Can a gradient vector field not be a conservative vector field?

The function f(x,y)=x43f(x,y)=x^{\frac{4}{3}} has a gradient vector field. 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?

EDIT: After further research, I've come to the conclusion that (1) a vector field is conservative if and only if it is a gradient of a function. (2) If a vector field is on a simply connected region, and P and Q have continuous first-order partial derivatives, and Py=Qx{\frac{\partial P}{\partial y}}={\frac{\partial Q}{\partial x}} throughout D, then the vector field is conservative. From (2), if all the conditions are not met, we cannot definitively say that the vector field is non-conservative.

#Calculus

Note by Dylan Yu
3 years, 2 months ago

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Comments

I have another inquiry. Why must we look at all first-order partial derivatives if only PyandQx{\frac{\partial P}{\partial y}} and {\frac{\partial Q}{\partial x}} matter?

Dylan Yu - 3 years, 2 months ago
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