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For a physics project I have to calculate the curvature of a certain particle which deflection dependent on it's current position. I've got three functions:

  • $\theta{(t)}$ which is the deflection at moment $t$;
  • $\phi{(t)}$ would be $\phi{(t)} = \alpha - \int_{dt}^{t} \theta{(p)}dp$ which is the total angle of deflection at moment $t$ in which case $\alpha$ is the starting angle of deflection;
  • \begin{equation} P{(t)}\begin{cases} x{(t)}=v \int_{0}^{t-dt} \cos{(\phi{(p)})} dp\\ y{(t)}=v \int_{0}^{t-dt} \sin{(\phi{(p)})} dp \end{cases} \end{equation} which is the formula for the position $P$ of the particle at moment $t$ and in which $v$ is constant. In addition, $P_{t=0}(0,0)$.

You probably already see what's wrong with this last formula. For example, a better way to write $x{(t)}$ would be $x{(t)}=v \int_{0}^{t} \cos{(\phi{(p)})} dp$, which we can do, as a surface with a width of $0$ would be $0$. However, this is not an option, as $\phi{(t)}$ depends on $\theta{(t)}$ and $\theta{(t)}$ depends on $x{(t)}$.

My question basically is, is there any way to rewrite this integral without having to make it an approximation?

EDT: I could clarify this using Riemann-sums: if we denote $\phi{(t)}$ as a Riemann-sum, for example, we would get $\phi{(t)}=\lim_{d \to 0} \sum_{k=0}^{\frac{t-d}{d}} d \cdot \phi{(d + k \cdot d)}$

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    $\begingroup$ What is $dp$? an infinitesimal length on the particle path or an infinitesimal time interval? $\endgroup$ – James Arathoon Feb 5 at 1:43
  • $\begingroup$ @JamesArathoon An infinitesmal time interval. $\endgroup$ – SuperSjoerdie Feb 5 at 8:14
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I managed to squeeze some mathematical beauty out of this with a fairly simple trick: if we first calculate the derivative of $x{(t)}$ as $$\frac{d}{dt} x{(t)} = \frac{d}{dt} v (\cos{( \Phi{( t-dt)})} - \cos{(\Phi{(0)})}) = v \cos{( \phi{( t-dt)})}$$ as $\phi{(0)}$ is not dependent on $t$. If we now calculate the antiderivative we get $$x{(t)}= v \int \cos{( \phi{( t-dt)})} dt$$ which is better than what we had as original. Of course it’s clearly still recursive, it’s just “more correct” now.

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