1
$\begingroup$

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?
$\endgroup$
  • $\begingroup$ Is the function $(xy)^{1/3}$ or $x\cdot y^{1/3}$ $\endgroup$ – Triatticus Mar 30 '18 at 3:08
  • $\begingroup$ 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. $\endgroup$ – VindictiV113025 Mar 30 '18 at 3:10
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.