Question

At what points does the curve r(t) = ti (4t − t2)k intersect the paraboloid z = x2 y2? (if an answer does not exist, enter dne.)

Answer 1

$\mathbf r(t)=x(t)\,\mathbf i+y(t)\,\mathbf j+z(t)\,\mathbf k=t\,\mathbf i+(4t-t^2)\,\mathbf k$
$\implies \begin{cases}x(t)=t\\y(t)=0\\z(t)=4t-t^2\end{cases}$

So the paraboloid is given by

$z(t)=x(t)^2y(t)^2\iff 4t-t^2=0$

which holds for $t^2-4t=t(t-4)=0$, i.e. when $t=0$ and $t=4$.

To make sure this is correct, plug in $t=0$ and $t=4$ to find out the value of $\mathbf r(t)$, and see if these points are on the paraboloid.

$\mathbf r(0)=0\,\mathbf i+0\,\mathbf j+0\,\mathbf k\implies (0,0,0)$
$0=0^2\times0^2$

$\mathbf r(4)=4\,\mathbf i+0\,\mathbf j+(4\times4-4^2)\,\mathbf k\implies (4,0,0)$
$0=4^2\times0^2$

Both of these hold, so the answer is correct.