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mathmari
Gold Member
MHB
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Hey! 
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)
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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