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What is the radius of curvature formula for an ellipse at slope = 1? 
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#1
Aug2413, 12:09 AM

P: 19

What is the radius of curvature formula for an ellipse at slope = 1?
I have found b^2/a, and a^2/b for the major and minor axis, but nothing for slope = 1. Thanks. 


#2
Aug2413, 09:06 AM

P: 543

MathWorld has a comprehensive entry on ellipses. Note that the measure of curvature given in that article, as is usual, is the reciprocal of the radius of curvature.



#3
Aug2413, 09:29 AM

P: 19

I read that page before I came here, but it didn't help with this specific problem. I don't need the radius from the center, or the focus. I need the radius of the line that is normal/perpendicular to slope =1. So, basically, a 45 degree angle from the edge of the ellipse to its center of curvature. I need the formula that describes this for any ellipse. I have come up with (b sqrt2)/((a^2/(b^2))+1) but it isn't accurate.



#4
Aug2413, 09:43 AM

P: 543

What is the radius of curvature formula for an ellipse at slope = 1?
Edit: or do you want the radius of curvature where y=x? again you can find the appropriate value of t and use in (59). 


#5
Aug2413, 10:41 AM

P: 19

Sorry, I'm not a mathematician (obviously) so I may be explaining it wrong. Let me try it with the help of a diagram.
I would like to know the formula that would give the length of the radius of curvature, for any ellipse, where the slope is at a 45 degree angle. In this diagram, it would be the formula for the length of the red line (if that line actually was the length of the radius). Please note that the radius may not (and most likely does not) end at the axis. I have been out of school for many decades, so this is not homework. And I am not asking for a specific answer. I am asking for the formula, in standard ellipse terms of a, b, c, e, etc. Thanks. 


#6
Aug2413, 10:59 AM

P: 19

R=b^2/a, and R=a^2/b describe the equivalent radius where the major and minor axis intersect with the ellipse. I would like to know the same sort of formula, but where the slope of the ellipse is at 45 degrees (instead of zero degrees, and 90 degrees).
(59) may contain the formula I need, somehow, but I don't have the math knowledge to distill it down to what I need from that. I also don't know what k or t are, and I don't know how something can use an exponent that is a fraction. I am probably working on a junior high math level at this point, which is why I came here for help. I understand R=b^2/a, and R=a^2/b, and want to know what their 45 degree equivalent would be, in the same simple nomenclature, if possible. Thanks for any help you can offer. 


#7
Aug2413, 11:37 AM

P: 543

Can I ask why you are interested in this answer?



#8
Aug2413, 01:12 PM

P: 97

the curvature is [tex]\kappa=\frac{x' y^"  y' x^" }{(x'^2+y'^2)^{3/2}}[/tex] for x = x(t) and y = y(t) [tex]R = {1\over \kappa}[/tex] R : the radius 


#9
Aug2413, 02:07 PM

P: 19

To janhaa: Sorry, that is completely over my head, and not in the form that I need it in (for my nonmath brain).
To anyone who may be able to help: The best way I know to describe what I am seeking is from my description above: R=b^2/a, and R=a^2/b are wellknow, common, and standard formulas for the radii where the major axis, and minor axis, intersect with the ellipse. I am seeking the equivalent formula, but where the slope of the ellipse is at 45 degrees, instead of zero degrees, or 90 degrees. It does seem like it should be possible, don't you think? Thanks to anyone and everyone for any help. 


#10
Aug2413, 02:48 PM

P: 97

If the blue point has coordinate: (x, 0)
The start of the red line has coordinate: (c, 0) theta, [tex]\theta[/tex]is the obtuse angle r is the radius (the red line) then [tex]\cos(\theta)=\frac{xc}{r}[/tex] 


#11
Aug2413, 04:10 PM

P: 19

Hi janhaa, thank you for your interest. Sorry, that was the best diagram I could find, but the blue point should be ignored. It is not part of the issue.
The issue is: if the red line (not drawn to scale) is the length of a radius, with its end point being on the ellipse (at the point where the ellipse has a slope of 45 degrees), and that same end is scribing the ellipse's radius of curvature (at slope = 45 degrees), how long is the red line? In other words, how long is the "red line" radius? The red line almost certainly will not end on the axis as shown in the diagram, that is just a coincidence. Thanks 


#12
Aug2413, 05:26 PM

P: 19

It is for part of a software program I am working on. I am 48 years old, so it is not for homework, if that is your concern. Have I described the problem well enough, in a way that is understandable/makes mathematical sense? Thanks 


#13
Aug2413, 07:02 PM

P: 543

The answer is going to lie somewhere between ## \frac{a^2}{b} ## and ## \frac{b^2}{a} ##, it would take about a dozen lines of maths to get it, and is almost certainly going to include terms such as ## (a^2+b^2)^{3/2} ## which you don't understand, so it would be a waste of time.
On top of that, I can't believe that in any program involving ellipses this is the only thing you are going to need. I suggest you find someone to work with on an ongoing basis  the underlying maths is not particularly hard but you won't achieve anything by writing code to solve equations without understanding it. 


#14
Aug2413, 07:09 PM

P: 543

And yes this is an adequate description of the problem:



#15
Aug2513, 12:32 AM

P: 19




#16
Aug2513, 04:02 AM

HW Helper
P: 7,171

Instead of using the parametric form the radius of curvature versus x =
r(x) =  ( 1+(y')^(2) ) ^(3/2) / (y'')  starting with equation for ellipse, and hoping I do the math right: (x^2/a^2) + (y^2/b^2) = 1 y = (b/a) ± sqrt*(a^2  x^2) y' = ± (b/a) x / sqrt(a^2  x^2) y'' = ± a b / (a^2  x^2)^(3/2) r(x) =  (1 + (b/a)^2 x^2 / (a^2  x^2) )^(3/2) / (a b / (a^2  x^2)^(3/2))  y' = 1 when x = ± a^2 / (sqrt(a^2 + b^2) plug this into r(x), which is a mess, but you'll have a formula. 


#17
Aug2513, 05:12 AM

P: 261

To help you out a little more:
As was said, [itex]R=\frac{(1+(y')^2)^{3/2}}{y''}[/itex] You want [itex](y')^2=1[/itex] As was in the previous post, [itex]y''=\frac{ab}{(a^2x^2)^{3/2}}[/itex] and [itex]x^2=\frac{a^4}{a^2+b^2}[/itex] Then [itex]y''=\frac{(a^2+b^2)^{3/2}}{a^2 b^2}[/itex] So the result is [itex]R= a^2 b^2 (\frac{2}{a^2+b^2})^{3/2}= a^2 b^2 \frac{2}{a^2+b^2} \sqrt{\frac{2}{a^2+b^2}}[/itex] I wrote the square root out separately if you don't like fractional exponents. 


#18
Aug2513, 01:12 PM

P: 19

To: rcgldr and chingel, thank you so much!! You guys rock! Your posts are very helpful. It really helps that you simplified it for me and put it in terms I can understand and relate to the things I know.
So, for the fractional exponent "3/2", the denominator is actually the root... so "2" is the square root in this case. And the numerator "3" is just the normal exponent. Is that correct? 


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