Question #94987
find the radius of curvature at the point (1,1) of the folium x^3+y^3=2yx
1
Expert's answer
2019-09-23T09:31:54-0400

y=2y3x23y22x,    y(1,1)=1.y'=\frac{2y-3x^2}{3y^2-2x},\;\;y'(1,1)=-1.

y=(18xy6y24x)y18xy+4y+6x2(3y22x)2,    y(1,1)=16.y''=\frac{(18xy-6y^2-4x)y'-18xy+4y+6x^2}{(3y^2-2x)^2},\;\;y''(1,1)=-16.

Radius of curvature:

R=1+y2)32y=28.R=|\frac{1+y'^2)^{\frac{3}{2}}}{y''}|=\frac{\sqrt{2}}{8}.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS