I'm wrestling with a problem from Calculus 3 and I would appreciate a slight push in the right direction. This is how the problem is stated:
The equations $x + xy + z^3 = 0$ and $\sin (xyz) = 0$ define a plane in $\mathbb{R}^3$.
a) Use implicit differentiation to determine if the equations define $y$ and $z$ as functions of $x$ in a neighborhood around the point $(1,0,−1)$ and find $(\frac{dy}{dx},\frac{dz}{dx})$ in that point.
b) Determine the curvature of the intersection of the planes at the point $(1,0,-1)$
Solution to a: The solution is a bit tedious to type in latex so I'll give you the cliffs.
Define $F(x,y,z) = x + xy + z^3$ and $G(x,y,z) = \sin (xyz)$. $y$ and $z$ are indeed functions of $x$ in the neighborhood of $(1,0,-1)$ and $\left.(\frac{dy}{dx},\frac{dz}{dx})\right|_{(1,0,-1)} = (0,-\frac{1}{3})$
Solution to b)
This is where I get stumped. I have no idea how to continue solving this problem. I understand that to find the curvature I need to transform something into a parametric equation, differentiate that etc but as I said - completely stumped. Any hint will be welcomed.
Thank you for taking the time to read this - I appreciate it!