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Equation of osculating circles

  • Thread starter oreosama
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  • #1
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Homework Statement



Find equations of the osculating circles of the ellipse 9x^2 + 4y^2 = 36 at the points
(2,0) (0,3)


as far as I understand I need to get T and N at some point on the curve, cross them to get B and use that + a point to write a equasion of plane

i dont know know how to even begin doing this. what do I do if I need to make something parametric? this isn't something I've ever been taught well.
 

Answers and Replies

  • #2
HallsofIvy
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An "osculating circle" at a point on curve is a circle that passes through the point, whose center is on the "concave" side of the curve, has radius perpendicular to the curve, and the length of the radius is equal to the "radius of curvature" of the curve- i.e. the reciprocal of the curvature.

The derivative of 9x^2+ 4y^2= 36 is given by 18x+ 8yy'= 0 or y= (9/4)(x/y). At (2, 0), of course, that does not exist which means a tangent line is vertical and the perpendicular is horizontal. What is the radius of curvature at (2, 0)? At (0, 3) y'= 0 so the tangent line is horizontal and the perendicular is vertical. What is the radius of curvature at (0, 3)?

"understand I need to get T and N at some point on the curve"

The points at which you need to get them are (2, 0) and (0, 3), of course. But how do you know that if you don't know how to get T and N?

The curvature, [itex]\kappa[/itex] is defined by
[tex]\frac{dT}{ds}= \kappa N[/tex]
where T is the unit tangent vector and N is the unit normal vector. Since T, being "unit" has constant length, 1, its derivative is perpendicular to it, the normal vector. [itex]\kappa[/itex] is the length of dT/ds.

There are any number of formulas for curvature of a curve in the plane- it is a common topic in multivariable Calculus.

As always- when you "dont know how to even begin doing this", review the definitions!
 

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