Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 122588 by bramlexs22 last updated on 18/Nov/20

$$Findthe\mathrm{arc}lengthofthecurve$$$$x=\frac{1}{4}{y}^2-\frac{1}{2}\ln \left(y\right)betweenthepoints$$$$withtheordinatesy=1andy=2.$$$$$$

Answered by MJS_new last updated on 18/Nov/20

$$\mathrm{arc}\mathrm{length}\mathrm{of}x=f\left(y\right)\mathrm{is}\mathrm{given}\mathrm{by}$$$$\int \sqrt{1+{\left(x^{\prime }\right)}^2}dy$$$$x=\frac{1}{4}{y}^2-\frac{1}{2}\ln y\mathrm{is}\mathrm{defined}\mathrm{for}y>0$$$$x^{\prime }=\frac{y}{2}-\frac{1}{2y}$$$$\sqrt{1+{\left(x^{\prime }\right)}^2}=\sqrt{\frac{{(y^2+1)}^2}{4y^2}}=\frac{y^2+1}{2y}[\mathrm{because}y>0]$$$$\int \frac{y^2+1}{2y}dy=\int \frac{y}{2}+\frac{1}{2y}dy=\frac{1}{4}{y}^2+\frac{1}{2}\ln y$$$$\Rightarrow \mathrm{arc}\mathrm{length}\mathrm{is}\frac{3}{4}+\frac{1}{2}\ln 2$$

Commented by bramlexs22 last updated on 18/Nov/20

$$thankyoubothsir$$

Answered by liberty last updated on 18/Nov/20

$$\ell =\underset{1}{\stackrel{2}{\int }}\sqrt{1+{\left({\mathrm{x}}^{\prime }\right)}^2}\mathrm{dy}$$$$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{1}{2}\mathrm{y}-\frac{1}{2\mathrm{y}}\Rightarrow \ell =\underset{1}{\stackrel{2}{\int }}\sqrt{1+{(\frac{1}{2}\mathrm{y}-\frac{1}{2\mathrm{y}})}^2}\mathrm{dy}$$$$\ell =\underset{1}{\stackrel{2}{\int }}\sqrt{1+\frac{1}{4}{\mathrm{y}}^2+\frac{1}{{4\mathrm{y}}^2}-\frac{1}{2}}\mathrm{dy}$$$$\ell =\underset{1}{\stackrel{2}{\int }}\sqrt{\frac{1}{4}{\mathrm{y}}^2+\frac{1}{2}+\frac{1}{{4\mathrm{y}}^2}}\mathrm{dy}$$$$\ell =\underset{1}{\stackrel{2}{\int }}(\frac{1}{2}\mathrm{y}+\frac{1}{2\mathrm{y}})\mathrm{dy}={[\frac{1}{4}{\mathrm{y}}^2+\frac{\ln \mathrm{y}}{2}]}_1^2$$$$\ell =\frac{3}{4}+\frac{\ln 2}{2}.▴$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com