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 $u(0)=1 , v(0)=0 and u^2(0)+v^2(0)=1$

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 !

  • 1
    $\begingroup$ How about $$ r(t)=\Big(1+t,t,(1+t)^2+t^2\Big)\;? $$ $\endgroup$
    – Kurt G.
    Sep 4 at 18:32
  • $\begingroup$ @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. $\endgroup$
    – GGG
    Sep 4 at 18:39
  • 1
    $\begingroup$ Can you calculate $r'(t)$ and then set $t$ to zero? If we did it right this should equal $w\,.$ $\endgroup$
    – Kurt G.
    Sep 4 at 18:51
  • 2
    $\begingroup$ 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. $\endgroup$
    – peek-a-boo
    Sep 4 at 18:54
  • $\begingroup$ @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 ! $\endgroup$
    – GGG
    Sep 4 at 22:55

1 Answer 1


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 $$ c(t) = (1, 0) + t(1, 1) = (1 + t, t). $$


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