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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?

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

2 Answers 2

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

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

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

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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.

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  • $\begingroup$ 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? $\endgroup$ Aug 10, 2015 at 5:15
  • $\begingroup$ $<\alpha(t),v>=\alpha_{x}v_{x}+\alpha_{y}v_{y}+\alpha_{z}v_{z}$ $\frac{d}{dt}<\alpha(t),v>=\frac{d}{dt}(\alpha_{x}v_{x})+\frac{d}{dt}(\alpha_{y}v_{y})+\frac{d}{dt}(\alpha_{z}v_{z})$ $\endgroup$
    – Tucker
    Aug 10, 2015 at 5:19
  • $\begingroup$ OH wait, it is verified, it's just that $v'(t)=0$ $\endgroup$ Aug 10, 2015 at 5:20
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    $\begingroup$ Yes, otherwise there would be another term from the product rule. $\endgroup$
    – Tucker
    Aug 10, 2015 at 5:21
  • $\begingroup$ Oh yes, oops, thanks @Tucker and Alexander. With the latex its \langle \rangle = $\langle \rangle$ $\endgroup$ Aug 10, 2015 at 5:22

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