radius of curvature


by hawaiidude
Tags: curvature, radius
hawaiidude
hawaiidude is offline
#1
Mar20-04, 06:21 AM
P: 45
how do you calculate the radius of curvature and the center p(h,k) of the circle with respect to the curve and how do you do improper integrals? kinda forgot improper integrals.
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Chen
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#2
Mar20-04, 07:22 AM
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P: 1,006
[tex]R = \frac{[1 + (\frac{dy}{dx})^2]^\frac{3}{2}}{\frac{d^2y}{dx^2}} = \frac{[1 + f'(x)^2]^\frac{3}{2}}{f''(x)}[/tex]
Chen
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#3
Mar20-04, 07:31 AM
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Do you mean like this?

[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]

hawaiidude
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#4
Mar20-04, 01:54 PM
P: 45

radius of curvature


yes...i was just wondering for the radius of curvature, why is it the sercond derivative? i can compute the curvature but i walsy wondered why the second derivatie, after all, i believe a book said the first was acceptable..
Chen
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#5
Mar20-04, 04:33 PM
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That's not the part that concerns me, I don't understand why the square root of [tex]1 + f'(x)^2[/tex] is raised to the power of three.

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?
Chen
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#6
Mar21-04, 11:14 AM
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P: 1,006
Here's a little drawing that might help you understand what I did... still want to know where my mistake is.
Attached Thumbnails
arcs.gif  
lethe
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#7
Mar21-04, 12:08 PM
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P: 657
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]
Chen
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#8
Mar21-04, 12:17 PM
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P: 1,006
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]
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? [:)]
lethe
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#9
Mar21-04, 12:22 PM
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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
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.

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

and how do you get to the formula you posted? [:)]
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 radical
Chen
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#10
Mar21-04, 01:28 PM
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P: 1,006
Ah-ha! I see now how to get the correct formula for the curvature. Thank you very much.
hawaiidude
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#11
Mar21-04, 04:54 PM
P: 45
thank you all always a great help! now i think i understand the concept of radius of curvature proving


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