calculus
I was wondering if I could use the derivative of a function to determine the average rate of change between two points, rather than one. I have a solution but I'm not sure whether it is valid or not: basically, to find the average rate of change between two points on any curve you must first know the rate of change at your maxima point, and subtract the rate of change of that point from the distance of it from your minima point in terms of x.
Thanks.
The average rate of change of $f(t)$ on the interval $a\le t\le b$ is just ${f(b)-f(a)\over b-a}$. So here---assuming I am interpreting your function correctly with the extra set of parentheses---you get:
$${f(4)-f(1)\over 4-1}\approx -0.373338.$$
To understand this geometrically, just visualize the secant line between the points $(1,f(1))$ and $(4,f(4))$ shown in black. The number above is the slope of this line. The original function is shown in blue. (Note how the axes are scaled though.)
I think a more fundamental way to approach the problem is by discussing geodesic curves on the surface you call home. Remember that the geodesic equation, while equivalent to the Euler-Lagrange equation, can be derived simply by considering differentials, not extremes of integrals. The geodesic equation emerges exactly by finding the acceleration, and hence force by Newton's laws, in generalized coordinates.
See the Schaum's guide Lagrangian Dynamics by Dare A. Wells Ch. 3, or Vector and Tensor Analysis by Borisenko and Tarapov problem 10 on P. 181
So, by setting the force equal to zero, one finds that the path is the solution to the geodesic equation. So, if we define a straight line to be the one that a particle takes when no forces are on it, or better yet that an object with no forces on it takes the quickest, and hence shortest route between two points, then walla, the shortest distance between two points is the geodesic; in Euclidean space, a straight line as we know it.
In fact, on P. 51 Borisenko and Tarapov show that if the force is everywhere tangent to the curve of travel, then the particle will travel in a straight line as well. Again, even if there is a force on it, as long as the force does not have a component perpendicular to the path, a particle will travel in a straight line between two points.
Also, as far as intuition goes, this is also the path of least work.
So, if you agree with the definition of a derivative in a given metric, then you can find the geodesic curves between points. If you define derivatives differently, and hence coordinate transformations differently, then it's a whole other story.
Best Answer
The average rate of change in the interval $[a,b]$ is $$\frac{f(b)-f(a)}{b-a}$$ if the function is differentiable you can think about it like the sum of all of the derivatives in the specific interval divided by its length: $$\frac{1}{b-a} \int_a^b f'(t)dt = \frac{f(b)-f(a)}{b-a}$$