Question

Question #b16d4

Answer 1

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