What is the angle between a position vector and a tangent?

Question

If r=f(θ) is a differentiable curve and P(r₀,θ₀) is a point on that curve, derive the "formula" for the angle between the position vector and the tangent to f at P.

Answer 1

saying Q is angle from x axis for the tangent

and θ is angle from x axis for the position vector

want Q-θ

know tanQ=dy/dx

know tanθ=y/x

tan(Q-θ)=[tanQ-tanθ]/[/1+(tanQ)(tanθ)]

have tan(Q-θ)=[(dy/dx)+y/x]/[1+((dy/dx)(y/x)]

=[xy'-y]/[x+yy'],,,y'=dy/dx

introducing r(θ)

dy/dx=[dy/dθ]/[dx/dθ]

dy/dθ=r'sinθ+rcosθ,,,r'=dr/dθ

dx/dθ=r'cosθ-rsinθ

x=rcosθ

y=rsinθ

now

tan(Q-θ)=[rcosθ(r'sinθ+rcosθ)+-rsinθ(r'cosθ-rsinθ)]/[rcosθ(r'cosθ-rsinθ)+rsinθ(r'sinθ+rcosθ)]

tan(Q-θ)=r/r' ,,,,,,r'=dr/dθ

example,,have curve y=x²

in polar coordinates,

r=tanθsecθ

r'=(tan²θ)secθ+sec³θ

r/r'=tanθ/[tan²θ+sec²θ]

at (2,4)

Q=tan^-1(dy/dx)=tan^-14,,, θ=tan^-1(y/x)=tan^-12

tan(Q-θ)=[4-2]/[1+2*4]=2/9

r=2sec(tan^-12)=2√5 or r=√(2²+4²)=2√5

θ=tan^-12

r'=4sec(tan^-12)+sec³(tan^-12)=4√5+5√5

tan(Q-θ)=r/r'=2/9

Q-θ=12.61 degrees at (2,4),,,(2√5,tan^-12)