Question

12. Find the radius of curvature of the
curve y2-3xy-AX24x34x4y+y5=0 at the
origin​

Answer 1

Answer:

Step-by-step explanation:

3x2+3y2y′=3ay+3axy′  

and isolating y’ yields

y′=3x2−3ay3ax−3y2=x2−ayax−y2

Then, to get y′′, we just differentiate y′ with respect to x:

y′′=(ax−y2)(2x−ay′)−(x2−ay)(a−2yy′)(ax−y2)2

(Exercise: show that y′′=2ax2y(ax−y2)3.)

Substituting x=y=3a2 gives us y′=−1, y′′=−323a and so

κ(3a2)=323a22–√=82–√3a

The radius of curvature is, of course, 1κ=3a82–√.

You can also do it using a parametrization and the usual curvature formulas for parametric space curves, but I’m not too thrilled about all those quotient rules. Then again, all things considered, it might be the same amount of work. For your information, the usual parametrization is

x=3at1+t3

y=3at21+t3

and the desired point is given by t=1