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I know that we can rotate a curve in $R^2$ about a linear axis, as is common for first year calculus problems involving solids of revolution. But has anyone come up with a general method to take a real valued function in $R^2$ and rotate about another function in $R^2$ that is not necessarily linear? I assume the generalized definition would take a point off the curve to the point on the other side of the curve the same distance off the curve along a line perpendicular to the curve at some point. The alternative definition I thought of was given in an answer below, but that is not what I'm looking for. I want to be able to rotate a curve about another curve geometrically with or without an established coordinate plane, which is why I assumed the definition above.

Trying to make this more precise: My definition takes a curve $C$ and finds the slope of the normal line at point $(x_o,y_o)$. Supposing the slope found is $m$, the normal line is $y=m(x-x_o)+y_o$. Find the point(s) $(x_o,y_o)$ for which this line intersects the point/curve to be rotated. Find the distance along this line between $C$ and the point to be rotated. Then, traversing the line in the opposite direction from $C$, find the point that same distance away from $C$. This yields the rotated point. Using the definition I give and use Curve $C$ as an axis: 1. Is the relation between a point in $R^2$ and its image after rotation a function? 2. Does this depend on whether $C$ represents a real valued function? For example, rotation about a circle is not a function while rotation about a parabola is? 3. Does this yield a well-defined surface of revolution? 4. Could such a rotation yield interesting results, e.g., transforming a smiley face into a sad face, or turning a one kind of conic into another? 5. Supposing this definition cannot yield a well-defined surface of revolution, as some have suggested, what definition could? 6. Are there helpful links or articles that address any of these issues?

A counterexample to the conjecture of(1): Take $C$ to be the unit circle centered at the origin and rotate $(0,2)$ about $C$.

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You need to define this a little bit more carefully in the sense that to get a well-definec surface of revolution, you need to give a rule telling which point of one curve is being rotated about which point of the other. Presumably you want their separation to be orthogonal to one of the curves or some such?? For example, you can rotate the unit circle about itself either by doing nothing (resulting in a zero volume solid of revolution), or you could rotate each point about the antipodal point giving you "self-intersecting donut". –  Jyrki Lahtonen Oct 22 '11 at 9:03

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As the previous answer said, your question is quite general. Assuming you want to stay in the same plane (you said rotate the function in $\mathbb{R}^2$), I will borrow a concept from differential geometry. In diff geom, when we construct a tube (a surface) around a parametrised curve $\gamma$ (which is your $g$, the function around which we rotate), we rotate a constant distance $r$ around $\gamma$. This rotation is carried over from o to $2\pi$. Here is the formula:

$$ T(u,v) = \gamma(u) + r(\mathbf{n}(u)\cos(v)+\mathbf{b}(u)\sin(v)) $$

where $\mathbf{n}$ and $\mathbf{b}$ are unit vectors pointing the in the normal and binormal ($\mathbf{t} \times \mathbf{n}$, note that $\mathbf{t} = \dot{\gamma}$).

Anyway, not to delve too far into differential geometry. In your case, two things need to be modified. The rotation only needs to be for $\pi$. There is no binormal as we are only in $\mathbb{R}^2$, and you can see that $\sin(\pi)$ knocks this part out. Another thing is that instead of rotating a constant distance, this is now a function $f$. So the formula for your question will be something like this:

$$ R(u) = g(u) - (f(u)-g(u)) \mathbf{n} $$

But this will not work since $f$ and $g$ could be travelling at different speeds ($\|f'\|$ and $\|g'\|$), so $f(u)$ would not necessarily be the point on $f$ encountered when extending the normal of the point $g(u)$. Say we are at a point $g_0=g(u_0)$. We need $f(u)$ in this case to be on the line passing through $g(u_0)$ and parallel to $\mathbf{n}$. Another way to say this is

$$ \tag{1} (f(u)-g_0) \cdot \mathbf{n} = \Vert f(u)-g_0\Vert $$ assuming that $\mathbf{n}$ is a unit vector. This means you will have

$$\tag{2} R(u)=g(u)- \{ (f(u)-g(u))\cdot\mathbf{n} \} \mathbf{n} $$

with the added requirement that at each particular point $u_0$, we need equation (1) to hold. This approach would still have problems, I think because you would need to do this for every point in that order--finding $\mathbf{n}(u_0)$ is not hard, and should be done first. Then finding $c$ such that $f(c)-g(u_0)$ is parallel to $\mathbf{n}(u_0)$; there might be more than one such $c$. Then taking

$$ \left(f(c)-g(u_0)\right) \cdot \mathbf{n} $$ gives you the length of the vector to reflect. This vector has direction $\mathbf{n}$. Whence we obtain equation (2). However, I can't think of a way to express this order of events explicitly.

I don't think this depends on the function being real-valued. It can be complex valued, but you'd need to be careful with some added structure from $\mathbb{C}$.

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Depending on the purpose you have in mind, rotation of the function $f$ about the function $g$ may be equivalent to the rotation of $f-g$ about the $x$-axis.

In particular, this works for finding the volume of the solid defined by the rotation of $f$ about $g$.

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I thought of this definition, but it doesn't suit my purpose. This is not for a class, but is a question I came up with on my own when randomly doodling. But I had not thought about the equivalency to rotating $f-g$ about the $x$-axis. –  Fingolfin Oct 22 '11 at 6:04

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