Work of a vector field along a curve

[DottZakapa]

Homework Statement

let $f : R^3 → R$ the function $f(x,y,z)=(\frac {x^3} {3} +y^2 z)$
let $\gamma$ :[0,$\pi$] $\rightarrow$ $R^3$ the curve $\gamma (t)$=(cos t, t cos t, t + sin t) oriented in the direction of increasing t.
The work along $\gamma$ of the vector field F=$\nabla f$ is

Relevant Equations

vector field integral

let $f : R^3 → R$ the function $f(x,y,z)=(\frac {x^3} {3} +y^2 z)$
let $\gamma$ :[0,$\pi$] $\rightarrow$ $R^3$ the curve $\gamma (t)$(cos t, t cos t, t + sin t) oriented in the direction of increasing t.
The work along $\gamma$ of the vector field F=$\nabla f$ is:

what i did is

$F = (x^2, 2yz, y^2)$

then i compute the integral

$\int_0^\pi F(\gamma (t)) (\gamma'(t) )dt$

but doing so becomes a very long integral. is there any short cut theorem, that shortens such evaluation? The curl of F is zero, could it be useful?

 

[member 587159]

Been a while since I did this vector field stuff, but I seem to recall that if the curl is $0$, then the vector field is conservative and the integral only depends on the end points, and not on the trajectory (but I could be wrong).

 

[DottZakapa]

Been a while since I did this vector field stuff, but I seem to recall that if the curl is $0$, then the vector field is conservative and the integral only depends on the end points, and not on the trajectory (but I could be wrong).

hmm i know that if the vector field is conservative, then is curl-free (curl is 0). But the opposite is not true. is possible to have a vector field which is curl-free but not conservative.

 

[Orodruin]

hmm i know that if the vector field is conservative, then is curl-free (curl is 0). But the opposite is not true.

Yes it is.

is possible to have a vector field which is curl-free but not conservative.

No.
(Not in $\mathbb R^3$.)

 

[Orodruin]

$\int_0^\pi F(\gamma (t)) (\gamma'(t) )dt$

What is the $t$ derivative of $f(\gamma(t))$?

 

[DottZakapa]

thank you all, i understood the point, the vector field is represented as the gradient of scalar function, there fore, if the vector field can be represented as the gradient of a scalar function, and the curl is zero, then the vector field is conservative, consequently the work does not depend from the path but only from the end points. If the end points are equal, then it becomes zero.

 

[Orodruin]

if the vector field can be represented as the gradient of a scalar function, and the curl is zero

If the vector field is the gradient of a function, then the curl is zero. The following statements are equivalent:

1. $\nabla \times \vec v = 0$.
2. There exists a scalar function $\phi$ such that $\vec v = \nabla \phi$.
3. $\vec v$ is a conservative vector field.

In other words, if any of those statements is true, then the other two are true as well. If any statement is false, then the others are false as well.

 

[vanhees71]

1. must be valid in a simply connected region. Then 2. follows for the simply connected region.

My favorite counterexample for the more general statement, leaving out the constraint of the region being simply connected is the famous "potential vortex"
$$\vec{v}=\frac{1}{x^2+y^2} \begin{pmatrix} -y \\ x \\0 \end{pmatrix},$$
which is defined on $\mathbb{R}^3 \setminus \{(x,y,z)|x^2+y^2=0 \}$, which is not simply connected. Everywhere in the domain is $\vec{\nabla} \times \vec{v}=0$, but it's not a conservative vector field, because the line integral along the circle $x^2+y^2=1$ is not $0$ ;-).

 

[Orodruin]

1. must be valid in a simply connected region. Then 2. follows for the simply connected region.

My favorite counterexample for the more general statement, leaving out the constraint of the region being simply connected is the famous "potential vortex"
$$\vec{v}=\frac{1}{x^2+y^2} \begin{pmatrix} -y \\ x \\0 \end{pmatrix},$$
which is defined on $\mathbb{R}^3 \setminus \{(x,y,z)|x^2+y^2=0 \}$, which is not simply connected. Everywhere in the domain is $\vec{\nabla} \times \vec{v}=0$, but it's not a conservative vector field, because the line integral along the circle $x^2+y^2=1$ is not $0$ ;-).

Based on the OP, the domain is $\mathbb R^3$, which is simply connected.

If we want to go beyond that I would say the vortex field has a well defined curl in the distributional sense and that curl is non-zero (it is a delta distribution). ;)