Question

Find the angle between the radius vector and tangent for the following polar curves
r^m=a^m(cosmteta +sinmteta )
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Answer 1

Answer:

We should be able to do this without having to go to Cartesian coordinates.

Let r^ be a unit vector in the radial direction and θ^ be an orthogonal unit vector so r^=(cos(θ),sin(θ)),θ^=(−sin(θ),cos(θ))

Now the position vector is

p=rr^

A tangent vector is

p′=r′r^+rr^′

=r′r^+rθ^

To find the angle ϕ between this tangent and the radius vector just use the dot product

p′⋅r^=cosϕ|p′||r^|

As r^ is of unit length

cosϕ=p′⋅r^p′⋅p′−−−−−√

cosϕ=r′r′2+r2−−−−−−√

So in this case r=a(1−cosθ) and r′=asinθ so

cosϕ=asinθa2sin2θ+a2(1−cosθ)2−−−−−−−−−−−−−−−−−−−√

=asinθa(√sin2θ+1−2cosθ+cos2θ)

=sinθ2−2cosθ−−−−−−−−√

Using double angle with θ=2α

=sin(2α)2−2cos(2α)−−−−−−−−−−√

=2sinαcosα2−2(1−2sin2α)−−−−−−−−−−−−−−√

=2sinαcosα2sinα

=cosα

So ϕ=θ2 .

We can see this answer works. If we plot the curve r = 1 - cos(th) we get a in black in the diagram below.

I’ve also plotted the radial line and tangent vectors at θ=π/2 (green), θ=3π/4 (red), and θ=π (blue). We observe it makes the angles π/4,3π/8 , and π/2.

Answer 2

Answer:

phi=π/4+m theta

Step-by-step explanation:

On differentiating the given equation on both sides, we get

m r^m-1 dr/dthetaa= a^m(mcos mtheta - m sin mtheta)

on cancelling m from both the sides we get

dr/ d theta = a^m(cos mtheta - sin mtheta) /r^m-1

tan phi = r d theta / dr

r * r ^ m-1/a^m(cos mtheta - sin mtheta)

on substituting for r^m

cancel a^m

divide the cos m thet) /1-tanπ/4*tan mtheta

on equating both sides

phi = π/4+m theta

= (1+tan mtheta) /(1-tan mtheta)

tanπ/4+tan mtheta