Question #b16d4

1 Answer
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=((d^2y)/(dx^2))/([1+((dy)/(dx))^2]^(3/2))

and for circle radius R

kappa=1/R

reverting to more economical prime notation and equating
so (y'')/([1+(y')^2]^(3/2)) = 1/R qquad star

R y'' = [1+(y')^2]^(3/2)

R^2 (y'')^2 = [1+(y')^2]^3

differentiate

R^2 2 y'' y''' = 3[1+(y')^2]^2 2 y' y''

cancel terms

R^2 y''' = 3[1+(y')^2]^2 y' qquad triangle

now from inverting star we can say that

([1+(y')^2]^(3/2))/(y'') = R/1

So
R^2 = ([1+(y')^2]^(3))/((y'')^2) qquad circ

pop circ into triangle

([1+(y')^2]^(3))/((y'')^2) y''' = 3[1+y'^2]^2 y'

cancel a few terms

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

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