Finding The Coordinates of The Center Of Curvature

[Bashyboy]

Homework Statement

Let C be a curve given by y = f(x). Let K be the curvature (K \ne 0) and let z = \frac{1+ f'(x_0)^2}{f''(x_0)}. Show that the coordinates ( \alpha , \beta ) of the center of curvature at P are ( \alpha , \beta ) = (x_0 -f'(x_0)z , y_0 + z)

Homework Equations

The Attempt at a Solution

I attached a picture of the solution. The portion of the solution with a half-box I would appreciate someone helping me with.

 

[LCKurtz]

So, have you abandoned this thread?

https://www.physicsforums.com/showthread.php?t=699782

 

[Bashyboy]

No, it is no longer that day in which I learn Linear Algebra. Today is devoted to reviewing multivariable calculus and electricity and magnetism. Tomorrow is when I resume Linear Algebra.

 

[LCKurtz]

With respect to your question in this thread, if f'' > 0 the curve is concave up, so the circle is above the curve. At the point shown, f'' < 0 so the curve is concave down and the circle is below the curve. That determines whether $\beta$ is less or greater than $y_0$.

 

[Bashyboy]

Ooh, you are very right. Thank you much for your help.