Question

find the center of curvature for the point P (2,0) for y=ln (x-1)

Answer 1

What does the center of curvature mean?

Answer 2

it means, i believe find the center of the circle that is tangent to (2,0) that has the same curvature

Answer 3

http://mathworld.wolfram.com/RadiusofCurvature.html

k is the usual name for curvature; the radius is 1/k

k = y''/(1+(y')^2)^(3/2)

y=ln(x-1)
y' = (x-1)^(-1)
y'' = -(x-1)^(-2)

\[k=-\frac{1}{(x-1)^2}*\frac{1}{\sqrt[2]{(1+(x-1)^2)}^3}\]
\[k=-\frac{1}{(2-1)^2}*\frac{1}{\sqrt[2]{(1+(2-1)^2)}^3}\]
\[k=-\frac{1}{\sqrt[2]{2}^3}\]
\[k=-\frac{1}{\sqrt[2]{8}}\]
\[k=-\frac{1}{2\sqrt{2}}\]
\[R=\frac{1}{k}=-2\sqrt{2}\]

Answer 4

R is spose to be an absolute value so ignore the negative :)

now if we now the tangent to this curve at the point; then we know that "normal" (th eperp to the tangent) at this point as well

y' = (x-1)^(-1); at x=2 we get a slope of 1

our normal to the curve is the slope -1. which is a 45 degree angle; the distance to the center is 2sqrt(2)

Answer 5

im gonna say the center is at 2 down and 2 more over from (2,0) -> (4,2)