Derivative and vector for start of curve are orthogonal to fixed vector, hence the curve is

Let $\alpha:I\to\Bbb R^3$ be a parameterized curve and let $v\in \Bbb R^3$ 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 $v,\forall t\in I$.

I have been having trouble solving this problem, and I have written it in math as follows:

$$v_1x'(t)+v_2y'(t)+v_3z'(t)=0=v_1x(0)+v_2y(0)+v_3z(0)$$

and using this I want:

$$v_1x(t)+v_2y(t)+v_3z(t)=0$$

Can I have a hint please?

• Is this from Do Carmo's text? Aug 10, 2015 at 5:18
• @Tucker Page 5, Q4 Aug 10, 2015 at 5:19

$$\alpha(t)-\alpha(0)=\int_{0}^{t}\alpha'(\tau)d\tau$$

Dotting both sides with $v$

$$v\cdot\alpha(t)-v\cdot\alpha(0)=\int_{0}^{t}v\cdot\alpha'(\tau)d\tau$$

The integrand is identically zero and so

$$v\cdot\alpha(t)-v\cdot \alpha(0)\equiv 0 \Rightarrow v\cdot\alpha(t)\equiv 0$$

Your condition that $\alpha'(t)\perp v$ for all $t$ is the same as saying that $\langle\alpha'(t),v\rangle =0$ for all $t$, but note also that $\frac{d}{dt}\langle\alpha(t),v\rangle = \langle\alpha'(t),v\rangle$ since the dot product is linear in the first coordinate. Therefore, $\langle\alpha(t),v\rangle$ is constant, and you can conclude.

• I haven't come across $\frac{d}{dt}\langle \alpha(t),v\rangle = \langle \alpha'(t),v\rangle$ before, can you verify that for me if it's not too much trouble? Aug 10, 2015 at 5:15
• $<\alpha(t),v>=\alpha_{x}v_{x}+\alpha_{y}v_{y}+\alpha_{z}v_{z}$ $\frac{d}{dt}<\alpha(t),v>=\frac{d}{dt}(\alpha_{x}v_{x})+\frac{d}{dt}(\alpha_{y}v_{y})+\frac{d}{dt}(\alpha_{z}v_{z})$ Aug 10, 2015 at 5:19
• OH wait, it is verified, it's just that $v'(t)=0$ Aug 10, 2015 at 5:20
• Yes, otherwise there would be another term from the product rule. Aug 10, 2015 at 5:21
• Oh yes, oops, thanks @Tucker and Alexander. With the latex its \langle \rangle = $\langle \rangle$ Aug 10, 2015 at 5:22