Question-211018

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

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

thank you mr frix

Answered by Frix last updated on 27/Aug/24

r = a e^{k θ}

x (θ) = a e^{k θ} \cos θ

y (θ) = a e^{k θ} \sin θ

Length of curve

L (p, q) = \underset{p}{\int^{q}} \sqrt{{(\overset{∙}{x})}^{2} + {(\overset{∙}{y})}^{2}} d θ

\overset{∙}{x} = \frac{d [x (θ)]}{d θ} = a e^{k θ} (k \cos θ - \sin θ)

\overset{∙}{y} = \frac{d [y (θ)]}{d θ} = a e^{k θ} (\cos θ + k \sin θ)

\sqrt{{(\overset{∙}{x})}^{2} + {(\overset{∙}{y})}^{2}} = e^{k θ} ∣ a ∣ \sqrt{k^{2} + 1}

L (0, 2 π) =∣ a ∣ \sqrt{k^{2} + 1} \underset{0}{\int^{2 π}} e^{k θ} d θ =

= \frac{∣ a ∣ \sqrt{k^{2} + 1}}{k} (e^{2 k π} - 1)

Commented by Frix last updated on 27/Aug/24

I know {there}^{'} s another formula for the

length of a curve given by r = r (θ) but I

{don}^{'} t remember it . .

Commented by Ghisom last updated on 27/Aug/24

L (α, β) = \underset{α}{\int^{β}} \sqrt{r^{2} + {(\overset{∙}{r})}^{2}} d θ =

[r = a e^{k θ} \land \overset{∙}{r} = a k e^{k θ}]

=∣ a ∣ \sqrt{k^{2} + 1} \underset{0}{\int^{2 π}} e^{k θ} d θ

\dots

Commented by Frix last updated on 27/Aug/24

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