2

Find a set of parametric equations for the tangent line to the curve of intersection of the surface $x^2 + z^2 = 2$ and the surface $x^2 + y^2 - z^2 = 1$ at the point $(1, 1, 1)$.

happywand
  • 63

1 Answers1

1

The line tangent to the intersection curve will be orthogonal to the vectors normal to both surfaces. Thus, we can find those two normal vectors, take their cross product, and obtain the direction we want for our line. The normal vectors can be obtained from the gradients, which are $\langle 2x , 0 , 2z \rangle$ and $\langle 2x , 2y , -2z \rangle$. At the point $(1,1,1)$, these are the vectors $\langle 2,0,2 \rangle$ and $\langle 2,2,-2\rangle$, respectively.

Thus we obtain the cross product: $\langle 2,0,2 \rangle\times \langle 2,2,-2\rangle = \langle -4,8,4\rangle$. Any scalar mutliple will do, so we can take the direction number for our line to be $1,-2,-1$, yielding parametric representation: $\langle x,y,z\rangle=\langle 1,1,1\rangle+t\langle 1,-2,-1\rangle$.

G Tony Jacobs
  • 31,218
  • Is there a method to find the tangent line from the curve of intersection? – happywand Nov 24 '17 at 16:38
  • If you can write the intersection curve parametrically, then you can take its derivative to obtain a vector tangent to it. The challenge would be writing that curve parametrically in the first place, but it can be done. – G Tony Jacobs Nov 24 '17 at 16:43
  • Hmm ok, I guess the question wants it to be answered like what you did. Thank you. – happywand Nov 24 '17 at 16:46
  • @happywand, If you set $x=t$, then we can take $z=\sqrt{2-t^2}$ and $y=\sqrt{3-2t^2}$. This parametrizes the part of the intersection curve that passes through $\langle 1,1,1\rangle$. If you take a derivative of this parametric curve, and then plug in $\langle 1,1,1\rangle$, you get the same answer. – G Tony Jacobs Nov 24 '17 at 17:02
  • Can i let $z=t$? – happywand Nov 25 '17 at 01:34
  • Sure, there are lots of ways to parametrize that curve. – G Tony Jacobs Nov 25 '17 at 01:36