Question #b16d4

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Answer 1 · 2016-07-11T17:51:35

see below

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