Question

busy kida?

kal v nahi kiti c
:(


ajj kida busy o spam ans hi kr rhe ho​

Answer 1

Answer:

Busy log..... xD

Explanation:

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.