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The following is a (may be simple) problem from Docarmo's Differential Geometry.

Let $\alpha\colon (a,b)\rightarrow \mathbb{R}^3$ be a parametrized curve which do not pass through origin. If $\alpha(t_0)$ is a point on the trace (image) of $\alpha$ closest to the origin and $\alpha'(t)\neq 0$, then $\alpha(t_0)$ is orthogonal to $\alpha'(t_0)$.

I do not understand "..closest to origin" in the statement. Does it mean that $|\alpha(t_0)|\leq |\alpha(t)|$ for all $t\in (a,b)$ ?

Further, I tried to prove that the dot product $\alpha(t_0).\alpha'(t_0)=0$. One way to prove this may be to show that the derivative of $\alpha(t).\alpha(t)$ at $t=t_0$ is zero. But I couldn't proceed further. Can anyone help me?

(Since this is my reading course, I couldn't contact with any teacher now, therefore, posting the (simple) problem here which I couldn't solve.)

Groups
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1 Answers1

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If $\alpha(t)$ is closest to the origin at $t=t_0$, then $\left|\alpha(t)\right|^2$, i.e., $\alpha(t)\cdot \alpha(t)$, is minimized at $t=t_0$. Now, differentiate!

Hope this helps!

Edit: Oh, I'm sorry! I should have added that there is a rule for differentiating the dot product of two vector-valued functions: if $f(t)$ and $g(t)$ are vector-valued functions in three-space $\mathbb{R}^3$, then can you prove that the derivative of $f(t)\cdot g(t)$ is $f'(t)\cdot g(t) + f(t)\cdot g'(t)$? Note that this is analogous to the usual product rule but here $\cdot$ signifies "dot product".

Amitesh Datta
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    Yes. This was the rule I was following, which is stated in the book just before the exercise. Also, $t\mapsto \alpha(t).\alpha(t)$ is differentiable (am I correct?), with minimum value at $t=t_0$, therefore, its derivative at $t=t_0$ must be zero. This proves (if my first statement is correct!) – Groups Sep 03 '14 at 09:13
  • Yes, @Groups, you are correct - this follows from the fact that sums and products of differentiable functions are differentiable, as well as the relevant rules for differentiating such sums and products! – Amitesh Datta Sep 03 '14 at 09:14
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    Thanks for nice answer. This helps too. – Groups Sep 03 '14 at 09:15
  • Sorry to revive a dead thread, but I am wondering how you know that "if $\alpha$(t) is closed to the origin at t=$t_0$, then |$\alpha$(t)|$^2$ is minimized at t = $t_0$ ? – jmoore00 Jan 31 '18 at 00:50
  • Hi @agra94, the expression $\left|\alpha(t)\right|^2$ is the square of the distance of $\alpha(t)$ to the origin. – Amitesh Datta Jan 31 '18 at 15:21