y′=2y−3x23y2−2x, y′(1,1)=−1.y'=\frac{2y-3x^2}{3y^2-2x},\;\;y'(1,1)=-1.y′=3y2−2x2y−3x2,y′(1,1)=−1.
y′′=(18xy−6y2−4x)y′−18xy+4y+6x2(3y2−2x)2, y′′(1,1)=−16.y''=\frac{(18xy-6y^2-4x)y'-18xy+4y+6x^2}{(3y^2-2x)^2},\;\;y''(1,1)=-16.y′′=(3y2−2x)2(18xy−6y2−4x)y′−18xy+4y+6x2,y′′(1,1)=−16.
Radius of curvature:
R=∣1+y′2)32y′′∣=28.R=|\frac{1+y'^2)^{\frac{3}{2}}}{y''}|=\frac{\sqrt{2}}{8}.R=∣y′′1+y′2)23∣=82.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments