Question Number 3215 by prakash jain last updated on 07/Dec/15
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{circles}\:\mathrm{perimeter}\:\mathrm{to} \\$$$$\mathrm{its}\:\mathrm{radius}\:\mathrm{is}\:\mathrm{constant}. \\$$$$\mathrm{Proof}\:\mathrm{should}\:\mathrm{not}\:\mathrm{utilize}\:\mathrm{a}\:\mathrm{result}\:\mathrm{based} \\$$$$\mathrm{on}\:\mathrm{the}\:\mathrm{above}\:\mathrm{fact}. \\$$
Answered by Yozzi last updated on 07/Dec/15
$${Define}\:{the}\:{curve}\:{y}=\sqrt{\boldsymbol{{r}}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\$$$${for}\:\mathrm{0}\leqslant{x}\leqslant\boldsymbol{{r}}\:\left(\boldsymbol{{r}}\in\mathbb{R}\right).\:{Graphically},\:{this}\: \\$$$${represents}\:{one}\:{quarter}\:{of}\:{a}\:{circle} \\$$$${of}\:{radius}\:\boldsymbol{{r}}\:{with}\:{centre}\:\left(\mathrm{0},\mathrm{0}\right).\:{The} \\$$$${form}\:{of}\:{this}\:{equation}\:{stems}\:{from}\: \\$$$${Phythagoras}'\:{Theorem}\:{and}\:{the} \\$$$${definition}\:{of}\:{the}\:{locus}\:{of}\:{points}\:{creating} \\$$$${a}\:{circle}.\:{We}\:{have}\:\frac{{dy}}{{dx}}=\frac{−\mathrm{2}{x}×\mathrm{0}.\mathrm{5}}{\:\sqrt{\boldsymbol{{r}}^{\mathrm{2}} −{x}^{\mathrm{2}} }} \\$$$$\frac{{dy}}{{dx}}=\frac{−{x}}{\:\sqrt{\boldsymbol{{r}}^{\mathrm{2}} −{x}^{\mathrm{2}} }}\Rightarrow\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} }{\boldsymbol{{r}}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\$$$$\therefore\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\frac{\boldsymbol{{r}}^{\mathrm{2}} −{x}^{\mathrm{2}} +{x}^{\mathrm{2}} }{\boldsymbol{{r}}^{\mathrm{2}} −{x}^{\mathrm{2}} }=\frac{\boldsymbol{{r}}^{\mathrm{2}} }{\boldsymbol{{r}}^{\mathrm{2}} −{x}^{\mathrm{2}} }. \\$$$${The}\:{differential}\:{form}\:{for}\:{arc}\:{length}\:{s} \\$$$${of}\:{a}\:{curve}\:{y}\:{dependent}\:{on}\:{x}\:{is}\:{given}\:{by} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{ds}}{{dx}}=\sqrt{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} }.\:\:\:\:\left(\ast\right) \\$$$${This}\:{form}\:{also}\:{stems}\:{from}\: \\$$$${Phythagoras}'\:{Theorem}.\: \\$$$${Think}\:{about}\:{a}\:{curve}\:{with}\:{a}\:{chord} \\$$$${of}\:{length}\:\delta{s}\:{joining}\:{two}\:{points}\:{on} \\$$$${a}\:{curve}\:{with}\:{vertical}\:{change}\:\delta{y}\:{and} \\$$$${horizontal}\:{change}\:\delta{x}.\:{Then},\:{we}\:{may} \\$$$${form}\:{a}\:{right}−{angled}\:{triangle}\: \\$$$${such}\:{that}\:\left(\delta{s}\right)^{\mathrm{2}} =\left(\delta{x}\right)^{\mathrm{2}} +\left(\delta{y}\right)^{\mathrm{2}} . \\$$$$\therefore\left(\frac{\delta{s}}{\delta{x}}\right)^{\mathrm{2}} =\mathrm{1}+\left(\frac{\delta{y}}{\delta{x}}\right)^{\mathrm{2}} \\$$$${As}\:\delta{x}\rightarrow\mathrm{0}\:{we}\:{obtain}\:{the}\:{differential} \\$$$${equation}\:\left(\frac{{ds}}{{dx}}\right)^{\mathrm{2}} =\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} . \\$$$${Taking}\:\frac{{ds}}{{dx}}>\mathrm{0}\:{we}\:{obtain}\:\left(\ast\right). \\$$$${Integrating}\:\left(\ast\right),\:{for}\:\mathrm{0}\leqslant{x}\leqslant\boldsymbol{{r}},\:{leads}\:{to}\: \\$$$${s}=\int_{\mathrm{0}} ^{\boldsymbol{{r}}} \sqrt{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} }{dx} \\$$$${s}=\int_{\mathrm{0}} ^{\boldsymbol{{r}}} \sqrt{\frac{\boldsymbol{{r}}^{\mathrm{2}} }{\boldsymbol{{r}}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{dx} \\$$$${s}=\int_{\mathrm{0}} ^{\boldsymbol{{r}}} \frac{\boldsymbol{{r}}}{\:\sqrt{\boldsymbol{{r}}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{dx}\:\:\:\:\:\left(\boldsymbol{{r}}>\mathrm{0}\right) \\$$$${s}=\boldsymbol{{r}}\left[{sin}^{−\mathrm{1}} \frac{{x}}{\boldsymbol{{r}}}\right]_{\mathrm{0}} ^{\boldsymbol{{r}}} \\$$$${s}=\boldsymbol{{r}}\left[{sin}^{−\mathrm{1}} \mathrm{1}−{sin}^{−\mathrm{1}} \mathrm{0}\right]=\boldsymbol{{r}}{sin}^{−\mathrm{1}} \mathrm{1} \\$$$${sin}^{−\mathrm{1}} \mathrm{1}\:{is}\:{a}\:{real}\:{constant}\:{since} \\$$$$\mid{sinx}\mid\leqslant\mathrm{1}\:\forall{x}\in\mathbb{R}\:{so}\:{that}\:{sin}^{−\mathrm{1}} \mathrm{1}\in\mathbb{R}. \\$$$${So},\:{the}\:{entire}\:{circle}\:{perimeter}\:{P}\:{is} \\$$$${given}\:{by}\:{P}=\mathrm{4}{s}=\mathrm{4}\boldsymbol{{r}}{sin}^{−\mathrm{1}} \mathrm{1}. \\$$$${So},\:\frac{{P}}{\boldsymbol{{r}}}=\mathrm{4}{sin}^{−\mathrm{1}} \mathrm{1}={constant}. \\$$
Commented by prakash jain last updated on 07/Dec/15
$$\mathrm{Great}. \\$$
Commented by Rasheed Soomro last updated on 08/Dec/15
$$\mathcal{EXCEL}{lent}!\:\mathcal{A}{lso}\:{like}\:{your}\:{explanation}! \\$$
Commented by 123456 last updated on 08/Dec/15
$$\mathrm{good}\:\mathrm{explanation} \\$$
Answered by 123456 last updated on 08/Dec/15
$$\mathrm{other}\:\mathrm{way}\:\left(???\right) \\$$$$\mathrm{lets}\:{l}:=\left\{\left({x},{y}\right)\in\mathbb{R}^{\mathrm{2}} ,{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \right\} \\$$$$\mathrm{and}\:\mathrm{take}\:\mathrm{the}\:\mathrm{line}\:\mathrm{integral}\:\mathrm{over}\:{l}\:\mathrm{with} \\$$$${f}\left({x},{y}\right)=\mathrm{1} \\$$$$\mathrm{integrating}\:{f}\:\mathrm{in}\:{l}\:\mathrm{give}\:\mathrm{the}\:\mathrm{lenght} \\$$$$\mathrm{P}=\underset{{l}} {\int}{f}\left({x},{y}\right){dl} \\$$$$\mathrm{we}\:\mathrm{have}\:\mathrm{that}\:\left(\Delta{l}\right)^{\mathrm{2}} =\left(\Delta{x}\right)^{\mathrm{2}} +\left(\Delta{y}\right)^{\mathrm{2}} ,\:\mathrm{taking} \\$$$$\Delta{x}\rightarrow\mathrm{0},\Delta{y}\rightarrow\mathrm{0}\:\mathrm{gives} \\$$$${dl}=\sqrt{\left({dx}\right)^{\mathrm{2}} +\left({dy}\right)^{\mathrm{2}} }\:\left(\mathrm{diferencial}\:\mathrm{element}\:\mathrm{of}\:\mathrm{line}\right) \\$$$$\mathrm{we}\:\mathrm{can}\:\mathrm{give}\:{l}\:\mathrm{as} \\$$$${x}={r}\:\mathrm{cos}\:{t} \\$$$${y}={r}\:\mathrm{sin}\:{t} \\$$$$\mathrm{0}\leqslant{t}<\mathrm{2}\pi \\$$$$\mathrm{note}\:\mathrm{that}\:\mathrm{its}\:\mathrm{give}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{since} \\$$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\$$$$\mathrm{so} \\$$$${dl}=\sqrt{\left({dx}\right)^{\mathrm{2}} +\left({dy}\right)^{\mathrm{2}} } \\$$$$=\sqrt{\left({dt}\right)^{\mathrm{2}} \left[\left(\frac{{dx}}{{dt}}\right)^{\mathrm{2}} +\left(\frac{{dy}}{{dt}}\right)^{\mathrm{2}} \right]} \\$$$$=\sqrt{\left(\frac{{dx}}{{dt}}\right)^{\mathrm{2}} +\left(\frac{{dy}}{{dt}}\right)^{\mathrm{2}} }{dt} \\$$$$\mathrm{so} \\$$$$\mathrm{P}=\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\sqrt{\left(\frac{{d}}{{dt}}{r}\mathrm{cos}\:{t}\right)^{\mathrm{2}} +\left(\frac{{d}}{{dt}}{r}\mathrm{sin}\:{t}\right)^{\mathrm{2}} }{dt} \\$$$$={r}\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\sqrt{\mathrm{sin}^{\mathrm{2}} {t}+\mathrm{cos}^{\mathrm{2}} {t}}{dt} \\$$$$={r}\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}{dt} \\$$$$=\mathrm{2}\pi{r} \\$$$$\frac{\mathrm{P}}{{r}}=\mathrm{2}\pi=\mathrm{const} \\$$$$\left(\mathrm{multivariable}\:\mathrm{calculus}\:\mathrm{version}\:\mathrm{of}\:\mathrm{above}\right. \\$$$$\left.\mathrm{proof}\right) \\$$
Commented by prakash jain last updated on 08/Dec/15
$$\mathrm{I}\:\mathrm{think}\:\mathrm{angle}\:\mathrm{going}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\mathrm{2}\pi\:\mathrm{is}\:\mathrm{dependent} \\$$$$\mathrm{on}\:\mathrm{the}\:\mathrm{definition}\:\mathrm{of}\:\mathrm{angle}\:\mathrm{as}\:\mathrm{radians}. \\$$$$\mathrm{Which}\:\mathrm{also}\:\mathrm{implies}\:\mathrm{perimeter}=\mathrm{2}\pi{r}. \\$$$$\mathrm{Am}\:\mathrm{i}\:\mathrm{correct}? \\$$
Commented by 123456 last updated on 08/Dec/15
$$\mathrm{hehehe},\:\mathrm{you}\:\mathrm{got}\:\mathrm{me}. \\$$$$\mathrm{s}\:\mathrm{is}\:\mathrm{lenght}\:\mathrm{of}\:\mathrm{arc}\:\mathrm{sector} \\$$$$\mathrm{radians}\:\rightarrow\:\theta=\frac{{s}}{{r}} \\$$$$\mathrm{degree}\:\rightarrow\:\mathrm{1}°=\frac{\mathrm{1}}{\mathrm{360}}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle} \\$$$$\mathrm{grade}\:\rightarrow\:\mathrm{1}\:\mathrm{gon}=\frac{\mathrm{1}}{\mathrm{400}}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle} \\$$$$\mathrm{2}\pi=\mathrm{360}°=\mathrm{400}\:\mathrm{gon} \\$$
Commented by Yozzi last updated on 08/Dec/15
$${That}\:{is}\:{why}\:{I}\:{hadn}'{t}\:{evalauted}\:{sin}^{−\mathrm{1}} \mathrm{1} \\$$$${in}\:{my}\:{proof}.\:\pi/\mathrm{2}\:{would}\:{have}\:{appeared} \\$$$${and}\:{I}\:{think}\:\pi\:{traditionally}\:{stems}\: \\$$$${from}\:{the}\:{definition}\:{of}\:{circumference} \\$$$${as}\:{stating}\:{that}\:{the}\:{ratio}\:{between}\:{the} \\$$$${circumference}\:{and}\:{diameter}\:{of}\:{a} \\$$$${circle}\:{is}\:{a}\:{constant}\:{denoted}\:{by}\:\pi. \\$$$${I}\:{thought}\:{that}\:{I}\:{needed}\:{to}\:{explain} \\$$$${the}\:{origin}\:{of}\:{every}\:{result}\:{I}\:{used}. \\$$
Commented by prakash jain last updated on 08/Dec/15
$$\mathrm{Once}\:\mathrm{you}\:\mathrm{got}\:\mathrm{to}\:\mathrm{sin}^{−\mathrm{1}} \mathrm{1}.\:\mathrm{The}\:\mathrm{proof}\:\mathrm{of}\:\mathrm{the}\:\mathrm{fact} \\$$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{is}\:\mathrm{constant}\:\mathrm{is}\:\mathrm{complete}.\:\mathrm{Evaluating}\:\mathrm{sin}^{−\mathrm{1}} \mathrm{1}\: \\$$$$\mathrm{after}\:\mathrm{that}\:\mathrm{point}\:\mathrm{keeps}\:\mathrm{the}\:\mathrm{proof}\:\mathrm{still}\:\mathrm{valid}. \\$$