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Question Number 11433 by FilupS last updated on 26/Mar/17

$$\mathrm{for}r=\frac{1}{\theta },\mathrm{show}\mathrm{that}\mathrm{the}\mathrm{arc}\mathrm{length}\mathrm{between}$$$$\theta =3{\pi }^{-1}\mathrm{and}\theta =n{\pi }^{-1}(\mathrm{where}n>3)\mathrm{is}\mathrm{aproxiately}$$$$\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{line}y=3{\pi }^{-1}$$$$\mathrm{between}\mathrm{the}\mathrm{same}\mathrm{bounds}.\mathrm{Or}\mathrm{show}\mathrm{otherwise}.$$$$$$

Commented by FilupS last updated on 26/Mar/17

Commented by @ANTARES_VY last updated on 26/Mar/17

$$\mathbf{what}\mathbf{is}\mathbf{the}\mathbf{name}\mathbf{of}\mathbf{the}\mathbf{program}..$$

Commented by FilupS last updated on 26/Mar/17

$$\mathrm{desmos}$$

Commented by mrW1 last updated on 26/Mar/17

$$y=r\sin \theta =\frac{\sin \theta }{\theta }$$$$\underset{\theta \to 0}{\lim }y=\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1$$$$thatmeansforsmall\theta thecurve$$$$r=\frac{1}{\theta }\approx theliney=1$$$$isthismaybethereasonforyour$$$$assumption?butitistrueonlyforsmall$$$$valuesof\theta .$$

Commented by FilupS last updated on 26/Mar/17

$$\mathrm{This}\mathrm{makes}\mathrm{sense}!\mathrm{Thanks}$$

Answered by mrW1 last updated on 26/Mar/17

$$curver=\frac{1}{\theta }:$$$$\frac{dr}{d\theta }=-\frac{1}{{\theta }^2}$$$$\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}=\sqrt{\frac{1}{{\theta }^2}+\frac{1}{{\theta }^4}}=\frac{\sqrt{1+{\theta }^2}}{{\theta }^2}$$$$L={\int }_{{\theta }_1}^{{\theta }_2}\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta ={\int }_{{\theta }_1}^{{\theta }_2}\frac{\sqrt{1+{\theta }^2}}{{\theta }^2}d\theta$$$${[-\frac{\sqrt{1+{\theta }^2}}{\theta }+\ln (\theta +\sqrt{1+{\theta }^2})]}_{{\theta }_1}^{{\theta }_2}$$$$=[\frac{\sqrt{1+{\theta }_1^2}}{{\theta }_1}-\frac{\sqrt{1+{\theta }_2^2}}{{\theta }_2}+\ln \left(\frac{{\vartheta }_2+\sqrt{1+{\theta }_2^2}}{{\theta }_1+\sqrt{1+{\theta }_1^2}}\right)]$$$$with{\theta }_1=3{\pi }^{-1}and{\theta }_2=n{\pi }^{-1}$$$$$$$$$$$$liney=3{\pi }^{-1}:$$$$\Rightarrow x=y\times \cot \theta =3{\pi }^{-1}\cot \theta$$$$x_1=3{\pi }^{-1}\cot {\theta }_1$$$$x_2=3{\pi }^{-1}\cot {\theta }_2$$$$L_1=\mid {\int }_{x_1}^{x_2}\sqrt{1+{\left(y^{\prime }\right)}^2}dx\mid =\mid {\int }_{x_1}^{x_2}dx\mid =x_1-x_2$$$$=3{\pi }^{-1}(\cot {\theta }_1-\cot {\theta }_2)$$$$=3{\pi }^{-1}[\cot \left(3{\pi }^{-1}\right)-\cot \left(n{\pi }^{-1}\right)]$$$$$$$$L\ne L_1$$

Commented by mrW1 last updated on 26/Mar/17

Commented by ajfour last updated on 26/Mar/17

$$\mathrm{length}\mathrm{of}\mathrm{curve}=\int \sqrt{{\left(\mathrm{rd}\theta \right)}^2+{\left(\mathrm{dr}\right)}^2}$$$$=\int \sqrt{{\mathrm{r}}^2+{\left(\frac{\mathrm{dr}}{\mathrm{d}\theta }\right)}^2}\mathrm{d}\theta$$

Commented by mrW1 last updated on 26/Mar/17

$$I=\int \frac{\sqrt{1+x^2}}{x^2}dx$$$$u=\sqrt{1+x^2}$$$$u^{\prime }=\frac{x}{\sqrt{1+x^2}}$$$$v=-\frac{1}{x}$$$$v^{{}^{\prime }}=\frac{1}{x^2}$$$$I=\int {uv}^{\prime }dx=uv-\int {vu}^{\prime }dx$$$$=-\frac{\sqrt{1+x^2}}{x}+\int \frac{xdx}{x\sqrt{1+x^2}}$$$$=-\frac{\sqrt{1+x^2}}{x}+\int \frac{dx}{\sqrt{1+x^2}}$$$$=-\frac{\sqrt{1+x^2}}{x}+\ln (x+\sqrt{1+x^2})+C$$

Commented by mrW1 last updated on 26/Mar/17

$$youareright!$$

Commented by ajfour last updated on 26/Mar/17

$$\int \sqrt{{\mathrm{r}}^2+{\left(\frac{\mathrm{dr}}{\mathrm{d}\theta }\right)}^2}=\int \sqrt{\frac{1}{{\theta }^2}+\frac{1}{{\theta }^4}}\mathrm{d}\theta$$$$=\int \frac{\sqrt{1+{\theta }^2}}{{\theta }^2}\mathrm{d}\theta$$

Commented by ajfour last updated on 26/Mar/17

$$\mathrm{and}\mathrm{if}{\varphi }^2=1+{\theta }^2$$$$\varphi \mathrm{d}\varphi =\theta \mathrm{d}\theta$$$$\int \frac{\sqrt{1+{\theta }^2}}{{\theta }^2}\mathrm{d}\theta =\int \frac{{\varphi }^2}{{({\varphi }^2-1)}^{3/2}}\mathrm{d}\mathrm{\varnothing }$$$$\mathrm{otherwise}\mathrm{if}\theta =\cot \varphi$$$$\int \frac{\sqrt{1+{\theta }^2}}{{\theta }^2}\mathrm{d}\theta =-\int \frac{\mathrm{cosec}{}^3\varphi }{\cot {}^2\varphi }\mathrm{d}\varphi$$$$=\int \frac{\sec {}^2\varphi }{\sin \varphi }\mathrm{d}\varphi \mathrm{unable}\mathrm{to}\mathrm{integrate}\mathrm{it}..$$

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