property of a curve $\alpha(t)$

Question

Find a parametrized curve $\alpha(t)$ whose trace is the circle $x^2 + y^2 = 1$ such that $\alpha(t)$ runs clockwise around the circle with $\alpha(0) = (0, 1)$.
A parametrized curve $\alpha(t)$ has the property that its second derivative $\alpha''(t)$ is identically zero. What can be said about $\alpha$ ?
Let $\alpha:I \to \mathbb{R}^3$ be a parametrized curve and let $v \in \mathbb{R}^3$ be a fixed vector. Assume that $\alpha'(t)$ is orthogonal to v for all $t\in I$ and that $\alpha(0)$ is also orthogonal to $v$. Prove that $\alpha(t)$ is orthogonal to $v$ for all $t\in I$.

My thoughts:-

1.the given curve will be $\alpha(t)=(\sin {t},\cos {t})$.

2.The given $\alpha $ is a equation of a straight line.

3.here by the given condition we have $$\alpha'(t).v=0$$ Integrating w.r.t $"t" $ from $t=0$ to $t=t$ we have $v.(\alpha(t)-\alpha(0))=0$
$\implies$ $v.\alpha(t)-v.\alpha(0)=0$ $\implies$ $v.\alpha(t)=0$ as [$\alpha(0)$ is also orthogonal to $v$ so $v.\alpha(0)=0$]

Am I right in these cases? Can someone please help me to verify these results.

$\alpha(t) = \alpha(0) + \int_0^t \alpha'(s) \mathrm{d}s$. Dot with $v$ on both sides. — Eric Towers, Feb 10 '14 at 10:31
These are exercises in chapter 1-2 in do Carmo's Differential Geometry of Curves and Surfaces. — self-learner, Mar 12 '16 at 04:24

score 2 · Answer 1 · answered Feb 10 '14 at 16:26

2

$\bigl(\alpha(t).v\bigr)'=\alpha'(t).v+\alpha(t).0=0$, i.e, $\alpha(t).v$ is constant. Now plug in $t=0$.

answered Feb 10 '14 at 16:26

Michael Hoppe

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How can we differentiate $v$ if it is just a vector and not a function? – user5826 Jan 18 '19 at 03:57
Component wise, as usual. – Michael Hoppe Jan 18 '19 at 07:05

score 0 · Answer 2 · answered Feb 10 '14 at 10:13

The first two points are what you need, yes.

Just a thought on the third, although I believe your solution is correct: the derivative of $\alpha(t)\cdot v$ is $\alpha'(t)\cdot v = 0$, meaning the value of $\alpha(t)\cdot v$ is constant over all $t$. Because $\alpha(0)\cdot v=0$, this means $\alpha(t)\cdot v=0$ for all $t$. I think this is a simpler way of proving what you proved, there is also less writing required if you want to write it rigorously.

property of a curve $\alpha(t)$

2 Answers2