| Posted on Sunday, 22 February, 2004 - 12:58 am: |
Q. Show that the curve with intrinsic equation s=y is a circle, radius 1
| Posted on Sunday, 22 February, 2004 - 11:46 am: |
The radius of curvature, R, is given by:
Therefore, the curve is a circle of radius 1. This is because the radius of curvature is the radius of the circle that best fits the curve at a given point (see http://www.ies.co.jp/math/java/calc/curve/curve.html ). We see, therefore, that the radius of the circle that best fits the curve at every point has radius 1. This shows it is a circle.
| Posted on Sunday, 22 February, 2004 - 02:50 pm: |
yes but is there a way to mathematically prove it?
| Posted on Sunday, 22 February, 2004 - 03:27 pm: |
Hi Umair, If s=y, coss=cosy dx/ds=cosy Therefore, dx/ds=coss sins=x+a sins=siny dy/dx=siny Therefore, dy/dx=sins -coss=y+b Therefore, (x+a)2+(y+b)2=1 which is the equation of a circle, radius 1. Hope that helps, Nicola