I have a cubic polynomial with a known arc length and a known starting point, and I'd like to find the x and y positions of the end point.
First, however, let me explain why I am doing this. I am looking to supply a length to a function in order calculate the $x$ or $y$ or both of the point that this distance ends at along a given cubic polynomial. I am searching for this point this way because I need the freedom to calculate this where and whenever, in hopes of using this data to calculate the slope at that point.
To make things clearer, say I have point $(0, 0)$ lying on polynomial $\frac 34x^3 + x$. If I had previously calculated a distance along this line, say $16$, where units are arbitrary, how would I find the coordinates at that position along the line, with the distance starting at $(0, 0)$?
Unfortunately, in this instance $x$ cannot be provided to calculate the $y$. I calculated the distance with an integral previously, and I am wondering if there is a way to kind of reverse the process by solving for an upper limit with an arc length.
I hope this makes some sort of sense.
Thanks!
First of all, "Welcome to the site !".
If I properly understood, you have $$L=\int_0^t \sqrt{1+\left(\frac {dy}{dx}\right)^2}\, dx$$ where you knwo the function $y(x)$ and the arc length $L$ and you search for $t$.
In most cases, this would be very difficult and, more than likely, you will need some numerical method.
Considering your case $y=\frac 34x^3 + x$, the resulting antiderivative will involve elliptic integrals of the first and second kinds (moreover with complex arguments - just ahve alook here) and the inversion of the formula is strictly impossible (notice that, if you precise bounds in the Wolfram Alpha window, it will not do it excpet if you have to Pro version).
But, in many cases, we can an estimate of $t$ since $$L > \int_0^t \left|\frac{dy}{dx}\right|\, dx$$ For your case $L=16$, this would give $t_0 \approx 2.61$ and this is an overestimate of the solution.
So, let $$F(t)=\int_0^t \sqrt{1+\left(\frac {dy}{dx}\right)^2}\, dx-L$$ be the function you look for the zero. Using Newton (or Taylor series), you will have $$F(t)\sim F(t_0)+ F'(t_0) (t-t_0) =0 \implies t_1=t_0-\frac {F(t_0)} {F'(t_0)}$$and you know $F'(t)$ since it is the integrand (fundamental theorem of calculus). The only problem is to be able to compute $F(t)$ for any $t$. Repeat the process until you reach the desired accuracy.
Applied to your example, the iterates will be $$\left( \begin{array}{cc} n & t_n \\ 0 & 2.613383734 \\ 1 & 2.588786071 \\ 2 & 2.588566341 \\ 3 & 2.588566323 \end{array} \right)$$
Edit
To give you an idea of the complexity, consider a parabola $y=ax^2+bx$ which seems to be simple. In fact, you would need to solve for $t$ the equation $$4aL+b\sqrt{b^2+1} +\sinh ^{-1}(b)=(2 a t+b)\sqrt{(2 a t+b)^2+1} +\sinh ^{-1}(2 a t+b)$$ Let $$k=4aL+b\sqrt{b^2+1} +\sinh ^{-1}(b)\qquad \text{and}\qquad z=2 a t+b$$ which makes the equation to be $$k=z\sqrt{z^2+1} +\sinh ^{-1}(z)$$