Given a curve $C$ formed by the intersection of two surfaces
(1) $z=x^2 + y^2$
(2) $x*y + 2*z/5 =0$
and a point $P =(2,-1,5)$, on the curve $C$.
I would like to find the tangent vector to the curve $C$ through point $P$.
Reading from an earlier answer to a related question in this forum, I think one recipe to find the tangent vector would be:
Substitute $z$ by $x^2 + y^2$ from equation (1) into equation (2) we get
(3) $x*y + 2*(x^2 + y^2)/5 =0$
Now implictly take the partial of y wrt x to get
(4) $y + x*dy/dx + 2*(2x + 2*y*dy/dx)/5 =0$
Substituting the point $P$ in the equation above we get
(5) $dy/dx = -1/2$
Differentiating equation (1) wrt x we get
(6) $dz/dx = 2*x + 2*y*dy/dx$
Substituting the values from point $P$ and $dy/dx$ computed from above we get
(7) $dz/dx = 5$
So the tangent vector for the curve $C$ at the point $P$ is given by
$(1,-1/2,5)$
Question: Is the the above reasoning correct?
Thank You
Yes, this looks good.
As a double check, you can confirm that the normal to both surfaces is perpendicular to $(1,-1/2,5)$. The normal to a surface is the gradient of the defining equation, so a normal to the first surface is $(-4,2,1)$. A normal to the second surface is $(-1,2,2/5)$.