I need to show that the equations: $$x^2y+xy^2+t^2=1$$ $$x^2+y^2 -2yt=0$$ is difinding a differentiable path $\vec{\gamma}=({x}_{(t)},{y}_{(t)})$ at the point $(x,y)=(-1,1)$.
after that I should find the tangent line for this path in this specific point.
So my idea is to use the implicit function theorem here but this is the first time I see expression with the more variables in the equation than the point I am asked to show. what I usually do is check if the point I have given is satisfy $F(x,y)=0$ but here I have the extra variable $t$. what I should do? I feel that I should use the implicit theorem here but not sure how to do it in this conditions.
Best Answer
Define $F: \Bbb{R}^3 \to \Bbb{R}^2$ by $F(t,x,y) = (x^2y + xy^2 + t^2, x^2 + y^2-2yt)$. We're going to apply the implicit function theorem to this function $F$. Now, notice that $F(1,-1,1) = (1,0)$. Also, the Jacobian matrix of partial derivatives is \begin{align} F'(t,x,y) &= \begin{pmatrix} 2t & 2xy + y^2 & x^2 + 2xy \\ -2y & 2x & 2y - 2t \end{pmatrix} \end{align} So, \begin{align} F'(1,-1,1) &= \begin{pmatrix} 2 & -1 & -1 \\ -2 & -2 & 0 \end{pmatrix} \end{align} Note that every $2 \times 2$ submatrix is invertible. So, by the implicit function theorem, you can solve for any two out of the three variables as a $C^{\infty}$ function of the third. In particular, you can solve for $x$ and $y$ as a function of $t$, in a neighbourhood of $t=1$, which is what you had to prove.
You're probably more comfortable with the lower-dimensional situation where $f: \Bbb{R}^2 \to \Bbb{R}$, and if there is a point $(a,b)$ where $\dfrac{\partial f}{\partial y}(a,b) \neq 0$, then in a neighbourhood of $(a,b)$, you can express $y$ as a smooth function of $x$. Well, this question is just a generalization to more variables.
Since you want to solve for $x,y$ as a function of $t$, you simply have to check if \begin{align} \det \begin{pmatrix} \frac{\partial F_1}{\partial x} & \frac{\partial F_1}{\partial y} \\ \frac{\partial F_2}{\partial x} & \frac{\partial F_2}{\partial y}\end{pmatrix} \neq 0 \end{align} (all the derivatives being evaluated at the appropriate point). And as I've shown above, this is indeed the case. For the statement of the theorem in the general case, take a look at the Wikipedia page.