Line integrals, gradient fields

[jonroberts74]

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?

$<\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z} >= <2xyze^{x^2},ze^{x^2},ye^{x^2}>$

 

[CAF123]

my book doesn't have a good example of a problem like this, am I looking for a potential?

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]

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?

$<\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z} >= <2xyze^{x^2},ze^{x^2},ye^{x^2}>$

$\frac{\partial}{\partial x} = 2xyze^{x^2} \Rightarrow \int 2xyze^{x^2}dx = yze^{x^2} + h(y,z)$

$\frac{\partial}{\partial y} = ze^{x^2} = ze^{x^2} + h(y,z) \Rightarrow h_{y}(y,z) = 0$

$\frac{\partial}{\partial z} = ye^{x^2} = ye^{x^2}+h(z) \Rightarrow h_{z}(z) = h'(z)$

$h'(z) = 0 \Rightarrow h(z) = k$ for a constant k, and because f(0,0,0) = 5 then

$f= yze^{x^2} + 5$

so $f(1,1,2) = 1(2)e^1+5 = 2e+5$

 

[ehild]

$\frac{\partial}{\partial x} = 2xyze^{x^2} \Rightarrow \int 2xyze^{x^2}dx = yze^{x^2} + h(y,z)$

$\frac{\partial}{\partial y} = ze^{x^2} = ze^{x^2} + h(y,z) \Rightarrow h_{y}(y,z) = 0$

$\frac{\partial}{\partial z} = ye^{x^2} = ye^{x^2}+h(z) \Rightarrow h_{z}(z) = h'(z)$

$h'(z) = 0 \Rightarrow h(z) = k$ for a constant k, and because f(0,0,0) = 5 then

$f= yze^{x^2} + 5$

so $f(1,1,2) = 1(2)e^1+5 = 2e+5$

Correct!

ehild