Question Number 122588 by bramlexs22 last updated on 18/Nov/20
$${Find}\:{the}\:\mathrm{arc}\:{length}\:{of}\:{the}\:{curve}\: \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{4}}{y}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({y}\right)\:{between}\:{the}\:{points} \\ $$$${with}\:{the}\:{ordinates}\:{y}=\mathrm{1}\:{and}\:{y}=\mathrm{2}. \\ $$$$ \\ $$
Answered by MJS_new last updated on 18/Nov/20
$$\mathrm{arc}\:\mathrm{length}\:\mathrm{of}\:{x}={f}\left({y}\right)\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$$\int\sqrt{\mathrm{1}+\left({x}'\right)^{\mathrm{2}} }\:{dy} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{4}}{y}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{y}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{y}>\mathrm{0} \\ $$$${x}'=\frac{{y}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}{y}} \\ $$$$\sqrt{\mathrm{1}+\left({x}'\right)^{\mathrm{2}} }=\sqrt{\frac{\left({y}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}{y}^{\mathrm{2}} }}=\frac{{y}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{y}}\:\left[\mathrm{because}\:{y}>\mathrm{0}\right] \\ $$$$\int\frac{{y}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{y}}{dy}=\int\frac{{y}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{y}}{dy}=\frac{\mathrm{1}}{\mathrm{4}}{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{y} \\ $$$$\Rightarrow\:\mathrm{arc}\:\mathrm{length}\:\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$
Commented by bramlexs22 last updated on 18/Nov/20
$${thank}\:{you}\:{both}\:{sir} \\ $$
Answered by liberty last updated on 18/Nov/20
$$\ell\:=\:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\sqrt{\mathrm{1}+\left(\mathrm{x}'\right)^{\mathrm{2}} }\:\mathrm{dy}\: \\ $$$$\frac{\mathrm{dx}}{\mathrm{dy}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}−\frac{\mathrm{1}}{\mathrm{2y}}\:\Rightarrow\ell=\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\sqrt{\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}−\frac{\mathrm{1}}{\mathrm{2y}}\right)^{\mathrm{2}} }\:\mathrm{dy} \\ $$$$\ell\:=\:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4y}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}}\:\mathrm{dy}\: \\ $$$$\ell\:=\:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\sqrt{\frac{\mathrm{1}}{\mathrm{4}}\mathrm{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4y}^{\mathrm{2}} }}\:\mathrm{dy}\: \\ $$$$\ell\:=\:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}+\frac{\mathrm{1}}{\mathrm{2y}}\right)\:\mathrm{dy}\:=\:\left[\frac{\mathrm{1}}{\mathrm{4}}\mathrm{y}^{\mathrm{2}} +\frac{\mathrm{ln}\:\mathrm{y}}{\mathrm{2}}\:\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\ell\:=\:\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}.\:\blacktriangle \\ $$