How do you find the slope of a secant line on a non-smooth curve?

Question

My background is engineering, not mathematics, so please forgive me if my terminology is not right. On a continuous curve a secant line is one formed by connecting two points on the curve. I'm trying to determine the angle of that line on a non-smooth curve. Think the angle of a bicycle that starts off on a level surface and then gets to a sudden linear down-slope. I'm trying to model the angle of the bicycle as it goes over the slope.

For a regular secant the slope is just $$ m = \frac{y_2-y_1}{x_2-x_1} $$ My problem is that x2 depends on the angle of rotation, so I'm not sure what to do.

If we supposed the function $$ f(x) = \begin{cases} x/2, & x < 0\\ 0 & x \ge 0 \end{cases} $$ How would we construct the function?

EDIT: I'm sorry. I didn't note that I know the value of x for one of the points, but I only know the distance between the points (the length of the bike) not the x value of the second point. (Also thank you gt6989b for formatting the math.)

Edit 2: Added image of the problem I'm trying to solve. I know (x1, y1) and I know L, but I can't figure out how to find (x2, y2)

Answer 1

As a hint, you framed the problem incorrectly, the question if your curve is smooth doesn't really matter, you would have exactly the same problem if your whole 'surface' was $y=x^2$, for example.

In your example, if you know $(x_1,y_1)$ , you are looking for a point $(x_2,y_2)$ with $x_2 \ge 0, y_2=0$ and the additional condition that

$$\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}=L.$$

Putting the information we have into this equation, we get

$$\sqrt{(x_1-x_2)^2 + y_1^2}=L,$$

which we can transform into

$$(x_1-x_2)^2 + y_1^2=L^2,$$

then

$$(x_1-x_2)^2 =L^2-y_1^2,$$

further to

$$x_1-x_2 =\pm\sqrt{L^2-y_1^2},$$

and finally to

$$x_2=x_1\pm\sqrt{L^2-y_1^2}.$$

Since $x_1$ is negative, $x_1-\sqrt{L^2-y_1^2} < 0$, so this solution is not valid.

You need to check the remaining solution

$$x_2=x_1+\sqrt{L^2-y_1^2}.$$

If $L^2-y_1^2<0$, the solution is not real so unusable. It correspondes to the point $(x_1,y_1)$ being 'too far away' from the origin, so distance $L$ is not 'reaching' to the positive $x$-axis. In addition, that $x_2$ might be negative, which is another manifestation of $(x_1,y_1)$ being 'too far away' from the origin.

If $x_2$ as calculated is positive, then you have found your point $(x_2,y_2=0)$ and can calculate the slope between $(x_1,y_1)$ and $(x_2,y_2)$ by the usual formula

$$m =\frac{y_2-y_1}{x_2-x_1}$$.

Answer 2

You can still use the same formula to find the secant line.

The only difference between the smooth and non-smooth curves is that a non-smooth curve doesn't have a defined tangent line at the "sharp turn."

Answer 3

All you need here is the angle between two lines. Let each line make angles $a$ and $b$ with respect to the horizontal. Then the angle between them is given by $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}.$$ Thus, from knowing the slopes of each of the lines forming the terrain over which your bicycle goes, you can find the slope of the angle it turns at the singularity, which gives you the angle of turn there.