Extra distance travelled along a sine wave path Let's say I have a blue line which is 10 metres long.  I then draw a single cycle of a sine curve along this line, in red which has a maximum distance of 0.1 meter from the line (so an amplitude of 0.1m).
If I walk along the sine curve, I will walk further than 10 metres.  But how do I calculate this new distance travelled?  
What about if I increase the frequency so that there are five full cycles along my 10 metre blue line?  How does the frequency affect it, in other words?
 A: HINT
This answer may not help directly for higher frequencies/waves but stating it nevertheless.., the following requires some imagination.
Sinusoidal waves are development of slantly cut cylinder intersections seen as surface development.
If a cylinder radius $a$ is cut at an angle $\alpha$ to radial plane the ellipse axes are $ (a, a \sec \alpha) $ then the ellipse has an eccentricity $ e= \sin \alpha$. 
$$ 2 \pi a = 10 m, \quad  a \tan \alpha = \dfrac{.01}{2} ; $$ 
Solve for $ (a,\alpha, e)$.
The perimeter is $ 4 a \, E(e)$ which a standard result using Elliptic integrals; it can be found by integration or google it for a start.
EDIT1:
Useful to remember the waveform equations of same amplitude $A$ for single and multiple frequencies $n$:
$$ y_1= A \sin \dfrac{2 \pi x}{\lambda}$$
$$ y_n= A \sin \dfrac{2 \pi x\cdot n}{\lambda}$$
A: You need the arc length formula for this: $\int\sqrt{1+f'(x)^2}dx$ on interval $[0,10]$. What you need to do is establish the formula of your curve, which is not hard to do. After all, you know the period (then use $b=\frac{2\pi}{period}$) and the amplitude as you stated. Anyway, the real problem is that the arc length formula is not going to come out "nice", in the sense that I greatly expect that you are going to deal with an integral that cannot be evaluated in terms of elementary functions. You are then left with an approximation using an TI for example
