Lineaire differentiaalvergelijking van de tweede orde

[ukster]

aadkr.
Klopt dat het niet klopt, de booglengte is..

booglengte 376 keer bekeken

Wat er in je boek staat weet ik niet!

[Professor Puntje]

Dat dat zo moet zijn zie je uit onderstaande schetsje:

\( (ds)^2 = (r d \varphi)^2 + (dr)^2 \)

[aadkr]

[ukster]

dit zou LaTex code moeten zijn

Code: Selecteer alles

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\author{}
\date{}

\begin{document}

\[\sum_{}^{}{\Delta L = \sum_{}^{}\sqrt{(\Delta x)^{2} + (\Delta y)^{2}} = \sqrt{1 + \left( \frac{\Delta y}{\Delta x} \right)^{2}}\Delta x}\]

\[L = \int_{x1}^{x2}{\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}dx}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}\frac{dx}{d\theta}d\theta}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{1 + \left( \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \right)^{2}}\frac{dx}{d\theta}d\theta}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2}}d\theta}\]

\[\frac{dx}{d\theta} = \frac{d\ }{d\theta}(rcos\theta) = \frac{dr\ }{d\theta}cos\theta - rsin\theta\]

\[\frac{dy}{d\theta} = \frac{d\ }{d\theta}\left( r\sin\theta \right) = \frac{dr\ }{d\theta}\sin\theta + r\cos\theta\]

\[\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2} = \left( \frac{dr\ }{d\theta}cos\theta - rsin\theta \right)^{2} + \left( \frac{dr\ }{d\theta}sin\theta + rcos\theta \right)^{2}\]

\[\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2} = \left( \frac{dr\ }{d\theta}cos\theta \right)^{2} - 2r\frac{dr\ }{d\theta}cos\theta sin\theta + (rsin\theta)^{2} + \left( \frac{dr\ }{d\theta}sin\theta \right)^{2} + 2r\frac{dr\ }{d\theta}cos\theta sin\theta + \left( r\cos\theta \right)^{2}\]

\[\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2} = r^{2} + \left( \frac{dr\ }{d\theta} \right)^{2}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{{r^{2} + \left( \frac{dr\ }{d\theta} \right)}^{2}}d\theta}\]

\end{document}

[Professor Puntje]

Zo doe je dat:

\[\sum_{}^{}{\Delta L = \sum_{}^{}\sqrt{(\Delta x)^{2} + (\Delta y)^{2}} = \sqrt{1 + \left( \frac{\Delta y}{\Delta x} \right)^{2}}\Delta x}\]

\[L = \int_{x1}^{x2}{\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}dx}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}\frac{dx}{d\theta}d\theta}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{1 + \left( \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \right)^{2}}\frac{dx}{d\theta}d\theta}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2}}d\theta}\]

\[\frac{dx}{d\theta} = \frac{d\ }{d\theta}(rcos\theta) = \frac{dr\ }{d\theta}cos\theta - rsin\theta\]

\[\frac{dy}{d\theta} = \frac{d\ }{d\theta}\left( r\sin\theta \right) = \frac{dr\ }{d\theta}\sin\theta + r\cos\theta\]

\[\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2} = \left( \frac{dr\ }{d\theta}cos\theta - rsin\theta \right)^{2} + \left( \frac{dr\ }{d\theta}sin\theta + rcos\theta \right)^{2}\]

\[\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2} = \left( \frac{dr\ }{d\theta}cos\theta \right)^{2} - 2r\frac{dr\ }{d\theta}cos\theta sin\theta + (rsin\theta)^{2} + \left( \frac{dr\ }{d\theta}sin\theta \right)^{2} + 2r\frac{dr\ }{d\theta}cos\theta sin\theta + \left( r\cos\theta \right)^{2}\]

\[\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2} = r^{2} + \left( \frac{dr\ }{d\theta} \right)^{2}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{{r^{2} + \left( \frac{dr\ }{d\theta} \right)}^{2}}d\theta}\]

[ukster]

Mooi.. wat deed ik niet goed?

[Professor Puntje]

Kop en staart moeten er af. En het moet niet als code gepost worden.

[ukster]

dus gewoon copy paste?

[Professor Puntje]

Als je mijn berichtje met je formules citeert zie je precies hoe het wel werkt.

[ukster]

Juist ja:

\[\sum_{}^{}{\Delta L = \sum_{}^{}\sqrt{(\Delta x)^{2} + (\Delta y)^{2}} = \sqrt{1 + \left( \frac{\Delta y}{\Delta x} \right)^{2}}\Delta x}\]

\[L = \int_{x1}^{x2}{\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}dx}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}\frac{dx}{d\theta}d\theta}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{1 + \left( \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \right)^{2}}\frac{dx}{d\theta}d\theta}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2}}d\theta}\]

\[\frac{dx}{d\theta} = \frac{d\ }{d\theta}(rcos\theta) = \frac{dr\ }{d\theta}cos\theta - rsin\theta\]

\[\frac{dy}{d\theta} = \frac{d\ }{d\theta}\left( r\sin\theta \right) = \frac{dr\ }{d\theta}\sin\theta + r\cos\theta\]

\[\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2} = \left( \frac{dr\ }{d\theta}cos\theta - rsin\theta \right)^{2} + \left( \frac{dr\ }{d\theta}sin\theta + rcos\theta \right)^{2}\]

\[\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2} = \left( \frac{dr\ }{d\theta}cos\theta \right)^{2} - 2r\frac{dr\ }{d\theta}cos\theta sin\theta + (rsin\theta)^{2} + \left( \frac{dr\ }{d\theta}sin\theta \right)^{2} + 2r\frac{dr\ }{d\theta}cos\theta sin\theta + \left( r\cos\theta \right)^{2}\]

\[\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2} = r^{2} + \left( \frac{dr\ }{d\theta} \right)^{2}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{{r^{2} + \left( \frac{dr\ }{d\theta} \right)}^{2}}d\theta}\]

[Professor Puntje]

Juist ja:

\[\sum_{}^{}{\Delta L = \sum_{}^{}\sqrt{(\Delta x)^{2} + (\Delta y)^{2}} = \sqrt{1 + \left( \frac{\Delta y}{\Delta x} \right)^{2}}\Delta x}\]

\[L = \int_{x1}^{x2}{\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}dx}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}\frac{dx}{d\theta}d\theta}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{1 + \left( \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \right)^{2}}\frac{dx}{d\theta}d\theta}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2}}d\theta}\]

\[\frac{dx}{d\theta} = \frac{d\ }{d\theta}(rcos\theta) = \frac{dr\ }{d\theta}cos\theta - rsin\theta\]

\[\frac{dy}{d\theta} = \frac{d\ }{d\theta}\left( r\sin\theta \right) = \frac{dr\ }{d\theta}\sin\theta + r\cos\theta\]

\[\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2} = \left( \frac{dr\ }{d\theta}cos\theta - rsin\theta \right)^{2} + \left( \frac{dr\ }{d\theta}sin\theta + rcos\theta \right)^{2}\]

\[\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2} = \left( \frac{dr\ }{d\theta}cos\theta \right)^{2} - 2r\frac{dr\ }{d\theta}cos\theta sin\theta + (rsin\theta)^{2} + \left( \frac{dr\ }{d\theta}sin\theta \right)^{2} + 2r\frac{dr\ }{d\theta}cos\theta sin\theta + \left( r\cos\theta \right)^{2}\]

\[\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2} = r^{2} + \left( \frac{dr\ }{d\theta} \right)^{2}\]

\[L = \int_{\theta 1}^{\theta 2}{\sqrt{{r^{2} + \left( \frac{dr\ }{d\theta} \right)}^{2}}d\theta}\]


En dan nog daar weer kop en staart (de quote tags) vanaf zodat je als hierboven sec de formules krijgt.

[aadkr]

bedankt voor de hulp maar eerst nog een ander probleem die som met y kwadraat=2.p.x

[Professor Puntje]

De integraal is in twee delen gesplitst en het tweede deel is al gelijk geïntegreerd.

[aadkr]

[aadkr]