# Determining Motion from a Derivative

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1. Jan 7, 2015

### friendbobbiny

1. The problem statement, all variables and given/known data
Given position function r(t) and r'(t) = c X r(t), where c is some constant vector, describe the path of the particle. In other words, describe r(t).

2. Relevant equations
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3. The attempt at a solution

a) r'(t) points in the direction of motion. If we can understand how r'(t) changes direction, we can understand how r(t) moves.
b) r'(t) is the cross product of c and r(t). Therefore, the function is perpendicular to both c and r(t).
c) If r'(t) is perpendicular to c, r'(t) lies on a plane that c is normal to.
d) ---- No other leads

2. Jan 7, 2015

### RUber

Consider iterating a few times with Newton's method to see what happens.
For starters, assume that C = Identity and evaluate the cross product.

3. Jan 10, 2015

### friendbobbiny

what does C = identity mean, sorry? a constant function?

if c is given by <C1,C2,C3> and I cross that by r(t) <R1,R2,R3>, I get a cross product.:
<C2(R3) - C3(R2), C1(R3) - C3(R1), C1(R2) - C2(R1)>

I then multiply that cross product by time interval t. This gives the position function's next approximate position.

\$<t[C2(R3) - C3(R2)], t[C1(R3) - C3(R1)], t[C1(R2) - C2(R1)]t>

Subsequent iterations become very messy. Are you sure about this? No conceptual way of understanding the problem?

4. Jan 10, 2015

### haruspex

Dot both sides with r(t) and solve. What does this tell you about |r(t)|?

5. Jan 10, 2015

### vela

Staff Emeritus
You might find it helpful to choose a coordinate system to make the analysis easier. Orient the coordinate system so that $\vec{c}$ points along the z-axis. That'll give you a much simpler expression for $\vec{c}\times\vec{r}$.