Question

Show that in the equiangular spiral r = ae^{θ cot α}, the tangent is inclined at a constant angle to the radius vector.

Answer 1

The given equation r = ae^θcotα 

On differentiation gives, dr/dθ = αe^θcotα . cot α = r cot α  

so that  tan φ = r dθ/ dr = tan α or φ = α 

Hence the angle φ, the angle between the tangent and the radius vector is constant