Find the angle between the surface: $z=x^2+2y^2$ and a curve $$x=t-1, y=-\frac2t, z=t^2-1,$$ at the point $(1,-1,3)$.
My attempt: I am first eliminating t from the given equation of curve. It is easy to see that $$z=(t-1)(t+1)=x(t+1)=x(1-\frac2y)\implies xy-yz-2x-0.$$ Let us write $$\Phi_1(x, y, z)=x^2+2y^2-z=0, \ \Phi_2(x, y, z)=xy-yz-2x=0.$$ Then angle between them is given by $$\cos\theta =\frac{\nabla \Phi _{1}\cdot\nabla \Phi _{2}}{\left | \nabla \Phi _{1} \right |\cdot \left | \nabla \Phi _{2} \right |}$$ Please evaluate it and explain whether I am doing it right or not?
I think $\Phi_2$ is no longer a curve but a surface, so I think it may probably not work (not sure). I think there is in fact no need to convert the curve into your $\Phi_2$.
To get the angle between the surface and the curve means to get the angle between the tangent plane of the surface and the tangent line of the the curve. The tangent line of the curve is just its derivative: $(1,\frac{2}{t^2},2t)$, so $(1,\frac{1}{2},4)$ at $t=2$. The tangent plane of the surface is harder to calculate but we can use the normal vector of the surface at $(1,-1,3)$ instead, which is $(2x,4y,-1)$ or $(2,-4,-1)$ at $(1,-1,3)$.
Hence what is left to do is to calculate the angle between $(1,\frac{1}{2},4)$ and $(2,-4,-1)$ and consider its complement angle.