Is the Curve a Line or a Point with Zero Acceleration?

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The discussion centers on the mathematical proof that a curve $\overrightarrow{\sigma}$ in $\mathbb{R}^3$ with zero acceleration must be either a straight line or a point. It is established that if the second derivative of the curve, represented as $a(t)=\overrightarrow{\sigma}''(x''(t),y''(t),z''(t))$, equals zero, then the parametric equations for the curve can be expressed as $x(t)=c_1t+c_2$, $y(t)=c_3t+c_4$, and $z(t)=c_5t+c_6$. The discussion confirms that cases must be considered for the coefficients $c_1$, $c_3$, and $c_5$ to determine if the curve is a straight line or a point.

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mathmari
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Hey! :o

Let $\overrightarrow{\sigma}$ a curve in $\mathbb{R}^3$ with zero acceleration. Show that $\overrightarrow{\sigma}$ is a straight line or a point.

Let $\overrightarrow{\sigma}(t)=(x(t),y(t),z(t))$ be the curve.

We have that $a(t)=\overrightarrow{\sigma}''(x''(t),y''(t),z''(t))=(0,0,0)$

So $x(t)=c_1t+c_2, y(t)=c_3t+c_4, z(t)=c_5t+c_6$

Do we have to take cases if $c_1=c_3=c_5=0$ or not to conclude that the curve is a straight line or a point?? (Wondering)
 
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Heya! (Mmm)

mathmari said:
Do we have to take cases if $c_1=c_3=c_5=0$ or not to conclude that the curve is a straight line or a point?? (Wondering)

Yep. (Nod)
 
I like Serena said:
Heya! (Mmm)
Yep. (Nod)
Great! Thanks! (Smile)
 

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