Multiplying Path Functions (Topology) Assuming I have two path functions $$f, g : R^+\to R^2$$
for example a sin curve: $$f(t) = (t, \sin(t)) t:[0, 1]$$
or a circle: $$g(t) = (\sin(t), \cos(t)) t:[0, 1]$$
What I need is a way to calculate $$h = f * g$$
so that the the result will look like it's a sin going in a circle, assume you take the X axis and bent it into a circle while the sin on it, same way bending sin over another sin and etc.
So this operation basically takes the first function and shape it over the second.
Anyone know how it's called and what is the operation? There must be something like this in topology.
Example:

applied on this:

will result in something like this (using coffeemath method):

the blue line is the sin going around the circle.
but this only works for functions like: $$f(t)=(t, v(t))$$
here is sin around sin around sin using same method:

This is good already but maybe there is a function for the general case?
Solution: (by coffeemath)

given
$$f(t)=(u(t),v(t))$$
$$g(t)=(x(t),y(t))$$
to draw f around g will be:
$$w(t)=g(t)+u(t)T(t)+v(t)N(t)$$
now i can draw circle around another circle:

 A: In your example you had a curve $f(t)=(t,\sin(t))$ to be "wrapped around" another curve $g(t)$. From the picture it looks like the important thing to use is the $\sin(t)$ part, so in the following I'll assume the special case of $f(t)=(t,h(t))$ where you want to wrap the function $h(t)$ around the other curve $g(t)$.
If the curve $g(t)$ is thought of as a vector valued function $g(t)=(x(t),y(t))$ then one way to go is to compute the unit normal vector $T(t)$ of $g(t)$. By this I mean we first find the tangent vector $(x'(t),y'(t))$ and rotate it 90 degrees so it becomes $(-y'(t),x'(t))$, and then divide this by its length to get a unit normal vector $T(t)$.
Then one possibility for the "wrapped" curve formula is $w(t)=g(t)+h(t)N(t)$. Here $h(t)$ is a scalar function which is multiplying the unit normal vector $N(t)$ in order to obtain the point on the wrapped curve $w(t)$, where near the point $g(t)=(x(t),y(t))$ we're using the  unit tangent and normal vectors to generate a coordinate system near $g(t)$.
I think this will work OK provided the curve $g(t)$ is not extremely curved, and provided the function $h(t)$ being wrapped doesn't get too large. These normal lines to the curve can typically intersect away from the curve, in which case the wrapped curve appearance won't be good.
A: Following my comment above, with $g$ defined as in your example, I would try defining $m: (x, y) \mapsto (2^y \sin x, 2^y \cos x)$ and $h = m \circ f$.
I don't know of a standard operation like this, and I'm not sure how to follow this approach for general $g$.
A: Given a smooth curve $\gamma:I\rightarrow \mathbb{R}^2$, You can define the Gauss map $n:I\rightarrow S^1\subset \mathbb{R}^2$ giving you a unit vector normal to your curve at every point in a smooth way. (Here this means it will stay on one side of the curve.)
If $f : I\rightarrow \mathbb{R}$ is a function, it seems you want to define a curve like $f\cdot\gamma(t) \equiv \gamma(t)+f(t)n(t)$, with addition in $\mathbb{R}^2$.
Is this what you are looking for?
