# Parametrized curve and tangent orthogonal to position vector at t0:A question

• Apr 16th 2012, 04:04 PM
x3bnm
Parametrized curve and tangent at t0 orthogonal to position vector at t0:A question
Do Carmo's book on "Differential geometry of curves and surfaces" has a problem that I can't find the answer of. It is on page 5(1st exercise problem#2) is:

Let $\alpha(t)$ be a parametrized curve which does not pass through the origin. If $\alpha(t_0)$ is the point of the trace of $\alpha$ closest to the origin
and $\alpha'(t_0) \neq 0$, show that the position vector $\alpha(t_0)$ is orthogonal to $\alpha'(t_0)$.

Here $\alpha'(t_0)$ is the derivative of $\alpha(t)$ at $t_0$.

How do I prove this? Is it possible to give some hints?
• Apr 17th 2012, 10:25 AM
HallsofIvy
Re: Parametrized curve and tangent orthogonal to position vector at t0:A question
The crucial point is "closest to the origin".

One way to derive the result is: If $\alpha(t)= x(t)\vec{i}+ y(t)\vec{j}+ z(t)\vec{k}$ then the distance to the origin is $D= \sqrt{x^2+y^2+ z^2}$ and the point will be closest to the origin when the derivative of that function is 0. $D'= (1/2)(x^2+ y^2+z^2)^{-1/2}(2xx'+ 2yy'+ 2zz')= \frac{xx'+ yy'+ zz'}{\sqrt{x^2+ y^2+ z^2}}= 0$. A fraction is equal to 0 if and only if its numerator is 0 so that is the same as $xx'+ yy'+ zz'= 0$ which is the same as the dot product of $\alpha(t)$ and its derivative.

Or, geometrically, the line from a point, (0, 0), to a curve, ( $\alpha(t)$), will be minimum if and only if the line from the point to the curve is perpendicular the curve. Here, the line tangent to the curve is in the direction of the tangent vector, $\alpha'(t)$, and the line from the origin to the point is in the direction of the "position vector", $\alpha(t)$.
• Apr 17th 2012, 03:38 PM
x3bnm
Re: Parametrized curve and tangent orthogonal to position vector at t0:A question
Thank you HallsofIvy.