Find the points where the curve $r(t) = \langle{t, t^2, t^3\rangle}$ intersects the surface $zx = 13y - 36$

Question

1) Rearrange the equation to : 13y - zx = 36

2) To find the point of intersection, we plug the parametric equations into equation for the plane;

13 (t^2) - 1(t^3) x 1(t) = 36

13t^2 - ( t^3 x t ) = 36

13t^2 - ( t^4 ) = 36

t^2 ( 13 - t^2 ) = 36

13 - ( t^2 ) = 36

Yea, I’m confusing myself because we can’t take the square root of a negative number — i square root of 23

3) Obtain t value (The zx variable is what conflicts me )

Answer 1

You have $t^4 -13t^2+36 = 0$ which factors as

$$(t^2-9)(t^2-4)=0.$$

So $t=\pm 3, \pm 2.$