How to determine if a field is a gradient field? What is a gradient field? If I have a scalar function : $$f(x,y)$$and the derivative of the scalar function   i.e. the gradient of the function is : $$ \nabla f $$Is it true that if $ \nabla f $ somehow equals the vector field only then can I call it a gradeint vector field?
I'm considerably new to this topic..so forgive me if this seems like a stupid question.
If I have a gradient that looks like $$<xy,y^2>$$ what would the original function look like? How do I know if the field is a gradient field.
Is it necessary that the fundamental theorem of line integral calculus : $$\int_C \overrightarrow {\nabla f}. \overrightarrow {dr} = f(P_1)-f(P_0)$$ where $P_1$ and $P_0$ are the final and initial points on the curve satisfies only if f is a gradient field.
 A: What you really want is the notion of a conservative vector field, which can be defined as a field such that the line integral of the field around any closed loop in the domain is zero. A smooth enough vector field is conservative if it is the gradient of some scalar function and its domain is "simply connected" which means it has no holes in it.
For a given smooth enough vector field, you can start a check for whether it is conservative by taking the curl: the curl of a conservative field is the zero vector.  However to ensure that it is conservative, you must additionally check that the domain is simply connected, otherwise line integrals around loops that contain the hole in the domain will not necessarily be zero.
The scalar function that a vector field is the gradient of is called the potential function of the vector field. It can be found by integrating the equations that define this relationship. For example, if I have the field $\langle y,x \rangle$, then a potential function $f$ must satisfy $\frac{\partial f}{\partial x} = y,\frac{\partial f}{\partial y}=x$. These equations can be integrated: the first one implies $f(x,y)=xy+g(y)$ (note that the "constant of integration" is just a function that doesn't depend on $x$). Similarly, the second one implies $f(x,y)=xy+h(x)$. Thus $g(y)-h(x)=0$, so $g$ and $h$ must both be equal to the same constant, and so $f(x,y)=xy+c$ for any number $c$.
