Math CHALLENGE Question?

Question

Consider the following differential equations written in parametric form:
r " – r(θ’)^2 + 2sinθ = 0
r(θ ”) + 2*(r ’)(θ’) + 2cosθ = 0
where (‘) = d/dt
subject to initial conditions
r(0) = 1
r ’(0) = 0
θ(0) = π/2
θ’(0) = -1

Show that the solution can be written in the form:
(rcosθ)^2 + rsinθ = 1

Answer 1

Nice one! But pretty easy.

Let x = r cos θ, y = r sin θ. Skipping the tedious bits we have
x'' = cos θ (r'' - r(θ')^2) - sin θ (2r'θ' + rθ'')
= cos θ (-2 sin θ) - sin θ (-2 cos θ) from the equations
= 0
and also
y'' = sin θ (r'' - r(θ')^2) + cos θ (2r'θ' + rθ'')
= sin θ (-2 sin θ) + cos θ (-2 cos θ)
= -2.
So we have x = at + b, y = -t^2 + ct + d for some constants a, b, c, d.
At t = 0 we have r = 0, θ = π/2 so x = 0, y = 1. Hence we get
x = at, y = -t^2 + ct + 1

Also, x' = r' cos θ - r θ' sin θ
and y' = r' sin θ + r θ' cos θ
so x'(0) = 0 - 1(-1) (1) = 1 = a
and y'(0) = 0 + 1(-1) (0) = 0 = c

So x = t, y = 1 - t^2
which we can write as x^2 + y = 1
<=> (r cos θ)^2 + (r sin θ) = 1.