Question: (i) Prove that the system of equations $$F_1(x, y, z) := 3x + e y - cos z + z = 0$$ and $$F_2(x, y, z) := sin x - x - y + cos y + z - 1 = 0$$ defines implicitly in a neighborhood of the origin, a curve with parametric equations $x = f_1(y)$ and $z = f_2(y)$.

(ii) Write down the parametric equations of the tangent line to the curve in (i) at the point $(0, 0, 0)$.

I've attempted to answer (i) but I'm new to the implicit function theorem and I'm not sure if I've completely miss used it. Also if somebody could show me how to answer (ii) that would be really appreciated.

Attempted Answer: Clearly the implicit function theorem needs to be used. So to start we note that $F_1(0,0,0)=0$ and $F_2(0,0,0)=0$. Next note that the Jacobian \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}equals

\begin{pmatrix} 3&e^y \\ \cos x -1& -1 -\sin x \end{pmatrix} which evaluated at $(0,0,0)$ has a non-zero determinant so by the implicit function theorem there exists $\alpha, \beta >0$ and two unique functions $f_1:B(y,\alpha) \rightarrow B(x,\beta)$ and $f_1:B(y,\alpha) \rightarrow B(z,\beta)$ such that $F_1(f_1(y'),y',f_2(y'))=0$ and $F_2(f_1(y'),y',f_2(y'))=0$ for all $y' \in B(y,\alpha)$.

  • $\begingroup$ What happened to the $z$ coordinate, in your matrix? $\endgroup$
    – user7530
    Commented Oct 25, 2013 at 14:43
  • $\begingroup$ It got differentiated out. $\endgroup$
    – Gottfried
    Commented Oct 25, 2013 at 14:46

1 Answer 1


There is a slight mishap in your attempt. You are supposed to get the curve $(x,z)=f(y)=(f_1(y),f_2(y))$, therefore you need to investigate the Jacobian $\partial F(x,y,z)/\partial (x,z)$ and not, as you have done, $\partial F(x,y,z)/\partial (x,y)$.

Once you corrected this you will get the tangent vector (from which the line will follow easily) by calculating $df/dy$, in a brute force approach just solve $0=\frac{dF}{dy}=\frac{\partial F}{\partial y}+\frac{\partial F}{\partial (x,z)}\cdot\frac{df}{dy}$ from which you get $\frac{df}{dy}=-(\frac{\partial F}{\partial (x,z)})^{-1}\frac{\partial F}{\partial y}$ (with some understanding which is a matrix and which is a vector) and finally you plug in the values of $x,y,z$ of your favorite point .


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