Work of a vector field along a curve

In summary, the function ##f(x,y,z)=(\frac {x^3} {3} +y^2 z)## and the curve ##\gamma (t)## are given. The work along ##\gamma## of the vector field ##F=(x^2, 2yz, y^2)## is computed by integrating ##F(\gamma (t)) (\gamma'(t) )##. If the vector field is conservative, then the curl is zero and the integral only depends on the end points. However, it is possible for a vector field to be curl-free but not conservative, as shown by the example of the "potential vortex" in ##\mathbb{R}^3 \set
  • #1
DottZakapa
239
17
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?
 
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  • #2
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).
 
  • #3
Math_QED said:
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.
 
  • #4
DottZakapa said:
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.

DottZakapa said:
is possible to have a vector field which is curl-free but not conservative.
No.
(Not in ##\mathbb R^3##.)
 
  • #5
DottZakapa said:
##\int_0^\pi F(\gamma (t)) (\gamma'(t) )dt##

What is the ##t## derivative of ##f(\gamma(t))##?
 
  • #6
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.
 
  • #7
DottZakapa said:
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.
 
  • #8
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## ;-).
 
  • #9
vanhees71 said:
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). ;)
 
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Related to Work of a vector field along a curve

What is a vector field?

A vector field is a mathematical concept that assigns a vector quantity to each point in a given space. This vector quantity can represent a physical quantity such as force or velocity.

What is the work of a vector field along a curve?

The work of a vector field along a curve is a measure of the total energy transferred by the vector field to a particle as it moves along the given curve. It is calculated by taking the dot product of the vector field and the tangent vector of the curve at each point.

How is the work of a vector field along a curve related to line integrals?

The work of a vector field along a curve is mathematically represented by a line integral. This means that the work can be calculated by integrating the dot product of the vector field and the tangent vector of the curve over the given curve.

What is the significance of the direction of the tangent vector in calculating the work of a vector field along a curve?

The direction of the tangent vector is important in calculating the work of a vector field along a curve because it determines the direction in which the vector field is acting on the particle. The dot product takes into account both the magnitude and direction of the vector field and the tangent vector.

Can the work of a vector field along a curve be negative?

Yes, the work of a vector field along a curve can be negative. This means that the vector field is doing work against the movement of the particle along the curve, rather than aiding it. This can happen if the vector field is acting in the opposite direction of the tangent vector at certain points along the curve.

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