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- Thread starter hawaiidude
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[tex]\int _1 ^\infty \frac{1}{x^2}dx = \lim _{t \rightarrow \infty} \int _1 ^t \frac{1}{x^2}dx = \lim _{t \rightarrow \infty} 1 - \frac{1}{t} = 1[/tex]

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I tried to find the radius myself. It is given by:

[tex]R = \frac{\mbox{arc}}{\theta}[/tex]

I found [tex]\theta[/tex] using the slopes of the tangent lines. The angle between them is given by the formula:

[tex]\tan \theta = \lim_{dx \rightarrow 0}\frac{m_1 - m_2}{1 + m_1m_2}[/tex]

And since [tex]\theta[/tex] is very small you can say that [tex]\tan \theta = \sin \theta = \theta[/tex]:

[tex]\theta = \lim_{dx \rightarrow 0}\frac{f'(x_1) - f'(x_2)}{1 + f'(x)^2}[/tex]

To find the arc length I just used the normal [tex]d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex] formula:

[tex]\mbox{arc} = \lim_{dx \rightarrow 0}\sqrt{(x_2 - x_1)^2 + (f(x_2) - f(x_1))^2} = [/tex]

[tex]\mbox{arc} = \lim_{dx \rightarrow 0}f(x_2) - f(x_1)[/tex]

From this R is:

[tex]R = \lim_{dx \rightarrow 0}\frac{1 + f'(x)^2}{\frac{f'(x_1) - f'(x_2)}{f(x_2) - f(x_1)}} = \frac{1 + f'(x)^2}{f''(x)}[/tex]

So where's my mistake?

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Originally posted by Chen

Here's a little drawing that might help you understand what I did... still want to know where my mistake is.

maybe here?

Originally posted by Chen

To find the arc length I just used the normal [tex]d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex] formula:

[tex]\mbox{arc} = \lim_{dx \rightarrow 0}\sqrt{(x_2 - x_1)^2 + (f(x_2) - f(x_1))^2} = [/tex]

[tex]\mbox{arc} = \lim_{dx \rightarrow 0}f(x_2) - f(x_1)[/tex]

i don't follow that last step, but that is not the normal formula for ds that i know. it should look like:

[tex]

ds=\sqrt{1+(f')^2}dx[/tex]

- #8

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Well, I just used the formula of distance between two points. I took two points on the graph [[tex]x_1, f(x_1)[/tex]] and [[tex]x_2, f(x_2)[/tex]] with [tex]x_2 - x_1[/tex] tending to zero.... how is that wrong, and how do you get to the formula you posted?Originally posted by lethe

i don't follow that last step, but that is not the normal formula for ds that i know. it should look like:

[tex]

ds=\sqrt{1+(f')^2}dx[/tex]

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yeah, dx tends to zero, but so does f(x2)-f(x1), so to zeroth order, ds is just zero. you can't lose one term but keep the other.Originally posted by Chen

Well, I just used the formula of distance between two points. I took two points on the graph [[itex]x_1, f(x_1)[/itex]] and [[itex]x_2, f(x_2)[/itex]] with [itex]x_2 - x_1[/itex] tending to zero.... how is that wrong

but in calculus, we are not interested in zeroth order approximations, we are interested in first order approximations.

multiply and divide the whole thing by x2-x1. do your division inside the radical (which means you are actually dividing by (x2-x1)^2) and your multiplication outside the radicaland how do you get to the formula you posted?

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Ah-ha! I see now how to get the correct formula for the curvature. Thank you very much.

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