Consider the curve C obtained by intersecting the surfaces defined by
$x^2+y^2+z^2=3$ and $x^2-y^2+z^2=1$
At the point $(1,1,1)$, what is the tangent vector to the curve?
I tried parameterizing the curve and got $(t,\pm1,\sqrt{2-t^2})$ but am lost on what to do next.
Thank you.
You have found both curves of intersection, clearly the one going through $(1,1,1)$ is $$s(t)=(t, 1, \sqrt{2-t^2})$$ note that $s(1)=(1,1,1)$ $$s'(t)=(1,0, \frac {-t}{\sqrt{2-t^2}}) $$ The unit tangent vector is given by $$\vec T = \frac{s'(1)}{||s'(1)||} $$