Curvature of curves on surfaces

Are there ways to know the curvature of a curve $\gamma$ that lives on a surface $\mathcal{S}$starting from the gaussian curvature of $\mathcal{S}$? In general, is it possible bound the curvature of this curve with a function that depends from the gaussian curvature?

No way the tangent curvature can be fixed .For example you can draw on a sphere a geodesic, any small circle which can be any displaced arbitrary parallel, all living on the sphere.

You can have them constant when deforming together isometrically.In bending both $$k_g, K$$ (geodesic and Gauss curvatures) remain constant.

But you cannot derive one from the other, they are independently defined or formed from the first fundamental form coefficients.

If a product is 12,factors can be (6,2) or (4,3) or (8, 3/2) etc. Anyway you can physically while embedding in 3-space broaden, narrow up or squeeze out a flexible surface isometrically in bending.

You can also, using a "concentric" geodesic polar coordinate system draw parallel circles of changing geodesic curvature. Even a constant geodesic curvature line (that would roll out to a circle if surface were a developable) ... can be drawn on a surface of of arbitrarily changing K.

But you cannot define or extract all lines $$\gamma$$ on a surface from given $$K$$.

To find geodesic curvature we need to further differentiate on arc the inclination of line to the lines of curvature through the second fundamental form.

For surfaces of revolution $$k_g$$ by Liouville's formula is:

$$k_g = \psi ^{\prime} + \sin \psi \sin \phi / r$$

and $$K$$ is:

$$K = \frac{d\phi}{ds \cos \psi}\cdot \frac {\cos \phi }{r}$$

which are independent.

No, and no.

Consider a torus with "core" radius $r$ and "tube radius" $s$, lying on a table like a donut. A point i=$P$ in contact with the table surface lies on a circle of radius $r$ and an orthogonal circle of radius $s$, both geodesic, so the Gaussian curvature at $P$ is $rs$.

Now fix $k$ and pick $r = k/s$. Then $rs = k$, so for any such $s$, we have the same gaussian curvature. But the "around the tube" curves through $P$ (for various values of $s$) will have curvature $1/s$, which is unbounded as $s$ grows small.