# X from a curved triangle

I'm having a hard time figuring out what to do here and it feels as though it involves converting the function into polar coordinates but I'm not a hundred percent sure.

Essentially, I have a triangle in which two sides are straight and one side is a curved function. Knowing only the length of one side, the angle between them, and the function, I need to either be able to find the other straight side or x. I do not need help with solving it but more so just guidance on how I would go about solving it. I've included a mspaint attempt at drawing it.

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Redbelly98
Staff Emeritus
Homework Helper
What is your application? Or is this a homework or self-study problem?

The application is towards image processing. I am a coop student and am deriving equations for a project so I cannot go into very much detail.
I've tried using polar coordinate system but my equation blows up when I approach 90 degrees, arc length seems like a reasonable approach but I am not fully certain on what to do.

Redbelly98
Staff Emeritus
Homework Helper
Okay, we are careful around here about helping people with homework. So it's good to know this is not a homework problem

.

I guess you've tried some obvious ways to do this, but let me just suggest the first thing that comes to my mind. There may be reasons why it won't work; if so I'll leave it to you to provide more info.

May I presume line L is in a known orientation (perhaps vertical)? That means the slop of the other straight side -- let's call it M -- is known.

From the slope of M and the coordinates of the common vertex of L and M, you can write an equation for M in y=mx+b form. Then you need to solve for the intersection of the line with f(x); there are standard techniques for doing this numerically.

Once you know the common vertex of L and M, as well as the intersection of M with f(x), you can get the length of M using the Distance Formula from geometry.

The application is towards image processing. I am a coop student and am deriving equations for a project so I cannot go into very much detail.
I've tried using polar coordinate system but my equation blows up when I approach 90 degrees, arc length seems like a reasonable approach but I am not fully certain on what to do.
Hmmm, when θ is 90°, the line is simply y=whatever (the y-coordinate of the vertex common to L and M. Not sure why that would be a problem.

Okay, we are careful around here about helping people with homework. So it's good to know this is not a homework problem
.

I guess you've tried some obvious ways to do this, but let me just suggest the first thing that comes to my mind. There may be reasons why it won't work; if so I'll leave it to you to provide more info.

May I presume line L is in a known orientation (perhaps vertical)? That means the slop of the other straight side -- let's call it M -- is known.

From the slope of M and the coordinates of the common vertex of L and M, you can write an equation for M in y=mx+b form. Then you need to solve for the intersection of the line with f(x); there are standard techniques for doing this numerically.

Once you know the common vertex of L and M, as well as the intersection of M with f(x), you can get the length of M using the Distance Formula from geometry.

Hmmm, when θ is 90°, the line is simply y=whatever (the y-coordinate of the vertex common to L and M. Not sure why that would be a problem.

Its understandable for your skepticism, and so thank you for your time. I guess I could be a little less vague. Your correct the orientation is vertical and also the function is an exponential function, we can say exp(x).

I just finished trying your method, it works to a point. When I equate the linear function with the known exponential function, it requires a 'Lambert W" (also known as "Product Log") function. I will look into these functions, but at the same time, are there any other suggestions you could make?

Redbelly98
Staff Emeritus
Homework Helper
Okay, glad I was pretty much on target with things.

You have to solve an equation that looks something like
Aekx = mx + b​
or
g(x) = Aekx - (mx + b) = 0​

You can use the Newton-Raphson method to do this numerically; it is described here:
You'll need an initial trial value for x. I suggest using a very large value for x to guarantee convergence to the desired solution. If theta is close to 90 degrees, or greater, than a small trial value of x will give problems. If you have some idea of the largest x could possibly be in practice, that would be a good initial value.

Hope that helps.

Okay, glad I was pretty much on target with things.

You have to solve an equation that looks something like
Aekx = mx + b​
or
g(x) = Aekx - (mx + b) = 0​

You can use the Newton-Raphson method to do this numerically; it is described here:
You'll need an initial trial value for x. I suggest using a very large value for x to guarantee convergence to the desired solution. If theta is close to 90 degrees, or greater, than a small trial value of x will give problems. If you have some idea of the largest x could possibly be in practice, that would be a good initial value.

Hope that helps.

Yes, it does help. I remember that sort of process from class when told to approximate the value of $$\sqrt{}2$$ with out a calculator. Never even crossed my mind to use it outside of that. Thank you very much for your help