1
$\begingroup$

Let $\alpha(t)$ be a regular curve. Suppose there is a point $a$ in $R^3$ such that $\alpha(t)-a$ is orthogonal to $T(t)$ for all t. Prove that $\alpha(t)$ lies on a sphere.

So, I let $\alpha(t)=(x(t), y(t),z(t))$ and $a=(a,b,c)$.

Since, $\alpha(t)-a$ is orthogonal to $T(t)$,

$<\alpha(t)-a,T(t)>=0$.

But I don't know how to solve this..

$\endgroup$
6
  • 1
    $\begingroup$ what is $T(t)$? $\endgroup$
    – Tsemo Aristide
    Commented Mar 22, 2018 at 13:56
  • $\begingroup$ @TsemoAristide $T(t)=\alpha'(t)$, clearly. $\endgroup$
    – SphericalTriangle
    Commented Mar 22, 2018 at 13:57
  • 1
    $\begingroup$ @JooE Compute the derivative of $\|\alpha(t)-a\|^2$ $\endgroup$
    – SphericalTriangle
    Commented Mar 22, 2018 at 13:59
  • $\begingroup$ @SphericalTriangle Actually, it should be $\frac{1}{\lVert \alpha'(t)\rVert}\alpha'(t)$, because it does not state that the curve is in arclength parametrization. Not that it matters too much. $\endgroup$
    – user228113
    Commented Mar 22, 2018 at 14:01
  • $\begingroup$ @G.Sassatelli Clearly, you don't know what you are talking about. $\endgroup$
    – SphericalTriangle
    Commented Mar 22, 2018 at 14:07

1 Answer 1

Reset to default
1
$\begingroup$

I believe that $T(t)=\alpha'(t)$, then compute the derivative of $\|\alpha(t)-a\|^2$, it is $2<\alpha(t)-a,\alpha'(t)>$.

$\endgroup$
1
  • $\begingroup$ I solved it! Thank you :) $\endgroup$
    – JooE
    Commented Mar 22, 2018 at 14:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .