0
$\begingroup$

I know $r^2 = (x-h)^2 + (y-k)^2$ (A)

Differentiate (A)

$0 = 2(x-h) + 2(y-k)^2y'$

$0 = x - h + (y-k)^2y'$ (B)

Differentiate B

$0 = 1 + 2(y-k)y'^2 + (y-k)^2y''$ (C)

At this point I can solve for h, solving for k I think I would have to expand the terms in (C). I am unable to get the answer: $[1 + y'^2]^3 = r^2(y'')^2$

Any help would be appreciated.

$\endgroup$

1 Answer 1

0
$\begingroup$

Use the formula (see here) for the radius of curvature :

$$R=\frac1r=\frac{(1+y'^2)^{3/2}}{y''}$$

Why is it necessary to square this relationship to have the final answer ? In order to take into account the upper and the lower part of the circles ( a circle has two cartesian equations).

$\endgroup$
1
  • $\begingroup$ Any comment, two days later ? $\endgroup$
    – Jean Marie
    Dec 12, 2023 at 23:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .