Vector Functions — Question 3

Question 3 – Solution

Strategy Substitute the curve’s coordinate functions into the surface equation.

Parameter equation We need $t^2+t^4=2$. Let $u=t^2\ge 0$; then $u^2+u-2=0$, so $(u-1)(u+2)=0$. Thus $t^2=1$ and $t=\pm 1$.

Points At $t=1$, $\vec r=(1,1,1)$; at $t=-1$, $\vec r=(-1,1,-3)$. Hence the intersections are $$\boxed{(1,1,1)\quad\text{and}\quad(-1,1,-3)}.$$

Verification Both points satisfy $x^2+y^2=2$.