The discussion revolves around a problem involving line integrals and gradient fields, specifically focusing on finding a potential function given its gradient. The original poster presents a gradient vector field and seeks to determine the potential function at a specific point.
The discussion includes attempts to derive the potential function from the gradient, with some participants providing steps and reasoning. There is an acknowledgment of the original poster's confusion regarding the problem setup, and guidance is offered on how to approach finding the potential function.
Participants note that the original poster's textbook lacks examples similar to the problem, which may contribute to the confusion regarding the approach to take. The discussion also reflects on the implications of the initial condition provided in the problem.
Yes, solve the three differential equations to find F(x,y,z) up to a constant. The given condition sets this constant, from which you can find F(x,y,z) for all x,y,z.jonroberts74 said:my book doesn't have a good example of a problem like this, am I looking for a potential?
##\frac{\partial}{\partial x} = 2xyze^{x^2} \Rightarrow \int 2xyze^{x^2}dx = yze^{x^2} + h(y,z)##jonroberts74 said:Homework Statement
##\nabla{F} = <2xyze^{x^2},ze^{x^2},ye^{x^2}##
if f(0,0,0) = 5 find f(1,1,2)Homework Equations
The Attempt at a Solution
my book doesn't have a good example of a problem like this, am I looking for a potential?
##<\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z} >= <2xyze^{x^2},ze^{x^2},ye^{x^2}>##
jonroberts74 said:##\frac{\partial}{\partial x} = 2xyze^{x^2} \Rightarrow \int 2xyze^{x^2}dx = yze^{x^2} + h(y,z)##
##\frac{\partial}{\partial y} = ze^{x^2} = ze^{x^2} + h(y,z) \Rightarrow h_{y}(y,z) = 0##
##\frac{\partial}{\partial z} = ye^{x^2} = ye^{x^2}+h(z) \Rightarrow h_{z}(z) = h'(z)##
##h'(z) = 0 \Rightarrow h(z) = k## for a constant k, and because f(0,0,0) = 5 then
##f= yze^{x^2} + 5##
so ##f(1,1,2) = 1(2)e^1+5 = 2e+5##