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=ae^(kθ) x(θ)=ae^(kθ) cos θ y(θ)=ae^(kθ) sin θ Length of curve L (p, q) =∫_p ^q (√((x^• )^2 +(y^• )^2 ))dθ x^• =((d[x(θ)])/dθ)=ae^(kθ) (kcos θ −sin θ) y^• =((d[y(θ)])/dθ)=ae^(kθ) (cos θ +ksin θ) (√((x^• )^2 +(y^• )^2 ))=e^(kθ) ∣a∣(√(k^2 +1)) L (0, 2π) =∣a∣(√(k^2 +1))∫_0 ^(2π) e^(kθ) dθ= =((∣a∣(√(k^2 +1)))/k)(e^(2kπ) −1)](https://www.tinkutara.com/question/Q211050.png)
$${r}={a}\mathrm{e}^{{k}\theta} \\ $$$${x}\left(\theta\right)={a}\mathrm{e}^{{k}\theta} \mathrm{cos}\:\theta \\ $$$${y}\left(\theta\right)={a}\mathrm{e}^{{k}\theta} \mathrm{sin}\:\theta \\ $$$$\mathrm{Length}\:\mathrm{of}\:\mathrm{curve} \\ $$$${L}\:\left({p},\:{q}\right)\:=\underset{{p}} {\overset{{q}} {\int}}\sqrt{\left(\overset{\bullet} {{x}}\right)^{\mathrm{2}} +\left(\overset{\bullet} {{y}}\right)^{\mathrm{2}} }{d}\theta \\ $$$$\overset{\bullet} {{x}}=\frac{{d}\left[{x}\left(\theta\right)\right]}{{d}\theta}={a}\mathrm{e}^{{k}\theta} \left({k}\mathrm{cos}\:\theta\:−\mathrm{sin}\:\theta\right) \\ $$$$\overset{\bullet} {{y}}=\frac{{d}\left[{y}\left(\theta\right)\right]}{{d}\theta}={a}\mathrm{e}^{{k}\theta} \left(\mathrm{cos}\:\theta\:+{k}\mathrm{sin}\:\theta\right) \\ $$$$\sqrt{\left(\overset{\bullet} {{x}}\right)^{\mathrm{2}} +\left(\overset{\bullet} {{y}}\right)^{\mathrm{2}} }=\mathrm{e}^{{k}\theta} \mid{a}\mid\sqrt{{k}^{\mathrm{2}} +\mathrm{1}} \\ $$$${L}\:\left(\mathrm{0},\:\mathrm{2}\pi\right)\:=\mid{a}\mid\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\mathrm{e}^{{k}\theta} {d}\theta= \\ $$$$=\frac{\mid{a}\mid\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}}{{k}}\left(\mathrm{e}^{\mathrm{2}{k}\pi} −\mathrm{1}\right) \\ $$
Commented by Frix last updated on 27/Aug/24
$$\mathrm{I}\:\mathrm{know}\:\mathrm{there}'\mathrm{s}\:\mathrm{another}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{a}\:\mathrm{curve}\:\mathrm{given}\:\mathrm{by}\:{r}={r}\left(\theta\right)\:\mathrm{but}\:\mathrm{I} \\ $$$$\mathrm{don}'\mathrm{t}\:\mathrm{remember}\:\mathrm{it}.. \\ $$
Commented by Ghisom last updated on 27/Aug/24
![L (α, β) =∫_α ^β (√(r^2 +(r^• )^2 ))dθ= [r=ae^(kθ) ∧r^• =ake^(kθ) ] =∣a∣(√(k^2 +1))∫_0 ^(2π) e^(kθ) dθ ...](https://www.tinkutara.com/question/Q211052.png)
$${L}\:\left(\alpha,\:\beta\right)\:=\underset{\alpha} {\overset{\beta} {\int}}\sqrt{{r}^{\mathrm{2}} +\left(\overset{\bullet} {{r}}\right)^{\mathrm{2}} }{d}\theta= \\ $$$$\:\:\:\:\:\left[{r}={a}\mathrm{e}^{{k}\theta} \wedge\overset{\bullet} {{r}}={ak}\mathrm{e}^{{k}\theta} \right] \\ $$$$=\mid{a}\mid\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\mathrm{e}^{{k}\theta} {d}\theta \\ $$$$… \\ $$
Commented by Frix last updated on 27/Aug/24
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