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3.8 Arc length calculation

Example.

Find the arc length along the logarithmic spiral r=eθr=e^{\theta}, from the point θ=0\theta=0, r=1r=1 to the point θ=π\theta=\pi, r=eπr=e^{\pi}.

Solution. Recall that x=r⁢cos⁡θx=r\cos\theta, y=r⁢sin⁡θy=r\sin\theta. Thus the logarithmic spiral is the parametrized curve given by x=eθ⁢cos⁡θ,x=\,{e^{\theta}\cos\theta,}y=eθ⁢sin⁡θ.y=\,{e^{\theta}\sin\theta.} Hence

d⁢xd⁢θ=eθ⁢(cos⁡θ-sin⁡θ), d⁢yd⁢θ=eθ⁢(cos⁡θ+sin⁡θ).\frac{dx}{d\theta}=\,{e^{\theta}(\cos\theta-\sin\theta),}\;\;\frac{dy}{d\theta% }=\,{e^{\theta}(\cos\theta+\sin\theta).}

Therefore (d⁢xd⁢θ)2+(d⁢yd⁢θ)2= 2⁢e2⁢θ.\left(\frac{dx}{d\theta}\right)^{2}+\left(\frac{dy}{d\theta}\right)^{2}=\,{2e^% {2\theta}.} We deduce that

L=∫0π2⁢eθ⁢d⁢θ=[2⁢eθ]0π=2⁢(eπ-1).L=\,{\int_{0}^{\pi}\sqrt{2}e^{\theta}\,d\theta}\,{=\left[\sqrt{2}e^{\theta}% \right]_{0}^{\pi}}\,{=\sqrt{2}(e^{\pi}-1).}