$\alpha(t)\cdot v=0$ for all $t$

Question

Let $\alpha:I\rightarrow R^3$ be a parametrized curve and let $v\in R^3$ be a fixed vector. Assume that $\alpha^\prime (t)$ 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 $t\in I$.

Here's what I came up with:

Since $v$ is a fixed vector, then $v^\prime=0$. We know that $\alpha^\prime (t)\cdot v=0$. Thus, $$\alpha^\prime(t)\cdot v=\alpha^\prime(t)\cdot v+\alpha(t)\cdot v^\prime =\frac{d}{dt}[\alpha(t)\cdot v]=0$$ for all $t\in I$. Integrating, $$\int \frac{d}{dt}[\alpha(t)\cdot v]=\alpha(t)\cdot v=C$$ fr all $t\in I$. For $t=0$, $$\alpha(0)\cdot v=0$$ Thus, $C=0$ and the final equation is $$\alpha(t)\cdot v=0$$ for all $t\in I$. QED.

Is this a valid proof? Is it missing any steps/explanations?

Thanks.

Answer 1

Looks good to me, with some typos:

• $\alpha'(t) \cdot v = \alpha'(t)\cdot v + = ...$ there is probably a $0$ missing here
• $\int\frac d{dt}(\alpha(t)\cdot v) = ...$ An integral should always end with $dx$, where $x$ is the variable you are integrating over. In this case, that's $dt$.

Also, I think you can avoid the integral overall. You have shown that $f(t) = \alpha(t)\cdot v$ has a constantly zero derivative. The only functions with a constantly zero derivative are constants, therefore, $f(t)$ is constant. Knowing that $f(0)=0$ gives you the result you need.