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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!

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1 Answer 1

up vote 1 down vote accepted

Note, the equations do not represent planes but just surfaces...

This is just an idea:

Let $\vec{r}(x) = (x, y(x), z(x))$. Then the curvature is

$\kappa(x) = \frac{\lvert \vec{r}\;'(x) \times \vec{r}\;''(x) \lvert}{\lvert \vec{r}\;'(x)\lvert^3}$

And you have $\vec{r}(1) = (1, 0, -1)$.

And for example $\vec{r}\;'(1) = (1, \frac{dy}{dx}, \frac{dz}{dx})\lvert_{(x,y,z) = (1,0,-1)} = (1, 0, \frac{1}{3})$ (derivatives from part (a))

You now "just" need to find the second derivatives to find the vector $\vec{r}\;''(1)$ to find the curvature... (Note that you just use implicit differentiation like before to find the second derivatives.)

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Sorry for the screw up. This is a good suggestion and it looks familiar. I will try it. Thank you! –  docjay Mar 6 '12 at 23:21

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