# How can one best link a circle to a straight line by an arc of continuously varying curvature?

Are there well known plane curves with equations of the form y=f(x), which specify an arc s in the Cartesian plane E-having distinct end points P1,P2-so that the following conditions may be satisfied?

Let R be a given positive real number. Let C be a circle in E having radius R, which does not intersect the x-axis and has its center on the positive y-axis. f(x) is a strictly decreasing function of x and the curvature of its graph s is a continuous and strictly decreasing function of arc length. P1 is the only intersection point of s and C. s and C have the same tangent at P1 and the curvature of s at P1 is 1/R. s is tangent to the x-axis at P2 and its curvature at P2 is zero.

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How is it possible that the graph of a strictly decreasing function is tangent to the x axis anywhere? – S. Carnahan Jun 2 '10 at 3:41
The function f(x)=-(x^3) is a strictly decreasing function of x that is tangent to the x-axis at x=0. It accomplishes this by crossing its tangent line at the point of tangency. In the situation I am considering, the only point of intersection of the graph s and the x-axis is an end-point of s. Consequently this problem doesw not arise. – Garabed Gulbenkian Jun 7 '10 at 14:30