Question

A aprticle moves along a curve y=x^2 with constant speed of 10m/s when the particle is at origin then the radius of curvature of path is

Answer 1

Answer:

Radius of curvature = 1/2

Step-by-step explanation:

A aprticle moves along a curve y=x^2 with constant speed of 10m/s when the particle is at origin then the radius of curvature of path is

Speed is Constant

Radius of curvature = { ( 1 +  (dy/dx)²)^(3/2) } /(d²y/dx²)

y = x²

dy/dx = 2x

d²y/dx² = 2

Radius of curvature = { ( 1 + (2x)²)^(3/2)}/2

={ ( 1 + 4x²)^(3/2) } / 2

at origion x = 0

Radius of curvature = 1/2 at origin

Answer 2

Answer:

Hey mate your answer is 1/2

And hope you will understand