Is this gradient vector field also a conservative vector field?

Question

The function $f(x,y) = x^{4/3} + y$ has a gradient vector field that is defined for all (x,y). The gradient vector field does not have continuous 1st order partial derivatives. Therefore,

1. is the gradient vector field a conservative vector field? and
2. Is a line integral of this vector field independent of path?

Answer 1

The answer to both questions is yes (1. follows from 2.) Let $\gamma(t)=(x(t),y(t))$, $a\le t\le b$, be a piecewise $C^1$ curve. Then $$ \int_\gamma\nabla f=\int_a^b\Bigl(\frac43\,x(t)^{1/3}\,x'(t)+y'(t)\Bigr)\,dt=x(a)^{4/3}-x(b)^{4/3}+y(a)-y(b)=f(\gamma(a))-f(\gamma(b)). $$

Answer 2

After further research I have concluded that any gradient vector field is, by definition, a conservative vector field if the potential function is defined everywhere. There is a theorem that requires continuous 1st order partial derivatives, but if that requirement is not met, it does not mean that the vector field is necessarily non-conservative.

https://mathinsight.org/conservative_vector_field_determine