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Question Number 211018 by peter frank last updated on 26/Aug/24

Commented by peter frank last updated on 27/Aug/24

$$\mathrm{thank}\mathrm{you}\mathrm{mr}\mathrm{frix}$$

Answered by Frix last updated on 27/Aug/24

$$r=a{\mathrm{e}}^{k\theta }$$$$x\left(\theta \right)=a{\mathrm{e}}^{k\theta }\cos \theta$$$$y\left(\theta \right)=a{\mathrm{e}}^{k\theta }\sin \theta$$$$\mathrm{Length}\mathrm{of}\mathrm{curve}$$$$L(p,q)=\underset{p}{\stackrel{q}{\int }}\sqrt{{\left(\stackrel{\bullet }{x}\right)}^2+{\left(\stackrel{\bullet }{y}\right)}^2}d\theta$$$$\stackrel{\bullet }{x}=\frac{d\left[x\left(\theta \right)\right]}{d\theta }=a{\mathrm{e}}^{k\theta }(k\cos \theta -\sin \theta )$$$$\stackrel{\bullet }{y}=\frac{d\left[y\left(\theta \right)\right]}{d\theta }=a{\mathrm{e}}^{k\theta }(\cos \theta +k\sin \theta )$$$$\sqrt{{\left(\stackrel{\bullet }{x}\right)}^2+{\left(\stackrel{\bullet }{y}\right)}^2}={\mathrm{e}}^{k\theta }\mid a\mid \sqrt{k^2+1}$$$$L(0,2\pi )=\mid a\mid \sqrt{k^2+1}\underset{0}{\stackrel{2\pi }{\int }}{\mathrm{e}}^{k\theta }d\theta =$$$$=\frac{\mid a\mid \sqrt{k^2+1}}{k}({\mathrm{e}}^{2k\pi }-1)$$

Commented by Frix last updated on 27/Aug/24

$$\mathrm{I}\mathrm{know}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{another}\mathrm{formula}\mathrm{for}\mathrm{the}$$$$\mathrm{length}\mathrm{of}\mathrm{a}\mathrm{curve}\mathrm{given}\mathrm{by}r=r\left(\theta \right)\mathrm{but}\mathrm{I}$$$${\mathrm{don}}^{\prime }\mathrm{t}\mathrm{remember}\mathrm{it}..$$

Commented by Ghisom last updated on 27/Aug/24

$$L(\alpha ,\beta )=\underset{\alpha }{\stackrel{\beta }{\int }}\sqrt{r^2+{\left(\stackrel{\bullet }{r}\right)}^2}d\theta =$$$$[r=a{\mathrm{e}}^{k\theta }\wedge \stackrel{\bullet }{r}=ak{\mathrm{e}}^{k\theta }]$$$$=\mid a\mid \sqrt{k^2+1}\underset{0}{\stackrel{2\pi }{\int }}{\mathrm{e}}^{k\theta }d\theta$$$$...$$

Commented by Frix last updated on 27/Aug/24

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