# Bee moving in spiral motion

1. Mar 19, 2014

### negation

1. The problem statement, all variables and given/known data

A bee goes out from its hive in a spiral path given in plane polar coordinates by r = bekt q = ct where b, k, and c are positive constants. Show that the angle between the velocity vector and the acceleration vector remains constant as the bee moves outward.
Solution
r(t) = bekt q(t) = ct

3. The attempt at a solution

r'(t) = bke^(kt)
r' ' (t) = bk^(2)e^(kt)

I'm lost as to what should I be doing with the information to make the jump to the proof.

2. Mar 19, 2014

### BvU

Not good to throw up your hands under 3). See guidelines.

How does one go about calculating an angle between vectors ?

3. Mar 19, 2014

### BvU

He, hello Negation!

Also some more explanation under 1) is needed. You should know better by now! I can deduce some things, so I can guess that $r = b \, e^{kt}$ and $\theta = q\, t$ but this way you make it too difficult for others. At least learn a bit about Go Advanced. It is a safe, easy environment. $\TeX$ is much more powerful, but also a lot tougher. Still worth the investment...

4. Mar 19, 2014

### BvU

Oh, and: if a vector in plane polar coordinates has two components, how come you only write down $\dot r$ and $\ddot r$ (this is shorthand for $dr\over dt$ and $d^2 r \over dt^2$) if you really need $\dot {\vec r}$ and $\ddot {\vec r}$ to say something about the angle ?

5. Mar 19, 2014

### negation

By using the dot product and which by definition states:

A.B = |A||B|cosΘ
Θ=arc cos[(A.B)/(|A||B|)]

I understand that there are 2 components. I trying to outline the idea first.
Suppose, if I can show that theta does not change, would I then have proven?

6. Mar 19, 2014

### BvU

Right! the dot product divided by the magnitudes should give you something constant. From now on, it is math all the way to the answer.

7. Mar 19, 2014

### BvU

Yes. Showing (A.B)/(|A||B|) is constant is sufficient.
Now, the exercise is written out in polar coordinates. You can do a lot of work transforming to cartesian, or you can stay with polar (r, Θ).

8. Mar 19, 2014

### negation

I'm a little disturbed by the amount of hassle having to make the conversion from polar to cartesian and vice versa. It's tedious and messy. Do you recommend learning lagrangian in first year physics or should I really focus on the existing math and physics in school first?

9. Mar 19, 2014

### BvU

Focus. But $x=r \cos\theta$ and $y = r \sin\theta$ is first year. So is the dot product in polar coordinates $(r_1, \theta_1)\cdot (r_2, \theta_2) = |r_1|\, |r_2|\, \cos(\theta_1 - \theta_2)$

There is no way to avoid math in phys, I would say. Philosophers might dream otherwise.