I have a particle in a 3D space that moves on a curve of the function $$r(x)=\begin{bmatrix}x \\ x\sin(x) \\ \exp(x^2)\end{bmatrix}$$

I know that there must be 1 degree of freedom left thus $S = 3N-P$ must lead to $1=3*1-P \Rightarrow P=2$ therefore there must only be two equations of constraint but i read that there are the three constraining equations

\begin{align}r_{1}-x&=0 \\ r_{2} -x\sin(x)&=0 \\ r_{3} - \exp(x^2) &= 0\end{align}

What is wrong there why do I have three constraints when it should only be two?


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  • $\begingroup$ One option is to view the first equation as a definition rather than a constraint... $\endgroup$ – Qmechanic May 3 at 10:33

Your first equation, $r_1=x$, isn't actually constraining anything. It just defines the symbol that you're going to be using for $r_1$ in the other two equations.

In order for an equation to be a constraint, it has to either:

  • Impose a relationship between two or more degrees of freedom, for example $5\sqrt{r_1}=2r_2^2$, or

  • Impose a relationship between one or more degrees of freedom and a constant, for example $r_1=5$.

Your equation $r_1=x$ does neither of these things. It relates a single degree of freedom to a non-constant parameter. As such, it's not a constraint by itself. You can actually see this relatively easily - by substituting $r_1=x$ into the other two equations, you get two equations that do satisfy the definition of constraint:

$$r_2-r_1\sin{r_1}=0$$ $$r_3-e^{r_1^2}=0$$

  • $\begingroup$ but what would this be like if the first part would be instead of $$r_{1} = x$$ $$ r_{1} = tan(x) $$ wouldnt this be a constraint then and therefore we would have 3 Constraints and thus 0 Degrees of freedom, which cant be but what am i missing there? $\endgroup$ – The Mastermage May 3 at 21:24
  • $\begingroup$ @TheMastermage No, it's still a relation between one degree of freedom and a non-constant parameter. It's still just a definition of $x$ as used in the other two equations, and you can still get two contstraints by substituting $x=\arctan{r_1}$ in the other two equations. $\endgroup$ – probably_someone May 4 at 8:54
  • $\begingroup$ ok cool, so you could basically say the ammount of parameters i have is the degrees of freedom i have if it is in this Parametrix Form. And also Thank you very much for answering my questions. $\endgroup$ – The Mastermage May 4 at 13:03

In your equations $x$ isn't one of coordinates, it's a parameter. It could as well be $w$, $q$, $k$.

In 3D geometry a curve is defined in two ways:

  • writing two equations among the three coordinates (not necessarily cartesian)

  • giving parametric equations, i.e. three equations of the form $$\eqalign{ \xi &= f(u) \cr \eta &= g(u) \cr \zeta &= h(u) \cr}$$ where $(\xi,\eta,\zeta)$ are coordinates - your $(r_1,r_2,r_3)$.


To find the constraint equations use the fact that constraints do no work, and thus the product of constraint forces $\vec{F}$ and velocity $\vec{v}$ must be zero in a vectorial form $$ \vec{F} \cdot \vec{v} = 0 $$

Where $\cdot$ is the vector inner product.

We only care about directions here, and the direction of velocity is found by differentiation $$ \vec{v} = \frac{\rm d}{{\rm d}t} \vec{r}(t) = \pmatrix{1 \\ t \cos(t) + \sin(t) \\ 2 t \exp(t^2)} \dot{t} $$

Where $\vec{r}(t) = \pmatrix{t \\ t \sin(t) \\ \exp(t^2) }$ is the path curve equation.

So the constraint equations are all the directions orthogonal to $\vec{v}$. For example

$$\vec{n}_1 = \vec{v} \times \pmatrix{0\\0\\1} \propto \pmatrix{t \cos(t)+\sin(t) \\ -1 \\ 0}$$ $$\vec{n}_2 = \vec{v} \times \pmatrix{0\\1\\0} \propto \pmatrix{-2t \exp(t^2)\\0\\1}$$

Where $\times$ is the vector cross product.

So the constraint forces are some kind of linear combination of the two constraint directions

$$ \vec{F} = \lambda_1 \pmatrix{t \cos(t)+\sin(t) \\ -1 \\ 0} + \lambda_2 \pmatrix{-2t \exp(t^2)\\0\\1} $$

You can prove that $\vec{F} \cdot \vec{v}$ is zero for all instances of $t$. So for ever $t$ there are two variables that describe the constraints, $\lambda_1$ and $\lambda_2$.


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