I have a 3 DOF robot arm and have been tasked with making it follow a path of $y=6-x^2$. Normally to compute the reference trajectory, I would simply create a linearly spaced vector of $x$ values and then compute a vector of $y$ values using the given equation. However, for this task, I need the reference trajectory to have a constant speed. Our textbook has given us this hint, but so far, I have been unable to do anything about it.

How to use the Leibnitz rule to solve for reference trajectory:

  1. Calculate the length of an arc of the parabola at a generic point whose horizontal position is $x(t)$;
  2. Differentiate the length of the arc of the parabola at time $t$ to deduce the translational velocity along the arc. This velocity is constant and equal to $V$. Thus, you can solve for $\dot{x}(t)$.
  3. Once you have $\dot{x}(t)$, you can deduce the reference position and velocity in $x$ and $y$.

This advice confuses me, mostly because I do not understand how an arc can have a length at one specific point.

Thank you for any help you can give me with this problem.


3 Answers 3


If the norm of the velocity shall remain constant along the arc of the parabola, then it holds $\sqrt{\dot{x}^2+\dot{y}^2}=V$.

Plugging $\dot{y}=-2x\cdot\dot{x}$ into the equation above, we end up with the nonlinear differential equation:

$$ \dot{x}\sqrt{1+4x^2}=V. $$

It is not possible to solve the system symbolically, but we can integrate it numerically by relying for example on the model below:


The simulation provides us with the trajectories $x\left(t\right)$, $y\left(t\right)$, $\dot{x}\left(t\right)$, $\dot{y}\left(t\right)$.

The following snippet plots the data (obtained for $V=0.1$ m/s):

% simulate the model
mdl = 'model';
cs = getActiveConfigSet(mdl);
simOut = sim(mdl, cs);

x = find(simOut.yout, 'x');
y = find(simOut.yout, 'y');
xdot = find(simOut.yout, 'xdot');
ydot = find(simOut.yout, 'ydot');

mdlWks = get_param(mdl, 'ModelWorkspace');
V = getVariable(mdlWks, 'V');

% plot the trajectory
figure('color', 'white');
tiledlayout(3, 2);

nexttile([3 1]);
stairs(x.Values.Data, y.Values.Data);
xlabel('x [m]');
ylabel('y [m]');

stairs(x.Values.Time, x.Values.Data);
stairs(y.Values.Time, y.Values.Data);
xlabel('time [s]');
legend({'$x$' '$y$'}, 'Interpreter', 'latex');

stairs(xdot.Values.Time, xdot.Values.Data);
stairs(ydot.Values.Time, ydot.Values.Data);
xlabel('time [s]');
legend({'$\dot{x}$' '$\dot{y}$'}, 'Interpreter', 'latex');

% verify the constant speed along the arc
vel = sqrt(xdot.Values.Data.^2 + ydot.Values.Data.^2);
stairs(xdot.Values.Time, vel);
ylim([0.9 1.1]*V.Value);
xlabel('time [s]');
legend({'$V$' '$\sqrt{\dot{x}^2+\dot{y}^2}$'}, 'Interpreter', 'latex');

enter image description here

As shown in the bottom right graph, the reconstructed velocity along the path is constant and equates to $V$.

You can find the model and the snippet hosted at https://github.com/pattacini/se-robotics-24695.


The arc length can by calculated by the following formula(section: finding arc length by integration): $$ s=\int_a^b \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx $$

So each segment in your trajectory has a length that can be calculated from the equation above.


Assuming the system sample interval $\Delta t$ is small enough, the elements of the $x$ and $y$ vectors are given by:

$$ x_n = x_{n-1} + \dot{x}_{n-1}\Delta t$$ $$ y_n = 6 - x_n^2$$

As observed in another answer:

$$ \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = V $$

To put in terms of just $dx/dt$:

$$\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} =-2x\frac{dx}{dt} $$

so that:

$$ \frac{dx}{dt} \sqrt{1 + 4x^2} = V $$

or $$ \frac{dx}{dt} = \frac{V}{\sqrt{1 + 4x^2}} $$

Substituting and going back to dot notation:

$$ x_n = x_{n-1} + \frac{V\Delta t}{\sqrt{1 + 4{x_{n-1}^2}}} $$

$$ y_n = 6 - x_n^2$$

We are essentially starting a uniform vector in $t$ not $x$ and going from there.


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