# Definition of 'average inclination'?

1. Mar 27, 2014

### dilloncyh

It's not a homework problem, just a problem that suddenly popped out of my mind.

1. The problem statement, all variables and given/known data

So, my question is : how to calculate, or how to define 'average inclination'? Suppose I am given an equation y=f(x) that resembles the shape of a section of a hill, and I want to calculate the average inclination (something I always see when I watch cycling or alpine skiing race), how do I do that? Let's use f(x) = x^2 as an example for the following discussion.

2. Relevant equations

I know the average of a function is defined as:
So I suppose the correct equation should be similar, with f(x) the function that gives the shape of the hill I am calculating?

3. The attempt at a solution

Here comes the problem: Do I use dx for ds? And for (b-a) in the image, I need to replace it with the actual length of the function from x=a to x=b, right?
Since I don't really know I want to calculate (to find the slope at each point of the function and add them together, and then divide it by the total length of the slope?), my question may seem very silly, but please give me some idea.

thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 27, 2014

### Simon Bridge

The average slope is usually just the rise-over-run for two positions on the slope.
$$\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ ... that would be what they did on the ski races.

You want to be a bit more detailed than that:

If $y(x)$ is the height of the slope at position $x$,
Then $$\bar y = \frac{1}{b-a}\int_a^b y(x)\;\text{d}x$$ would be the average height of the hill in a<x<b.

The slope (gradient) function would be g(x)=dy/dx ... tells you the gradient at position x.

3. Mar 27, 2014

### dilloncyh

I'm still a bit confused.

average slope = detla y / delta x seems very legit, but take y=x^2 and y=x as example.

If I want to calculate the average slope between x=0 and x=1, then by calculating the difference in height and horizontal displacement of the two end points, then both should give the same result (slope=1), but the length of the curve of y=x^2 is obviously longer than y=x (which is just sq root of 2), so by calculating the the sine of the 'triangle' if I straighten the curve section of y=x^2, I will get difference result.

4. Mar 27, 2014

### haruspex

It's a matter of how you choose to define the average. The usual would be to define it as average over horizontal distance, but as you note you could instead chose to define it as average over path distance.

5. Mar 28, 2014

### Simon Bridge

I'm with haruspex - there is no reason that two different averaging methods should produce the same value.
They are just producing a different kind of average.