Find trajectory

By symmetry, they must meet at the center (lets call it O). Suppose we look at the motion of A.

Taking OA as the initial line and the fact that OA = a√3, the distance r of the particle A from O at any angle θ, is given by the relation

r = a√3 e ^{&ndash √3 θ}

which gives the trajectory in the polar coordinate

sir can u post the solution :)

Fact Is That The Solution To This Problem Gives An Interesting Result . A Curve Traced Out By A

Particle In Which The Radius Vector Forms A Constant Angle With The Tangent , Is Known As

LOGARITHMIC SPIRAL . This In Fact Was Discovered By Jacob Bernoulli In The Year 1693 . Its

General Equation Is given By -

r = a e ^{b θ}

Where " a " And " θ " Are As Shown By Kaymant Sir In His Diagram . The Factor " b " Is Given By ,

b = tan α ,

Whence , " α " = 60 ^o ------ In This Case .

So , b = √ 3 .

And " a " = a√ 3 ------ Again As Shown By Kaymant Sir .

So , The Trajectory Is -

r = a√ 3 e ^{√ 3 θ}

- dadt = v - v cos 2 π3 ,

So , - dadt = 3 v2 , not √3 v2 .

But Here , r = a√3 .

So , - drdt = √ 3 v2 .

-\frac{\mathrm{dr}}{\mathrm {dt}}=\frac{\sqrt {3}v}{2} \\ rd\theta\cos 60^0 =vdt \\ \\ \frac{\mathrm{d\theta}}{\mathrm {dt}}=\frac{2v}{r}

-\int_{\frac{a}{\sqrt{3}}}^{r}{dr}=\frac{\sqrt{3}v}{2}\int_{0}^{t}{dt} \\ \frac{a}{\sqrt{3}}-r=\frac{\sqrt{3}vt}{2} \\ r=\frac{2a-3vt}{2\sqrt{3}} \\ \\ d\theta=\frac{4\sqrt{3}vdt}{2a-3vt} \\ -\theta =\frac{4}{\sqrt{3}}\int_{0}^{t}\frac{d(-3vt)}{2a-3vt} \\ -\frac{\sqrt{3}}{4}\theta=\ln{\left( 1-\frac{3vt}{2a}\right)} \\ e^{-\frac{\sqrt{3}}{4}\theta}=1-\frac{2a-2\sqrt{3}r}{2a} \\

e^{-\frac{\sqrt{3}}{4}\theta}=1-\frac{2a-2\sqrt{3}r}{2a} \\ e^{-\frac{\sqrt{3}}{4}\theta}=\frac{\sqrt{3}r}{a} \\ \\ r=\frac{a}{\sqrt{3}}e^{-\frac{\sqrt{3}}{4}\theta}

damn it missed by a factor of 4 in exponent :( :( [17]!!

We can take up a generalized case: A particle starting from a distance r₀ from a fixed point (O in the following figure) moves with a constant speed v such that the velocity vector always forms an angle α (0<α<Ï€/2) with the line joining the point O to the current location of the particle.

In a time dt, the particle moves an infinitesimal distance v dt. It's easy to see that
r dθ = v dt sin α ------- (1)
and
|dr| = v dt cos α
In our case, dr < 0, so that |dr| = - dr. Hence
&ndash dr = v dt cos α ------ (2)
Dividing (2) by (1), we get
&ndash dr r dθ =cot α
i.e. &ndash dr r =cot α dθ
Integrating and using the initial condition i.e. at θ=0, r = r₀, we get
r = r₀ e^{&ndash θ cot α}

@sonne
yes, now its okay.