Question

Prove that the radius of curvature at a point (a cos3 e’a sinn3 ) on the curve x ^(2/3)+ y ^(2/3)= a^(2/3) is 3a sin theta cose theta ​

Answer 1

Answer:

Step-by-step explanation:

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