Find the radius of the curvature at any point of `y= sqrt(4ax)`

A particle is projected with speed u at angle theta to the horizontal. Find the radius of curvature at highest point of its trajectory

The radius of curvature at any point of r=ae^(theta cot alpha) is :

Find radius of curvature of common surface

A particle is projected with a velocity vecv=ahati+bhatj . Find the radius of curvature of the trajectory of the particle at the (f) point of projection (ii) highest point.

A particle is projected from ground at an angle theta with horizontal with speed u. The ratio of radius of curvature of its trajectory at point of projection to radius of curvature at maximum height is -

The sum of the intercepts made on the axes of coordinates by any tangent to the curve sqrt(x)+sqrt(y)=sqrt(a) is equal to

Find the equations of the tangent and the normal to the given curve at the indicated point : y^(2) = 4ax " at " ((a)/(m^(2)), (2a)/(m))

The sum of the intercepts made on the axes of coordinates by any tangent to the curve sqrt(x)+sqrt(y)=2 is equal to

A spherical soap bubble of radius 2 cm attached to the outside of a spherical bubble of radius 4 cm. Find the radius of curvature of the common surface.

The radius of curvature of the curve y^2=4x at the point (1, 2) is