Conservative Vector Field Potential

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SUMMARY

The vector field \(\vec{E}(\vec{r}) = \frac{\vec{r}}{r^2}\) is proven to be conservative by demonstrating that its curl \(\nabla \times \vec{E} = \vec{0}\). Consequently, \(\vec{E}\) can be expressed as the gradient of a scalar potential function \(f\). The scalar potential is determined to be \(f = \log r\), where \(r = \sqrt{x^2 + y^2 + z^2}\). This conclusion is derived through implicit differentiation and confirms the conservative nature of the vector field.

PREREQUISITES
  • Understanding of vector calculus concepts, specifically conservative vector fields.
  • Familiarity with the curl operator and its implications in vector fields.
  • Knowledge of scalar potential functions and their relationship to vector fields.
  • Proficiency in implicit differentiation and logarithmic functions.
NEXT STEPS
  • Study the properties of conservative vector fields in depth.
  • Learn about the implications of the curl operator in three-dimensional vector fields.
  • Explore the applications of scalar potential functions in physics and engineering.
  • Investigate advanced topics in vector calculus, such as Green's Theorem and Stokes' Theorem.
USEFUL FOR

This discussion is beneficial for students in vector calculus, mathematicians focusing on vector fields, and professionals in physics and engineering who require a solid understanding of conservative fields and potential functions.

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Homework Statement



Let [itex]\vec{E}[/itex]([itex]\vec{r}[/itex]) = [itex]\vec{r}[/itex]/r2, r = |[itex]\vec{r}[/itex]|, [itex]\vec{r}[/itex] = x[itex]\hat{i}[/itex] + y[itex]\hat{j}[/itex] + z[itex]\hat{k}[/itex] be a vector field in ℝ3. Show that [itex]\vec{E}[/itex] is conservative and find its scalar potential.

Homework Equations



All of the above.

The Attempt at a Solution



[itex]\vec{\nabla}[/itex] [itex]\times[/itex] [itex]\vec{E}[/itex] = [itex]\vec{0}[/itex] [itex]\Rightarrow[/itex] [itex]\vec{E}[/itex] is conservative [itex]\Rightarrow[/itex] [itex]\vec{E}[/itex] = [itex]\vec{\nabla}[/itex]f, where f is a scalar function.

xf = x/r2
yf = y/r2
zf = z/r2

r2 = x2 + y2 + z2

By implicit differentiation:

2r[itex]\frac{∂r}{∂x}[/itex] = 2x [itex]\Rightarrow[/itex] [itex]\frac{∂r}{∂x}[/itex] = x/r

And as follows:

[itex]\frac{∂r}{∂y}[/itex] = y/r, [itex]\frac{∂r}{∂z}[/itex] = z/r

[itex]\frac{x}{r}[/itex] [itex]\cdot[/itex] [itex]\frac{1}{r}[/itex] = r-1[itex]\frac{x}{r}[/itex] = r-1[itex]\frac{∂r}{∂x}[/itex] = [itex]\frac{∂}{∂x}[/itex](log r)

[itex]\Rightarrow[/itex] f = log r


This is an extra credit question in my Vector Calculus class. I think this is the correct solution, but I'm not positive.
 
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It looks fine to me.
 

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