# Is this gradient vector field also a conservative vector field?

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?
• Is the function $(xy)^{1/3}$ or $x\cdot y^{1/3}$ – Triatticus Mar 30 '18 at 3:08
• x * y^(1/3). I think a more appropriate function would be x^(4/3), because the gradient vector field is defined for all x,y. – VindictiV113025 Mar 30 '18 at 3:10

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 $$\int_\gamma\nabla f=\int_a^b\Bigl(\frac43\,x(t)^{1/3}\,x'(t)+y'(t)\Bigr)\,dt=x(a)^{4/3}-x(b)^{4/3}+y(a)-y(b)=f(\gamma(a))-f(\gamma(b)).$$