# A problem from science fiction - intersecting curves

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1. May 6, 2013

### Emspak

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

A spaceship is traveling on a curved path, f(t) = (t, t2)

(We'll assume that the path isn't affected by gravity, this is a math problem, not physics :-) )

It has to release a pod to intersect a space station that has an orbit described by the following:

g(t) = (4+cos($\frac{πt}{8})$, 8-sin($\frac{πt}{8}$))

The pod will travel at a tangent from the curved path f(t) and I want to know at what time s I should release it to get to the space station and how long it will take.

3. The attempt at a solution

OK, so I know that since I am releasing the pod at time s, my first vector to add will be (s, s2). And those numbers will be constant.

I know that f'(t) = (1, 2t)

So that means that a pod will be zooming off the curve in a path described by (s+t, s2+2t). (correct?)

So to see if the pod will hit the space station and when, I need

(s+t, s2+2t) = (4+cos($\frac{π(s+t)}{8})$, 8-sin($\frac{π(s+t)}{8}$))

But after that I am stuck, and I feel like there is some simple step I am missing. I know that the time s of release can't be more than s=3 and less than 0 because if s=0 the line is horizontal and never hits the circle described, and if s≥3 it will miss the circle full stop. So there are some constraints here. I am almost there, I feel like.

2. May 6, 2013

### Simon Bridge

Not quite.

For the spacecraft, v(t)=(1,2t)
At time t=s, the speed of the spacecraft, and thus the pod, will be... ?
... and thus, the position-time equation of the pod will be... ?

3. May 6, 2013

### Emspak

A time t=s the speed of the spacecraft is f'(s) or (1,2s) so that's the speed of the pod.

The position of the pod would have to be (s+t, s2+(2s)t) after that, yes?

So you'd want (s+t, s2+(2s)t) = (4+cos$\frac{(s+t)π}{8}$, 8-sin$\frac{(s+t)π}{8}$)

That's still pretty ugly when you want to solve it, no?

4. May 6, 2013

### Simon Bridge

It's usually nastier than that because the pod will be travelling along a conic section as well.
And then - real orbits are seldom that tidy.

You want to solve for s - you'll find there are lots of solutions from the restraints you have specified.

5. May 6, 2013

### Emspak

OK, but when I try to do that I get s = 4-t+cos($\frac{(t+s)π}{8}$)

And I still have a lot of possibilities for both s and t. I do know that -1<s+t-4<1, b/c of the cosine function. And I know that s>0, and t>0 so for this to work 3< s+t < 5. Am I getting it right so far?

Further, s2+2st = 8-sin($\frac{(t+s)π}{8}$)

which gets me s(1+2t) = 8-sin($\frac{(t+s)π}{8}$)

or 8-s(1+2t)=sin($\frac{(t+s)π}{8}$)

since -1<8-s(1+2t) <1 7<s(1+2t)<9, yes?

6. May 6, 2013

### Simon Bridge

Yes you do - why would you expect any different?

You are using t as the elapsed time between release and capture.
You should find an equation for this time in terms of the release time and the speed of the pod.

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