Question

At what points does the curve \(r(t) = t\;i + (2t - {t^2})k\)intersect the paraboloid\(z = {x^2} + {y^2}\)?

Short answer

The paraboloid and curve intersect at points \((0,0,0),(1,0,1)\)

Step-by-step solution

Parametric equations

From \(r(t) = t\;i + (2t - {t^2})k\) we get parametric equations

\(x = t,y = 0,z = 2t - {t^2}\)

Plug the parametric equations into the paraboloid equation

\(\begin{array}{c}z = {x^2} + {y^2}\\2t - {t^2} = {t^2} + 0\\2t - 2{t^2} = 0\\2t(1 - t) = 0\\t = 0,1\end{array}\)

Intersection points

The t values found will give us the points of intersection

\(\begin{array}{l}x = 0\,\;\;\;\;\;\;y = 0\;\;\;\;\;z = 2(0) - {0^2} = 0 \to A(0,0,0)\\x = 1\,\;\;\;\;\;\;y = 0\;\;\;\;\;z = 2(1) - {1^2} = 1\;\;\; \to A(1,0,1)\end{array}\)