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Let C represent the curve generated by the position function r(t)=((t^2)-2t, t+1,(t^2)+t-2) for -infinty

Mathematics
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is \[ \frac{d\overrightarrow{r}(t)}{dt}\]a constant or a function of t? let's find out. \[\frac{d\overrightarrow{r}(t)}{dt}= (\frac{dx}{dt} , \frac{dy}{dt} , \frac{dz}{dt})\] \[=(2t-2 , 1 ,2t+t)\]which is still a function of t, so the slope of the curve is changing in the x direction and the z direction. because of this the curve cannot be a plane curve
woops, the z component of the derivative should be 2t + 1

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