Trajectory of a 2D constant jerk motion

Solution 1:

Your second rotation won't work. There is a sense in which a rotation can work, but that rotation does not take place in the $\Delta\xi$-$\Delta\eta$ plane. There is a "higher dimensional" (generic, projective) version of your problem which simplifies the way you want by means of a rotation in the higher dimensional space. Honestly, though, this is not the most productive way for you to proceed.

A better way (than solving for $T$, or solving for the combination $kT$, in which $T$ always appears) is to recognize that you can go directly from (5) to its solution. In $$ \Delta \xi_k^r = k T \dot{\xi_0^r} + \frac{(kT)^2}{2}u_{\xi}^r \text{,} $$ the terms on the right-hand side are \begin{align*} k T \dot{\xi_0^r} &= \left( \frac{\dot{\xi_0^r}}{\dot{\eta_0^r}} \right) k T \dot{\eta_0^r} = \left( \frac{\dot{\xi_0^r}}{\dot{\eta_0^r}} \right) \Delta \eta_k^r \\ \frac{(kT)^2}{2}u_{\xi}^r &= \left( \frac{u_{\xi}^r}{2 \left(\dot{\eta_0^r} \right)^2}\right) \left( k T \dot{\eta_0^r} \right)^2 = \left( \frac{u_{\xi}^r}{2 \left(\dot{\eta_0^r} \right)^2}\right) \left( \Delta \eta_k^r \right)^2 \end{align*} and we obtain $$ \Delta \xi_k^r = \left( \frac{\dot{\xi_0^r}}{\dot{\eta_0^r}} \right) \Delta \eta_k^r + \left( \frac{u_{\xi}^r}{2 \left(\dot{\eta_0^r} \right)^2}\right) \left( \Delta \eta_k^r \right)^2 \text{.} $$ A very mechanical way to obtain this is through repeated polynomial division (with remainder) with respect to the variable $kT$ (to eliminate all "$kT$"s). \begin{align*} \Delta \xi_k^r &= \frac{k T \dot{\xi_0^r} + \frac{(kT)^2}{2}u_{\xi}^r}{\Delta \eta_k^r} \Delta \eta_k^r \\ &= \frac{k T \dot{\xi_0^r} + \frac{(kT)^2}{2}u_{\xi}^r}{k T \dot{\eta_0^r}} \Delta \eta_k^r \\ &= \left(\frac{\dot{\xi_0^r}}{\dot{\eta_0^r}} + \frac{kTu_{\xi}^r}{2\dot{\eta_0^r}} \right)\Delta \eta_k^r \\ &= \left(\frac{\dot{\xi_0^r}}{\dot{\eta_0^r}} + \frac{\frac{kTu_{\xi}^r}{2\dot{\eta_0^r}}}{\Delta \eta_k^r} \Delta \eta_k^r \right)\Delta \eta_k^r \\ &= \left(\frac{\dot{\xi_0^r}}{\dot{\eta_0^r}} + \frac{\frac{kTu_{\xi}^r}{2\dot{\eta_0^r}}}{k T \dot{\eta_0^r}} \Delta \eta_k^r \right)\Delta \eta_k^r \\ &= \left(\frac{\dot{\xi_0^r}}{\dot{\eta_0^r}} + \frac{u_{\xi}^r}{2\left(\dot{\eta_0^r}\right)^2} \Delta \eta_k^r \right)\Delta \eta_k^r \text{,} \end{align*} then distribute the $\Delta \eta_k^r$ to finish.

If we directly apply this method to your (6), there is an immediate problem -- the remainder on the first division contains $kT$ to the first power. To avoid this, we start with $\left(\Delta \xi_k^r\right)^2$ and perform the repeated division. Much algebra later, we obtain $$0 = -9 \left(\ddot{\eta_0^r}\right)^3 \left(\Delta\xi_k^r\right)^2 \\ {}+ \left( -6 \dot{\eta_0^r} \left(2u_\xi^r \left(\dot{\eta_0^r}\right)^2 + 3 \dot{\xi_0^r} \left( \ddot{\eta_0^r} \right)^2 -3 \dot{\eta_0^r} \ddot{\eta_0^r} \ddot{\xi_0^r}\right) -18 \ddot{\eta_0^r} \left( u_\xi^r \dot{\eta_0^r}-\ddot{\eta_0^r}\ddot{\xi_0^r} \right) \Delta\eta_k^r \right) \Delta\xi_k^r \\ {}+ \left( \left( 6 \dot{\xi_0^r}\left( 2 u_\xi^r \left( \dot{\eta_0^r}\right)^2 + 3 \dot{\xi_0^r} \left( \ddot{\eta_0^r}\right)^2 - 3 \dot{\eta_0^r} \ddot{\eta_0^r} \ddot{\xi_0^r} \right) \right) + \left( 3 \left( 4 u_\xi^r \dot{\xi_0^r} \ddot{\eta_0^r} + 2 u_\xi^r \dot{\eta_0^r} \ddot{\xi_0^r} - 3 \ddot{\eta_0^r} \left( \ddot{\xi_0^r} \right)^2 \right) \right) \Delta\eta_k^r + \left( 2 \left(u_\xi^r\right)^2 \right) \left( \Delta\eta_k^r \right)^2 \right) \Delta\eta_k^r \text{.} $$ This result is cubic in $\Delta\eta_k^r$ and quadratic in $\Delta\xi_k^r$, about the best that can be expected from eliminating between a cubic and quadratic, so this is a cubic curve.

If we go back to system (1) and raise $\xi_k$ and $\eta_k$ to the powers $1$, $2$, and $3$, then eliminate as much as we can between the cubes, then eliminate as much as we can by including the squares, then eliminate as much as we can by including the linear powers, we obtain the equation, where we have used the centered $\xi_k^c = \xi_k - \xi_0$ and $\eta_k^c = \eta_k - \eta_0$, $$ \left( 1, \eta_k^c, \left( \eta_k^c \right)^2, \left( \eta_k^c \right)^3 \right) \cdot \begin{pmatrix} 0 & 6 \dot{\eta_0}\left( 3 \ddot{\xi_0}^2 \dot{\eta_0} u_\eta -3 \ddot{\eta_0} \ddot{\xi_0} \dot{\xi_0} u_\eta +2 \dot{\xi_0}^2 u_\eta^2 -3 \ddot{\eta_0}\ddot{\xi_0}\dot{\eta_0}u_\xi +3 \ddot{\eta_0}^2 \dot{\xi_0} u_\xi -4 \dot{\eta_0}\dot{\xi_0}u_\eta u_\xi +2 \dot{\eta_0}^2 u_\xi^2 \right) & 3 \left( -3 \ddot{\eta_0}^2 \ddot{\xi_0} u_\eta +4 \ddot{\xi_0} \dot{\eta_0} u_\eta^2 +2 \ddot{\eta_0} \dot{\xi_0} u_\eta^2 +3 \ddot{\eta_0}^3 u_\xi -6 \ddot{\eta_0}\dot{\eta_0}u_\eta u_\xi \right) & 2 u_\eta^3 \\ -6 \dot{\xi_0} \left( 3 \ddot{\xi_0}^2 \dot{\eta_0} u_\eta -3 \ddot{\eta_0} \ddot{\xi_0} \dot{\xi_0} u_\eta +2 \dot{\xi_0}^2u_{\eta}^2 -3 \ddot{\eta_0} \ddot{\xi_0}\dot{\eta_0}u_\xi +3 \ddot{\eta_0}^2 \dot{\xi_0} u_\xi -4 \dot{\eta_0}\dot{\xi_0}u_\eta u_\xi +2 \dot{\eta_0}^2 u_\xi^2 \right) & -6 \left( -3 \ddot{\eta_0}\ddot{\xi_0}^2 u_\eta +3 \ddot{\xi_0} \dot{\xi_0}u_\eta^2 +3 \ddot{\eta_0}^2 \ddot{\xi_0} u_\xi + \ddot{\xi_0}\dot{\eta_0}u_\eta u_\xi - \ddot{\eta_0}\dot{\xi_0}u_\eta u_\xi -3 \ddot{\eta_0} \dot{\eta_0} u_\xi^2 \right) & -6 u_\eta^2 u_\xi & 0 \\ - 3 \left( 3 \ddot{\xi_0}^3u_\eta - 3 \ddot{\eta_0}\ddot{\xi_0}^2 u_\xi - 6 \ddot{\xi_0} \dot{\xi_0} u_\eta u_\xi + 2 \ddot{\xi_0} \dot{\eta_0} u_\xi^2 + 4\ddot{\eta_0} \dot{\xi_0} u_\xi^2 \right) & 6 u_\eta u_\xi^2 & 0 & 0 \\ -2 u_\xi^3 & 0 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ \xi_k^c \\ \left( \xi_k^c \right)^2 \\ \left( \xi_k^c \right)^3 \end{pmatrix} = 0 $$ The "$0$"s in the lower-right triangle are evidence that we had successful cancellation of terms with total degree exceeding $3$. The $0$ in the upper-left entry shows that there is no constant term (which is what we expect from the centered variables, providing weak evidence that any errors in the above are transcriptional, not arithmetical).