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Let $C=\{(x,y,z) \in \mathbb(R)^3| \sin x + \sin^2 y + \sin^4 z=0 \ \text{and} \ (x-z)^2=4\pi^2\}.$

By the implicit function theorem, we have that $C$ can be parametrized as a smooth curve in the form of a graph, in a neighborhood $I$ of the point $(0,0,-2\pi)$, in the form of $t \mapsto \gamma(t)=(x(t),t,z(t))$, with $t \in I$, where I is an open interval around $0$ in $\mathbb(R)$ og $t \mapsto x(t)$ and $t \mapsto z(t)$ are smooth functions on $I$.

Find $\gamma'(0)$ and $\gamma''(0).$

We see that $C=\{(x,y,z) \in \mathbb(R)^3| F(x,y,z)=(0,0)\}$ where $F(x,y,z)=(\sin x + \sin^2 y + \sin^4 z, (x-z)^2-4\pi^2)$.

Since $\gamma$ is a reparametrization on $I$, it must have the same properties as $F$ on $I$, so if i derive $F$ and insert $0$, i guess i would obtain $\gamma ' (0)$ ? if this is not correct, im lost.

Edit: \begin{align*} DF(x,y,z)=\begin{pmatrix} \cos x & 2\sin y \cos y & 4 \sin^3 z \cos z \\ 2(x-z) & 0 & -2(x-z) \end{pmatrix}. \end{align*} \begin{align*} DF(0,0,0)=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. \end{align*}

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    $\begingroup$ You could accept more answers. Please read about accepting answers here and here. $\endgroup$ – Git Gud May 17 '14 at 10:38
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    $\begingroup$ If I understood you correctly, your idea is good. What's keeping you from making progress? $\endgroup$ – Git Gud May 17 '14 at 10:47
  • $\begingroup$ Because i was unsure, and would like to get my idea verified. But i will proceed right away, and edit it in asap. $\endgroup$ – SuddenlyNotHorrific May 17 '14 at 10:50
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Since the values of second derivatives are requested it is not sufficient to look at $DF$. I suggest the following:

The second constraint defines the two planes $z=x\pm 2\pi$. Given the initial condition $(x_0,y_0,z_0)=(0,0,-2\pi)$ it follows that $z=x-2\pi$, so that we are left with the single constraint $$\sin x+\sin^2 y+\sin^4 x=0\ ,\tag{1}$$ connecting the variables $x$ and $y$. Now we want the curve $\gamma$ in the form $$\gamma: \quad t\mapsto \bigl(x(t), t, z(t)\bigr)\ .$$ Plugging the "Ansatz" $$x(t):=a t+ b t^2+?t^3$$ into $(1)$ gives $$ (at +bt^2)+t^2+?t^3=0\ ,$$ from which we conclude $a=0$, $\>b=-{1\over2}$. As $z(t)=x(t)-2\pi$ we finally obtain $$\gamma'(0)=(0,1,0),\quad \gamma''(0)=(-1,0,-1)\ .$$

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  • $\begingroup$ Wouldn't it be easier to find the hessian matrix of F? $\endgroup$ – SuddenlyNotHorrific May 17 '14 at 11:30
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    $\begingroup$ You wouldn't know what to do with it; nor would I, for that matter. $\endgroup$ – Christian Blatter May 17 '14 at 11:31
  • $\begingroup$ okay thanks for the help. this is really useful. $\endgroup$ – SuddenlyNotHorrific May 17 '14 at 11:34

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