# How to find a curve $r(t)=R(u(t),v(t))$ ,where $R: \mathbb{R}^2\rightarrow \mathbb{R}^3$ a paraboloid without knowing $u(t),v(t)$

The problem bellow is a subquestion from an exercize of a previous exam in university of Athens.

Let $$R(u,v)=(u,v,u^2+v^2)$$ be a paraboloid. Show that $$w=(1,1,2)$$ is tangent to $$R(q)$$ ,$$q=(1,0)$$.Also find the curve $$r(t)=R(u(t),v(t))$$ such as $$r(0)=R(q)=(1,0,1)$$ and $$r'(0)=w$$\

What I have done so far :
$$W$$ is in a tangent vector at $$q$$ because it belongs to $$q$$'s tangent space $$T_qR$$ ,since we can observe that $$w=R_u(q)+R_v(q)$$.
For the rest of the question. I've tried solving the problem in the form of of a differential equation but of as expected it lead nowhere since I dont have enough data to proceed or that's what I think at least.That's where im stuck right now : $$r(t)=(u(t),v(t),u^2(t)+v^2(t))$$, so $$u(0)=1 , v(0)=0 and u^2(0)+v^2(0)=1$$

$$r'(t)=R_{u(t)}u'(t)+R_{v(t)}v'(t)=...=(u'(t),v'(t),2u(t)u'(t)+2v(t)v'(t))$$
and hence $$u'(0)=1,v'(0)=1$$

My guess is that's a wrong way to proceed but it was the only thing that came to mind .Any help or tip would be really appreciated ,thank you in advance !

• How about $$r(t)=\Big(1+t,t,(1+t)^2+t^2\Big)\;?$$ Sep 4 at 18:32
• @kurtG It works for sure ,but is there a way to prove that's r(t). I'm not talking about just checking the values given ,but a way to smoothly prove that's the curve in question.
– GGG
Sep 4 at 18:39
• Can you calculate $r'(t)$ and then set $t$ to zero? If we did it right this should equal $w\,.$ Sep 4 at 18:51
• In general, there are infinitely many such curves, so you just have to find one and show it works, and you’re done. THe easiest way is essentially what was suggested above. Sep 4 at 18:54
• @peek-a-boo Yeah that's what was troubling me .My mind was stuck to the thought that there was a way and some attribute or lemma that I wasn't aware of that would give a unique solution .Thank you for your time !
– GGG
Sep 4 at 22:55

The paraboloid is described using the "Monge patch" $$R(u, v) = (u, v, u^{2} + v^{2})$$. Finding a path with initial position $$r(0) = R(1, 0) = (1, 0, 1)$$ and initial velocity $$r'(0) = w = (1, 1, 2) = DR(1, 0)(1, 1)$$ is consequently equivalent to finding a path $$c(t) = \bigl(u(t), v(t)\bigr)$$ satisfying
• $$c(0) = (1, 0)$$ and
• $$c'(0) = (1, 1)$$,
and setting $$r(t) = (R \circ c)(t) = R\bigl(c(t)\bigr)$$. The fact $$w$$ is tangent to the paraboloid ensures the third component of $$r'(0)$$ is $$2$$. (Why?)
Kurt's comment gives the unique affine choice of $$c$$, namely $$c(t) = (1, 0) + t(1, 1) = (1 + t, t).$$