# property of a curve $\alpha(t)$

1. 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 $\alpha(0) = (0, 1)$.

2. A parametrized curve $\alpha(t)$ has the property that its second derivative $\alpha''(t)$ is identically zero. What can be said about $\alpha$ ?

3. 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 $v.\alpha(0)=0$]

• $\alpha(t) = \alpha(0) + \int_0^t \alpha'(s) \mathrm{d}s$. Dot with $v$ on both sides. Feb 10, 2014 at 10:31
• These are exercises in chapter 1-2 in do Carmo's Differential Geometry of Curves and Surfaces. Mar 12, 2016 at 4:24

$\bigl(\alpha(t).v\bigr)'=\alpha'(t).v+\alpha(t).0=0$, i.e, $\alpha(t).v$ is constant. Now plug in $t=0$.
• How can we differentiate $v$ if it is just a vector and not a function? Jan 18, 2019 at 3:57
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.