9 contact of a circle with a curve curvature the

Info icon This preview shows pages 349–352. Sign up to view the full content.

View Full Document Right Arrow Icon
9. Contact of a circle with a curve. Curvature. * The general equation of a circle, viz. ( x - a ) 2 + ( y - b ) 2 = r 2 , (1) contains three arbitrary constants. Let us attempt to determine them so that the circle has contact of as high an order as possible with the curve y = f ( x ) at the point ( ξ, η ), where η = f ( ξ ). We write η 1 , η 2 for f 0 ( ξ ), f 00 ( ξ ). Differentiating the equation of the circle twice we obtain ( x - a ) + ( y - b ) y 1 = 0 , (2) 1 + y 2 1 + ( y - b ) y 2 = 0 . (3) If the circle touches the curve then the equations (1) and (2) are satisfied when x = ξ , y = η , y 1 = η 1 . This gives ( ξ - a ) 1 = - ( η - b ) = r/ p 1 + η 2 1 . If the contact is of the second order then the equation (3) must also be satisfied when y 2 = η 2 . Thus b = η + { (1 + η 2 1 ) 2 } ; and hence we find a = ξ - η 1 (1 + η 2 1 ) η 2 , b = η + 1 + η 2 1 η 2 , r = (1 + η 2 1 ) 3 / 2 η 2 . The circle which has contact of the second order with the curve at the point ( ξ, η ) is called the circle of curvature , and its radius the radius of curvature . The measure of curvature (or simply the curvature ) is the reciprocal of the radius: thus the measure of curvature is f 00 ( ξ ) / { 1 + [ f 0 ( ξ )] 2 } 3 / 2 , or d 2 η 2 1 + 2 3 / 2 . * A much fuller discussion of the theory of curvature will be found in Mr Fowler’s tract referred to on p. 324 .
Image of page 349

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
[VII : 151] ADDITIONAL THEOREMS IN THE CALCULUS 334 10. Verify that the curvature of a circle is constant and equal to the reciprocal of the radius; and show that the circle is the only curve whose curvature is constant. 11. Find the centre and radius of curvature at any point of the conics y 2 = 4 ax , ( x/a ) 2 + ( y/b ) 2 = 1. 12. In an ellipse the radius of curvature at P is CD 3 /ab , where CD is the semi-diameter conjugate to CP . 13. Show that in general a conic can be drawn to have contact of the fourth order with the curve y = f ( x ) at a given point P . [Take the general equation of a conic, viz. ax 2 + 2 hxy + by 2 + 2 gx + 2 fy + c = 0 , and differentiate four times with respect to x . Using suffixes to denote differen- tiation we obtain ax + hy + g + ( hx + by + f ) y 1 = 0 , a + 2 hy 1 + by 2 1 + ( hx + by + f ) y 2 = 0 , 3( h + by 1 ) y 2 + ( hx + by + f ) y 3 = 0 , 4( h + by 1 ) y 3 + 3 by 2 2 + ( hx + by + f ) y 4 = 0 . If the conic has contact of the fourth order, then these five equations must be satisfied by writing ξ , η , η 1 , η 2 , η 3 , η 4 , for x , y , y 1 , y 2 , y 3 , y 4 . We have thus just enough equations to determine the ratios a : b : c : f : g : h .] 14. An infinity of conics can be drawn having contact of the third order with the curve at P . Show that their centres all lie on a straight line. [Take the tangent and normal as axes. Then the equation of the conic is of the form 2 y = ax 2 + 2 hxy + by 2 , and when x is small one value of y may be expressed ( Ch. V , Misc. Ex. 22) in the form y = 1 2 ax 2 + ( 1 2 ah + x ) x 3 , where x 0 with x . But this expression must be the same as y = 1 2 f 00 (0) x 2 + { 1 6 f 000 (0) + 0 x } x 3 , where 0 x 0 with x , and so a = f 00 (0), h = f 000 (0) / 3 f 00 (0), in virtue of the result of Ex. lv . 15. But the centre lies on the line ax + hy = 0.]
Image of page 350
[VII : 152] ADDITIONAL THEOREMS IN THE CALCULUS 335 15. Determine a parabola which has contact of the third order with the ellipse ( x/a ) 2 + ( y/b ) 2 = 1 at the extremity of the major axis.
Image of page 351

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 352
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern