Intro

Sorry that I didn’t notice you’re asking about the contour. My solutions are mostly about the vector field figure instead of the contour.

Reference. Ste16. 16.1. P1072 (E-Ver.)

Below is an intro for the contour of a vector field.

Here are some further explanations for TA Guan (if you need) - welcome for any correction.

“Actually, a contour is a definition for a R n \mathbb{R}^n Rn to R \mathbb{R} R function.”

Reference. Sample Midterm 2-1 with solutions

Consider the function f : R 2 \mathbb{R}^2 R2 → R \mathbb{R} R defined by
f ( x , y ) = 4 x 1 + x 2 + 2 y 2 f(x,y)=\frac{4x}{1+x^2+2y^2} f(x,y)=1+x2+2y24x​
d) Sketch the contours of f f f for the levels k ∈ { 0 , 1 , 2 , 2 } k \in \{0,1,\sqrt2,2\} k∈{0,1,2 ​,2}
The answer should be

Notice the form. We only obtain the contour when k is a constant, which means that f(x,y) is 1-dimensional, viz. the form for the contour should be from R n \mathbb{R}^n Rn to R \mathbb{R} R.

Reference. Sample Midterm 2-2 with solutions

This strengthen our conclusion that the form for the contour should be from R n \mathbb{R}^n Rn to R \mathbb{R} R.

“But a vector field is a R n \mathbb{R}^n Rn to R n \mathbb{R^n} Rn function …”

Reference. Ste16. 16.1 Vector Fields.

From the definition in Ste16, we can find F ⃗ ( x , y ) \vec{F}(x,y) F (x,y) is a vector in the R 2 \mathbb{R^2} R2 plane. This is pretty easy to understand: assume you are discussing a motion in a plane, then iff you use vectors in 2 dimensions, you may express the motion without information loss. Also, a 3-dimentional motion needs a vector that can span for the space.

… so except a 1-D vector field, there is no contour for the vector field.

Pretty easy to comprehend now - to express the “motion” without info loss, you must use the vector with exactly the same number of dimensions. Thus, only 1-dimensional vector field assure a contour (but it’s somewhat meaningless, you know).

I think what you mean is probably a contour for the potential function of a vector field.

Reference. Ste16. 16.3. P1089 (E-Ver.)

As shown above, the potential function is relevant to the vector field. Notice that in the formula above, F is a vector, but f is a function (called 势函数 in Chinese) - you may link this to “magnetic (or electronic, if you like) potential energy” in physics. Assume F is the magnetic force, then
R ↬ f ( x , y ) ⇔ U c o n s e r v a t i v e ( x , y ) ↫ R 1 = ∫ F ⃗ ∙ d r \mathbb{R}\looparrowright f(x,y) \Leftrightarrow U_{conservative}(x,y)\looparrowleft \mathbb{R1}\\=\int \vec{F}\bullet d\bold{r} R↬f(x,y)⇔Uconservative​(x,y)↫R1=∫F ∙dr

By definition, we know that a potential function is like the energy stored from a motion, viz. the potential function is always in R \mathbb{R} R, cf. what we found about contour, the potential function is more likely to fit in your question.

However, if a vector field is not conservative, then it does not have a potential function.

[Again] Reference. Ste16. 16.3. P1089 (E-Ver.)

See the same picture above (with annotations). From the context we conclude that,
F ⃗ c o n s e r v a t i v e ≡ ∇ f = F ⃗ \vec{F}_{conservative}\equiv \nabla f=\vec{F} F conservative​≡∇f=F
(because the description is “that is”, which is equiv. to “iff” - this means a definition).

viz. if F is not conservative, then there does not exist a function f(which is the potential function).

So there would be no contour for the potential function of the vector field.

Refer to discussion in the last part, this part seems pretty clear now.

Hope these helps and welcome for any correction.