# Constrained Curve in 3 Dimensions [closed]

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?

## closed as off-topic by Kyle Kanos, Jon Custer, Rory Alsop, ZeroTheHero, stafusaMay 5 at 23:37

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• One option is to view the first equation as a definition rather than a constraint... – 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:

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

• 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? – The Mastermage May 3 at 21:24
• @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. – probably_someone May 4 at 8:54
• 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. – 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 $$\lambda_2$$.