When will shapes with these characteristics be closed?
For any function, f(x), let x equal the arc length of the line which creates the shape, and f(x) be the direction that the tangent line of the shape points towards for any particular corresponding arc length (or x value). The direction will be measured in radians.
My question is, is there a way to tell if a function will create a shape that is close when graphed in this form, without actually drawing/graphing it?
This may not be a full answer.
I take this to be a question about differential geometry of curves (see my comment to the question).
I assume we can think of this "shape" as a curve in $\mathbb{R}^2$. The curve is specified by your function $f: \left] a,b \right[ \to \mathbb{R}$ where the independent variable $x\in \left] a,b \right[$ can be thought of as the arc length as measured along the curve, and the function value $f(x)$ is a measure, in radians, of the "head" direction of the tangent vector to the curve, at $x$.
$f$ cannot be too pathological, you should probably assume that it is continuous or something like that.
If $\gamma: \left] a,b \right[ \to \mathbb{R}^2$ is a parameterization, also by arc length $x$, of the curve we are interested in, then the derivative $\gamma'$ with respect to $x$ gives the tangent vector. Since we parameterize by arc length, we can also consider $x$ as the time if we say the speed is the constant $1$. So $|\gamma'| \equiv 1$.
Then I think: $$\gamma'(x) = \begin{pmatrix} \cos f(x) \\ \sin f(x) \end{pmatrix}$$ for all $x\in \left] a,b \right[$, since this is just the unit vector in the direction of $f(x)$.
Then the curve should just be the integral of this: $$\gamma(x) = \begin{pmatrix} \int\cos f(x)\;\mathrm{d}x \\ \int\sin f(x)\;\mathrm{d}x \end{pmatrix}$$ where the indefinite integrals involve arbitrarily choosing integration constants. For example if $0\in\left] a,b \right[$, we can just set $\gamma(0)=\begin{pmatrix} 0\\ 0 \end{pmatrix}$, without loss of generality.
Then, if we know $\gamma(x)$ for all $x$, it still remains to determine whether the curve is "closed". So what do you mean by closed?
Clearly if $\gamma(x_1)=\gamma(x_2)$ for $x_1 \ne x_2$ (i.e. $\gamma$ is not injective), it will mean that the curve self-intersects (crosses over itself). If furthermore $f(x_1)=f(x_2)$ for the same $x_1,x_2$, it could appear that the curve repeats itself (but who knows if it will keep doing so)? If furthermore you can prove that $f$ is periodic with period $(x_2-x_1)$, then the curve is "closed" and periodic.
Maybe if the interval of definition is closed, $[a,b]$, you can check if $\gamma(a)=\gamma(b)$, and if needed also check if the restriction $\left. \gamma \right| _{\left[ a,b \right[ }$ is injective.
Doing this in practice will require specific knowledge of $f$.
Found with Google: Conditions for a parametric curve to avoid self-intersection?