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*0^2`


`\mathbf r(4)=4\,\mathbf i+0\,\mathbf j+(4*4-4^2)\,\mathbf k\implies (4,0,0)`

`0=4^2*0^2`

Both of these hold, so the answer is correct.