# Find a curve

1. Sep 18, 2014

### Shackleford

Eh. I'm not quite sure how to find the curve γ(t). I think that the problem is probably a bit easier being given the parametrization of M. I do know that the point P lies in the xy-plane.

2. Sep 18, 2014

### BvU

Et tu, Shackle: Did you notice PF has a template ?

I have no clue what you mean with $v =(\frac{7}{2},2,3) \in T_p(M)$.
Enlighten me, and all those others who might want to help you...

3. Sep 18, 2014

### Shackleford

It's the tangent plane of M at P.

4. Sep 18, 2014

### BvU

Good. While you are explaining anyway, is the v in x(u,v) = (u, v, u2 - v2) also a point in this tangent plane ?

5. Sep 18, 2014

### Shackleford

The v is in the tangent plane, not necessarily in the surface, right?

6. Sep 18, 2014

### BvU

My guess is that there are two v floating around in the problem statement, and you need to make a distinction between them. Nice opportunity to catch up with the requirement to make use of the template!

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

2. Relevant equations

3. The attempt at a solution

7. Sep 18, 2014

### Shackleford

Yes, the v is a vector in the tangent plane, not the parameter v.

8. Sep 18, 2014

### BvU

Brilliant evasion of the chore to fill in the template. As if we're among real experts.
So now you have $\gamma(0)$ as a given, yielding you u(0) and v(0).
Your turn for $\gamma'(0)$