Question

x2/3 +y2/3 = a2/3  find the radius of curvature at any point on the curve

Answer 1

[tex] x^{2/3} + y^{2/3} = a^{2/3} \\ \\ let\ x = a^3 cos^3\ \alpha,\ \ \ y = a^3\ sin^3\ \alpha \\ \\ Differentiate\ wrt\ \alpha\ \\ \\ \frac{dx}{d\alpha} = a^3\ 3cos^2\ \alpha (-sin\ \alpha) \\ \\ \frac{dy}{d\alpha} = a^3\ 3 sin^2\ \alpha (cos\ \alpha) \\ \\ \frac{dy}{dx} = - tan\ \alpha \\ \\ \frac{d^2y}{dx^2} = - sec^2\ \alpha\ *\ \frac{d\alpha}{dx} \\ \\ \ \ \ = \frac{-sec^2\ \alpha}{-a^3\ 3 cos^2\ \alpha * sin\ \alpha} \\ \\ [/tex]
[tex]\frac{1}{3a^3\ cos^4\ \alpha\ * \ sin\ \alpha} \\ \\ NOW,\ \ \ R\ =\ Radius\ of\ curvature\ at\ (x,y)\ =\ \ \frac{(1+tan^2\ \alpha)^3/2}{\frac{1}{3a^3\ cos^4\ \alpha\ * \ sin\ \alpha}} \\ \\ \ \ \ \ =\ sec^3\ \alpha\ *\ 3a^3\ *\ cos^4\ \alpha\ *\ sin\ \alpha \\ \\ \ \ \ \ \ = 3a^3\ cos\ \alpha\ sin\ \alpha \\ \\ 3a^3\ (xy)^{\frac{1}{3}} / a^2\\ \\ 3a (xy)^{\frac{1}{3}} \\ \\ [/tex]

Answer 2

Answer:

On comparing the given equation by

sin²t+ cos²t=1 we get,

x= a*sin³t

y= a*cos³t

dx/dt = 3a*sin²t*cost

dy/dt= -(3a* cos²t* Sint)

dy/dX= -(cot t)

d²y/dx² = cosec²t * dt/dx

now,

ρ = (1+cot²t)³/²

cosec²t *dt/dx

= cosec³t * 3a*sin²t*cost.

cosec²t

= 3a*Sint* cost

= 3a *x¹/³ * y¹/³

a¹/³* a¹/³

= 3(a * x * y) ¹/³