still slightly confused with finding a potential function of a vector field?

Question

how do i find the potential function of this vector field, which i have already proved to be conservative?

F(x, y, z) = (-2xy^2z^3, -2x^2yz^3, -3x^2y^2z^2 - 4z)

I thought it was -3x^2y^2z^3 -2z^2 but thats wrong. Any help?

Answer 1

No. It is -x^2 y^2 z^3-2 z^2.
You want a function whose partial derivative with respect to x is
-2xy^2z^3,
whose partial derivative with respect to y is
-2x^2yz^3,
and whose partial derivative with respect to z is
-3x^2y^2z^2-4z.
If you integrate with respect to z, you get
-x^2y^2z^3-2z^2
which satisfies all three conditions. Usually, you would still have to determine a 'constant of integration' which involves x and y.

Answer 2

Sorry, I screwed up the initial answer. Here's the rework: Wow, I had to look this up in my calc book. Not entirely sure I remember how it works but it looks to me like you construct a vector function T that moves a particle from the origin A (0,0,0) to a point B (x1, y1, z1) where the parameter is t. T(t) = (x1*t, y1*t, z1*t), 0 <= t <= 1 Then you do the dot product of F with T, and integrate it from A to B: f(x1, y1, z1) = ∫ from A to B : F · T ds = ∫ from A to B : 2xy dx + (x² + z²) dy + 2yz dz substituting in the parameterized variables, x = x1t, dx = x1dt, y = y1t, dy = y1dt, z = z1t, dz = z1dt: = ∫from 0 to 1 : 2(x1)(y1)t²(x1 dt) + (x1² +z1²)t²(y1 dt) + 2(y1)(z1)t² (z1dt) = ∫ from 0 to 1: (x1²y1 + y1z1² + 2x1²y1 + 2y1z1²)t² dt = 1/3 (3x1²y1 + 3y1z1²)t³, evaluated from t=0 to t=1 = x1²y1 + y1z1² But since (x1,y1,z1) is an arbitrary point, the overall function would just be: f(x,y,z) = x²y + yz² I don't know if that's 100% right, but that's what I think my textbook says. Sorry for the initial mistake.

Answer 3

here f =potential function
f(x)=-2xy^2z^3
and integrate to get
f= -x^2y^2z^3+c(y,z)
take derivative w.r.t. y
-2x^2yz^3+cy(y,z)=f(y)= -2x^2yx^3
cy(y,z)=0
integrate w.r.t. y to get
c(y,z)=0+c(z)
f= -x^2y^2z^3 +c(z)
take derivative w.r.t. z
-3x^2y^2z^2 +c'(z)=f(z)
= -3x^2y^2z^2-4z
c'(z)= -4z
integrate
c(z) = -2z^2

f= -x^2y^2z^3 -2z^2

therefore, potential function

f(x,y,z)= -x^2y^2z^3 - 2z^2

you've probably used a minus
instead of a plus in your algebra-
cheers

i hope that this helps

Answer 4

You think the same as me, best of luck!