Calculating endpoint for a function so arc length equals l. I'm trying to simulate a line hanging from a given point using a quadratic function. The line is located at point (x_0, y(x_0)) where y is my quadratic function.

Jaelyn Payne

Jaelyn Payne

Answered question

2022-10-23

Calculating endpoint for a function so arc length equals l.
I'm trying to simulate a line hanging from a given point using a quadratic function.
The line is located at point ( x 0 , y ( x 0 ) ) where y is my quadratic function.
Now, the line has length l, and I simulate it hanging and swinging from side to side by changing the quadratic and linear coefficient. This isn't really important in the problem, but I mention it just to give You a full picture.
Now, the problem is, that when I change the coefficients of the formula, the length of the line changes as well, and I have to pick a point x 1 in which the line will end so the length can stay the same.
Of course, the simplest way is just to pick a point x 1 such that distance from ( x 0 , y ( x 0 ) ) to ( x 1 , y ( x 1 ) ) is equal to l, but this actually calculates the straight-line distance, not the arc length, so for certain coefficeints this becomes really inacurate.
So, the proper way to find x 2 would be solving this equation:
x 0 x 1 1 + f ( t ) 2 d t = l
For a proper x 1 . However, after integrating it's really complicated to "extract" x 1 from the equation.
Thinking about this, I have found a stupid idea that doesn't work, but I dont know why, and this is the main point of my question.
So, for this integral exists a antiderivative F(x), so this whole equation can be represented as:
F ( x 1 ) F ( x 0 ) = l
So:
F ( x 1 ) = l + F ( x 0 )
Now, l is a constant, and x 0 is set (I'm just solving for x 1 ), so F ( x 0 ) is a constant as well. This means, that I just can differentiate the whole equation over x 1 and get:
F ( x 1 ) = 0 => 1 + f ( x ) 2 = 0
Which is obviously wrong, because it doesn't depend at all on values of l and x 1 .
So now, I have two questions: 1. Why is my differentiaton step wrong? 2. Is there any way to simplify solving for this integral?

Answer & Explanation

Kason Gonzales

Kason Gonzales

Beginner2022-10-24Added 15 answers

Step 1
For question 1:
This is the same as attempting to "solve" a quadratic equation as follows:
x 2 + 3 x + 2 = 0 d d x ( x 2 + 3 x + 2 ) = d d x 0 2 x + 3 = 0 x = 3 2
One can easily verify that 3/2 is not a solution of the first equation.
Step 2
The problem is the following:
When we write x 2 + 3 x + 2 = 0, we do not mean that the written equation is an identity. That is, it is not true that x 2 + 3 x + 2 = 0 for all x R .
What we are writing is that there's some x such that x 2 + 3 x + 2 = 0. So it does not make sense to differentiate both sides with respect to x since it does not hold for "variable" x. (If you recall, the derivative of a function is a limiting operation, whereas the above equation holds only for two "discrete" points.)
duandaTed05

duandaTed05

Beginner2022-10-25Added 6 answers

Step 1
For simplifying things, what I suggest is consider your hanging point the origin of your coordinate system, Hence ( x 0 , y 0 ) = ( 0 , 0 )
Now, let the general quadratic passing through the point be of the form
y = a x 2 + b x
y = 2 a x + b
Step 2
Now, for curve length, we have
0 x 1 + ( 2 a x + b ) 2 d x = L
Any further simplification is difficult, and you'd mostly end up using Lambert W functions to reverse solve for x∗.

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