Question

the formula for radius of curvature in Cartesian coordinates is​

Answer 1

The formula for radius of curvature in Cartesian coordinates is $\boxed{\rho=\dfrac{(1+y_{1}^{2})^{\frac{3}{2}}}{y_{2}}}$.

More information:

For Cartesian equation $y=f(x)$, we have written the formula above for radius of curvature.

For parametric equation :

For parametric equation $x=\phi(t),y=\psi(t)$, the formula for radius of curvature is

$\quad\boxed{\rho=\dfrac{(x'^{2}+y'^{2})^{\frac{3}{2}}}{x'y''-y'x''}}$

For polar equation :

For polar equation $r=f(\theta)$, the formula for radius of curvature is

$\quad\boxed{\rho=\dfrac{(r_{2}^{2}+r_{1}^{2})^{\frac{3}{2}}}{r^{2}-2r_{1}^{2}-rr_{2}}}$

For tangential polar equation :

For tangential polar equation $p=f(\psi)$, the formula for radius of curvature is

$\quad\boxed{\rho=p+\dfrac{d^{2}p}{d\psi^{2}}}$

Answer 2

Cartesian form

A cartesian coordinate system is a coordinate system in a plane, that points out each value differently by a pair of coordinates of numerical form, which are signed distances to the value from two fixed perpendicular oriented lines, for which same unit length

For the radius of curvature in Cartesian coordinates, the formula is

• Let us assume C is a curve, which is defined by a real function and can be differentiated twice.

• Now let us assume that C is embedded in a cartesian plane.

By these parameters, The radius of curvature, denoted by  ρ, of C at a point $P=(x,y)$ is given by:

           ρ=   $\frac{(1+y^{2})^{\frac{3}{2} } }{ly"l}$

in which $y'=\frac{dy}{dx}$ is the  first  derivative of y with respect to x at P

             $y"=\frac{d^{2}y}{dx^{2} }$ is the 2nd derivative.

Hence For the radius of curvature in Cartesian coordinates, the formula is,

                    ρ=  $\frac{(1+y^{2})^{\frac{3}{2} } }{ly"l}$