{f}}^{\boldsymbol{{l}}} \left(\frac{\boldsymbol{\pi}}{\mathrm{3}}\right)=\:−\frac{\boldsymbol{{cos}}\frac{\boldsymbol{\pi}}{\mathrm{3}}}{\left(\mathrm{1}+\boldsymbol{{sin}}\frac{\boldsymbol{\pi}}{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$=−\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{4}}}=−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$$=−\frac{\mathrm{2}\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)}{\mathrm{49}−\mathrm{48}}=−\mathrm{14}+\mathrm{8}\sqrt{\mathrm{3}\:} \\ $$$$\boldsymbol{{a}}=−\mathrm{14}\:\:,\:\boldsymbol{{b}}=\mathrm{8},\:\:\boldsymbol{{c}}=\sqrt{\mathrm{3}} \\ $$$$ \\ $$
Commented by Spillover last updated on 27/Mar/23
$$\mathrm{thanks} \\ $$
Answered by som(math1967) last updated on 27/Mar/23
$${b}\:\:{x}={a}\left(\theta+{sin}\theta\right)\:{y}={a}\left(\mathrm{1}−{cos}\theta\right) \\ $$$$\:\frac{{dx}}{{d}\theta}=\:{a}\left(\mathrm{1}+{cos}\theta\right)\:\: \\ $$$$\frac{{dy}}{{d}\theta}={asin}\theta \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{sin}\theta}{\left(\mathrm{1}+{cos}\theta\right)}=\frac{\mathrm{2}{sin}\frac{\theta}{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}} \\ $$$$=\:{tan}\frac{\theta}{\mathrm{2}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}×{sec}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}×\frac{{d}\theta}{{dx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×{sec}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}×\frac{\mathrm{1}}{{a}\left(\mathrm{1}+{cos}\theta\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\frac{{sec}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{2}{acos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}} \\ $$$$=\frac{{a}^{−\mathrm{1}} {sec}^{\mathrm{4}} \frac{\theta}{\mathrm{2}}}{\mathrm{4}} \\ $$
Commented by Spillover last updated on 27/Mar/23
$$\mathrm{thanks} \\ $$