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Is there a math field that deals with quadratic, cubic etc. vectors? Or a non-linear equivalent of a vector? If so, why are they so much less common than linear vectors?

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What do you consider a linear vector? For many mathematicians, vectors are just elements of vector spaces. – rschwieb May 3 '13 at 14:35
@rschwieb A vector that can be represented as a segment of a line. – user73232 May 3 '13 at 14:36
So what is a quadratic vector? A curve bent like the graph of a quadratic equation is? – rschwieb May 3 '13 at 14:51
Maybe you could think of a parametrized curve as being a curved vector space? – Loki Clock May 3 '13 at 22:41

Formally, a vector is an element of a vector space. Other than the usual Euclidean spaces, we have other vector spaces, such as the space of all continuous functions on the unit interval.

What you have learnt in school is probably this: visualise say $(1,2)$ as an arrow pointing from the point $(3,4)$ to the point $(4,6)$. Now instead of thinking about this $(1,2)$ as an arrow moving one unit right and two units up, we associate it with the point $(1,2)$ in a horizontal plane. Then the set of all these vectors is simply associated with the set of all points in the plane ($2$-dimensional Euclidean space).

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Sorry we just started vectors and I have no idea what a vector space is. Could you please simplify your answer? – user73232 May 3 '13 at 14:38

From your response to my comment in your OP, I'll talk about why vectors are always considered "rays" geometrically.

The most important interpretation of a vector is as a list of numbers (we'll say from $\Bbb R^n$.) The list of numbers determines a point in $\Bbb R^n$, and if you imagine connecting this to the origin with a straight line and adding an arrowhead at the end with the point, you have a "ray" representing the vector. Actually you can slide this ray (without changing the direction) all over space and still have the same vector. Strictly speaking it is not a segment because it has a fixed direction. If you put the arrowhead on the other end, it is a different vector.

Since the vector only carries the information about a single point, it would does not determine anything more complex than a line. In order to "curve" the vector, you'd need to say more about the path it follows on its way from start to finish.

Another reason is that the points lying on the line determined by a vector through the origin are a subspace of the big space. Subspaces "aren't curved" in some sense. This property causes $\alpha v,\beta v, (\alpha+\beta)v$ to all lie on the same straight line.

Multiplying a vector (ray) in a real vector space by $\lambda \in \Bbb R$ just stretches or contracts the vector. If the vector were curved somehow, then scaling it would change the curviture of the shape. It does not seem desirable that a geometric property such as curviture is ruined merely by scaling the vector, so it doesn't seem very likely to be useful.

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What you are considering to be a vector can be abstracted as follows: Place the line segment so that its one end is on the origin and the other end is somewhere in the Cartesian plane. The coordinates of the other end can be identified with this straight line and so may be called a vector. So for example $(2,3)$ is a vector.

This can be generalized to $3$ dimensional space, and in fact to $n$-dimensional space with the notion of a straight line joining the origin to the point. (You can abstract it even further in the idea of an arbitrary vector space.)

Now your question, if I am understanding it correctly, is as to why we don't consider curves instead of straight lines to geometrically represent the vector. We could, but that would unnecessarily complicate things: The curve would have to be carefully chosen so that it is defining an invertible function on $\mathbb{R}$ with an appropriate parameter to ensure all points could be represented. Straight lines are pretty simple to handle and satisfy this criteria and hence are used.

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Vectors in the plane or in space are not lines; they are straight-line motions. You can add two motions to get a third, or scale a motion to get a larger or smaller motion in the same direction. But they do not have a start or an end point the way a line segment does.

In more advanced mathematics, we consider more abstract vectors. The fundamental abstract property of vectors is that they can be added together, and they can be scaled, and that vector properties are suitably preserved by these "linear" operations of addition and scalar multiplication: If $u$ and $v$ are vectors, so is $u+v$, and so is $cv$ for any scalar $c$.

So we generalize the idea of a vector from the idea of linear motions in space to the idea of any collection of objects with a notion of sum and scalar multiplication. Looked at this way, all sorts of things can be treated as vectors, even things that have nothing to do with motions or straight lines. The benefit of doing this is that our understanding of vectors can be transported to these other domains to give insight about the other sorts of objects.

One supremely important example of such a domain is that we can treat functions of various sorts as vectors. A function like $f(x)$ can be added to another function to give a third, and can be multiplied by a scalar. There is a zero function, just as there is a zero vector; it is the function whose value is 0 everywhere. Even the concept of vector dot product can be transferred appropriately to function spaces. One can find an analogue of a basis for such a vector space, and show that every function can be represented as a sum of the basis functions. This is the fundamental idea of the branch of mathematics called Fourier analysis, which is one of the most fruitful in mathematics, essential for understanding communications and signal transmission, digitization of music, and all sorts of other things.

For a less awesome but very typical example, see this answer where I showed how to find a polynomial $p$ satisfying $p\left(\sqrt2+\sqrt3\right) = 0$ by considering a certain vector space generated by the "vectors" $\left(\sqrt2+\sqrt3\right)^i$. Once I had cast the problem in this light, I was able to conclude immediately that $\left(\sqrt2+\sqrt3\right)^4$ would be expressible in terms of $\left(\sqrt2+\sqrt3\right)^i$ for $0\le i\le 3$ because the vector space was only 4-dimensional and so no set of 5 vectors could be independent.

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