# Question #b16d4

Jul 11, 2016

see below

#### Explanation:

i dont see the point of this but i can blag it for you

we start with the definition of curvature which I lift from here

$\kappa = \frac{\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}}{{\left[1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}\right]}^{\frac{3}{2}}}$

and for circle radius R

$\kappa = \frac{1}{R}$

reverting to more economical prime notation and equating
so $\frac{y ' '}{{\left[1 + {\left(y '\right)}^{2}\right]}^{\frac{3}{2}}} = \frac{1}{R} q \quad \star$

$R y ' ' = {\left[1 + {\left(y '\right)}^{2}\right]}^{\frac{3}{2}}$

${R}^{2} {\left(y ' '\right)}^{2} = {\left[1 + {\left(y '\right)}^{2}\right]}^{3}$

differentiate

${R}^{2} 2 y ' ' y ' ' ' = 3 {\left[1 + {\left(y '\right)}^{2}\right]}^{2} 2 y ' y ' '$

cancel terms

${R}^{2} y ' ' ' = 3 {\left[1 + {\left(y '\right)}^{2}\right]}^{2} y ' q \quad \triangle$

now from inverting $\star$ we can say that

$\frac{{\left[1 + {\left(y '\right)}^{2}\right]}^{\frac{3}{2}}}{y ' '} = \frac{R}{1}$

So
${R}^{2} = \frac{{\left[1 + {\left(y '\right)}^{2}\right]}^{3}}{{\left(y ' '\right)}^{2}} q \quad \circ$

pop $\circ$ into $\triangle$

$\frac{{\left[1 + {\left(y '\right)}^{2}\right]}^{3}}{{\left(y ' '\right)}^{2}} y ' ' ' = 3 {\left[1 + y {'}^{2}\right]}^{2} y '$

cancel a few terms

$\left(1 + y {'}^{2}\right) y ' ' ' = 3 y ' {\left(y ' '\right)}^{2}$

can't guarantee that's the most efficient, it just worked out first time