Answer to Question #271916 in Differential Geometry | Topology for Angel Nodado

Question

Find the curvature, the radius of curvature and the center of curvature having a parametric equations at the given point:

y=e^x , (0,1)

Answer 1

Solution;

$y=e^x$

$P(x_0,y_0)=(0,1)$

$y'=e^x$

$y'(0)=1$

$y''=e^x$

$y''(0)=1$

Radius of curvature;

$R=\frac{(1+y'(x_0)^2)^{\frac32}}{y''(x_0)}$

By substitution ;

$R=\frac{(1+1)^{\frac32}}{1}=\sqrt{2^3}=2\sqrt2$

The curvature;

$k=\frac1R=\frac{1}{2\sqrt2}$

The centre of curvature;

Let P be the center of curvature , given as (a,b)

$a=x_o-\frac{\frac{dy}{dx}[1+(\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}$

$a=0+\frac{e^{2x}}{e^x}=e^0=1$

$b=y_o+\frac{1+(\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}$

$b=1+\frac{1+e^{2x}}{e^x}=1+2=3$

Therefore;

P=(1,3)