Expert: Sherry Wallin - 5/31/2009

Question
In a mathematical model of the growth of a sunflower head,
it is assumed that a time, t, seed number, n, has polar co-
ordinates r=(t-n)^1/2 and θ=nα, where α is constant. Find
the cartesian co-ordinates at time t of seeds n=0 and n=q,
where q is constant. Then use the distance formula to show
that the square of the distance between seeds n=0 and n=q
is equal to 2t-q-2(t^2-qt)^1/2cos9(qα).

Thanks

Answer
Hi Ntokozo~
    I think the problem is written incorrectly. I think the answer needs to be 2t -q-2(t^2-qt)^(1/2)

When n = 0  x1 = t^(1/2) and y1 = 0 --> (t^(1/2), 0)
When n = q  x2 = ((t-q)^(1/2)cos(qα)) y2 = ((t-q)^(1/2)sin(qα))

Using the square of the distance from the distance formula means we have
(x2-x1)^(1/2) + (x2-y2)^(1/2)
=((t-q)^(1/2)cos(qα)-t^(1/2))^2 + ((t-q)^(1/2)sin(qα))^2
= (t-q)cos^2(qα)-2t^(1/2)(t-q)^(1/2)+ t + (t-q)sin^2(qα)
= (t-q)(cos^2(qα)+ sin^2(qα))-2t^(1/2)(t-q)^(1/2)+ t
= (t-q)(1)-2t^(1/2)(t-q)^(1/2)+ t
= 2t - q -2t^(1/2)(t-q)^(1/2)
= 2t - q -2(t^2-qt)^(1/2)

Math Prof