Q.227. Calculus - Radius of curvature.

Question 227.
Show that the radius of curvature at the point θ on the curve
x = 3a cosθ - acos3θ, y = 3a sinθ - a sin3θ  is l 3asinθ l.

Answer 227.
x = 3acosθ - acos3θ
=> dx/dθ
= - 3asinθ + 3asin3θ
= 3a (sin3θ - sinθ)
= 6a cos2θsinθ

y = 3asinθ - asin3θ
=> dy/dθ
= 3acosθ - 3acos3θ
= 3a (cosθ - cos3θ)
= 6a sin2θ sinθ

dy/dx
= (dy/dθ) / (dx/dθ)
= (6a sin2θ sinθ) / (6a cos2θ sinθ)
= tan2θ

d^2y/dx^2
= d/dx(dy/dx)
= d/dθ dy/dx) * dx/dθ
= d/dθ (tan2θ) * [1/(6a cos2θ sinθ)]
= 2 sec^2 2θ / 6a cos2θ sinθ
= sec^3 2θ / (3a sinθ)

Radius of curvature to the curve y = f(x) at any point is given by

[1 + (dy/dx)^2]^(3/2) / l d^2y/dx^2 l
= [1 + tan^2 2θ]^(3/2) / l [sec^3 2θ / (3a sinθ)] l
= (sec^2 2θ)^(3/2) / l [sec^3 2θ / (3a sinθ)] l
= l 3asinθ l.

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