find the radius of the curvature at (acostheta^3,asintheta^3) on the curve x^2/3+y^2/3=a^2/3
radius of the curvature:
R=(1+(dy/dx)2)3/2∣d2y/dx2∣R=\frac{(1+(dy/dx)^2)^{3/2}}{|d^2y/dx^2|}R=∣d2y/dx2∣(1+(dy/dx)2)3/2
2x/3+2yy′/3=02x/3+2yy'/3=02x/3+2yy′/3=0
y′=−xyy'=-\frac{x}{y}y′=−yx
y′′=−y−xy′y2y''=-\frac{y-xy'}{y^2}y′′=−y2y−xy′
at point (acosθ3,asinθ3):(acos \theta^3 ,asin \theta ^3):(acosθ3,asinθ3):
y′=−cotθ3y'=-cot\theta^3y′=−cotθ3
y′′=−sinθ3−cos2θ3/sinθ3sin2θ3y''=-\frac{sin\theta^3-cos^2\theta^3/sin\theta^3}{sin^2\theta ^3}y′′=−sin2θ3sinθ3−cos2θ3/sinθ3
R=sin2θ3(1+cot2θ3)3/2sinθ3−cos2θ3/sinθ3R=\frac{sin^2\theta ^3(1+cot^2\theta^3)^{3/2}}{sin\theta^3-cos^2\theta^3/sin\theta^3}R=sinθ3−cos2θ3/sinθ3sin2θ3(1+cot2θ3)3/2
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments