I need to solve this math problem and it has been giving me lots of trouble.
Find all values of t such that r'(t) is parallel to the xy-plane.
r(t)=<sqrt(t+1), cos(t), t^4-8t^2>
Thanks!
The normal vector to the xy-plane is $\displaystyle \hat{k}$.
Whe the normal of the plane is perpenduclar to $\displaystyle r'(t)$
the vector will be parallel to the plane.
So when $\displaystyle \hat{r}'(t)\cdot \hat{k}=0 \iff 4t^3-16t=0$.