How to find equation for a position vector function on a parabaloid?

[Thadis]

Homework Statement

Velocity vector given by r'(t)=<cos(t), -sin(t), -2sin(t)>
Surface that the position vector needs to lie on: z=x²+y²

Homework Equations

Integral of r'(t) will give position function

The Attempt at a Solution

I know that the integral of r'(t) will give me the position function down to a constant so I know the vector is something like <sin(t)+C_1, cos(t)+C_2, 2cos(t)+C_3> I just do not know how to figure out the different C's for the components. I have tried putting the components into the surfaces equation but it didnt turn out correct. Anyone have any advice of how to solve this problem.

 

[LCKurtz]

Homework Statement

Velocity vector given by r'(t)=<cos(t), -sin(t), -2sin(t)>
Surface that the position vector needs to lie on: z=x²+y²

Homework Equations

Integral of r'(t) will give position function

The Attempt at a Solution

I know that the integral of r'(t) will give me the position function down to a constant so I know the vector is something like <sin(t)+C_1, cos(t)+C_2, 2cos(t)+C_3> I just do not know how to figure out the different C's for the components. I have tried putting the components into the surfaces equation but it didnt turn out correct. Anyone have any advice of how to solve this problem.
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Try plugging the components of that vector into the equation $z=x^2+y^2$ and see if you can pick values of your three constants to make it work.

 

[HallsofIvy]

The real difficulty is that the statement in the problem isn't true. Two object "miles" apart can have exactly the same velocity vector. An object having that velocity is not automatically costrained to lie on that surface. Doing what LCKurtz suggest would show that it is possible for an object having that velocity to stay on that surface. It would not show that it "needs to lie" on it.