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

Question

Let C represent the curve generated by the position function r(t)=((t^2)-2t, t+1,(t^2)+t-2) for -infinty<t<infinity. Is C  a plane curve? Justify your answer analytically.

anonymous · Answer

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

anonymous · Answer

woops, the z component of the derivative should be 2t + 1