Question

Show that the radius of curvature at the point (a cos^3theta,a sin^3theta) on the curve x^2/3+y^2/3=a^2/3 is 3a sin theta cos theta

Answer 1

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The given parametric equations are

x = a cosθ ...(i)

y = a sinθ ...(ii)

Taking derivatives, we get

x' = - a sinθ (θ') ...(iii)

y' = a cosθ (θ') ...(iv)

Again, taking derivatives, we get

x" = - a cosθ (θ')² - a sinθ (θ") ...(v)

y" = - a sinθ (θ')² + a cosθ (θ") ...(vi)

Now, the radius of curvature (ρ)

= {(x'² + y'²)^(3/2)}/(x' y" - y' x")

= [{(a²sin²θ + a²cos²θ)(θ')²}^(3/2)]/{a²sin²θ (θ')³ - a² sinθ cosθ (θ') (θ") + a²cos²θ (θ')³ + a² sinθ cosθ (θ') (θ")}

= {(a² θ'²)^(3/2)}/(a²θ'³)

= (a³ θ'³)/(a² θ'³)

= a

Answer 2

Answer:

The given parametric equations are

x = a cosθ ...(i)

y = a sinθ ...(ii)

Taking derivatives, we get

x' = - a sinθ (θ') ...(iii)

y' = a cosθ (θ') ...(iv)

Again, taking derivatives, we get

x" = - a cosθ (θ')² - a sinθ (θ") ...(v)

y" = - a sinθ (θ')² + a cosθ (θ") ...(vi)

Now, the radius of curvature (ρ)

= {(x'² + y'²)^(3/2)}/(x' y" - y' x")

= [{(a²sin²θ + a²cos²θ)(θ')²}^(3/2)]/{a²sin²θ (θ')³ - a² sinθ cosθ (θ') (θ") + a²cos²θ (θ')³ + a² sinθ cosθ (θ') (θ")}

= {(a² θ'²)^(3/2)}/(a²θ'³)

= (a³ θ'³)/(a² θ'³)

= a