x2/3 +y2/3 = a2/3 find the radius of curvature at any point on the curve
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[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]
[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]
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4
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
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