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$y(t)$ is a path contained in two surfaces: $x^2+y^4+z^6=3$, $x+y^2=y+z^2$
also $y(0)=(1,1,1)$ and $||y'(0)||=1$
Need to find the vectors $-y'(0)$ and $+y'(0)$

To be honest, I'm not sure how to begin and would appreciate any help.

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up vote 1 down vote accepted

HINT: Let $\pi_1$ and $\pi_2$ be the planes tangent to the two surfaces $\cal S_1$ and $\cal S_2$ at the point $P=(1,1,1)$. Since the path $\gamma$ is contained in $\cal S_1\cap\cal S_2$, its tangent vector at time $t=0$, i.e. at the point $P$ will belong to $\pi_1\cap\pi_2$.

So the problem boils down to finding the unitary vectors in $\pi_1\cap\pi_2$.

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As stated, you cannot solve the problem. There is no way to distinguish $y'(0)$ and $-y'(0)$, you can only find two vectors of unit length in $\pi_1 \cap \pi_2$ – Blah May 5 '12 at 11:39
True, but the pair of vectors is well-defined. In order to distinguish one from one may introduce an orientation in the picture, e.g. requiring that the ordered triple $(\vec n_1,\vec n_2,\gamma^\prime(0))$ is positively oriented, where $n_1$ and $n_2$ (to be assumed linearly independent) are vectors normal to the surfaces. – Andrea Mori May 5 '12 at 11:53

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