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- Thread starter makovx
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[tex]

\int dt\;\sqrt{(\dot{x}(t))^2 + (\dot{y}(t))^2}.

[/tex]

Now you have the path defined with a formula [itex]\theta=f(r)[/itex]. A systematic way to find the length of the path is to somehow return it to the parametrization in Cartesian coordinates. This is how it can be done. We think of the [itex]r[/itex] as the parameter that replaces the [itex]t[/itex] in the previous formula, and then define the Cartesian components with formulas

[tex]

x(r) = r\cos(f(r))

[/tex]

[tex]

y(r) = r\sin(f(r))

[/tex]

Then the length of the path is given by

[tex]

\int dr\; \sqrt{(\dot{x}(r))^2 + (\dot{y}(r))^2} = \cdots

[/tex]

and the task left to be done is to compute [itex]\dot{x}(r)[/itex] and [itex]\dot{y}(r)[/itex] with the usual derivation rules.

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- #4

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To write equations, write for example [ tex ] 1+1=2 [ /tex ], but don't put the spaces after and before characters "[" and "]". If you click an equation with the mouse button, you can see the code that has been used for it.

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i still can't write equations. i tried to use that [ tex ] but it turned out that it can't read the symbols i used such as the integral sign with the upper and lower limit as theta 2 and theta 1.

in words, the equation is like this.

S= integral (upper limit: theta 2, lower limit: theta 1) of square root (1+ (r squared)*(dr/dtheta)^2) dr.

in words, the equation is like this.

S= integral (upper limit: theta 2, lower limit: theta 1) of square root (1+ (r squared)*(dr/dtheta)^2) dr.

Last edited:

- #6

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[tex]

\int\limits_{\theta_1}^{\theta_2} \sqrt{1 + r^2\big(\frac{dr}{d\theta}\big)^2} dr

[/tex]

(Click the equation to see how it can be typed)

You are integrating with respect to r, but the integration limits are thetas. The derivative [tex]dr/d\theta[/tex] seems to be of a function [tex]r(\theta)[/tex], but we don't have this kind of function now.

In any case, this is very close to the expression I derived couple of minutes ago, so I'm guessing that you just made some mistakes when concentrating in the latex problems. It could be we are talking about the same equation still.

- #7

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[tex]

x(r) = r\cos(f(r))

[/tex]

[tex]

y(r) = r\sin(f(r))

[/tex]

[tex]

\int dr\; \sqrt{(\dot{x}(r))^2 + (\dot{y}(r))^2} = \cdots

[/tex]

will lead. I'm sure that this is one way to derive the formula you are talking about.

According to the usual derivation rules we get

[tex]

\dot{x}(r) = \cos(f(r)) - r f'(r) \sin(f(r))

[/tex]

Then derive same kind of formula for [tex]\dot{y}(r)[/tex], and compute what

[tex]

(\dot{x}(r))^2 + (\dot{y}(r))^2 = \cdots

[/tex]

is.

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