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

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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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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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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