$y(t)$ 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.
$y(t)$ 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.
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 $\pi_1\cap\pi_2$.