Vector Calculus: Level curves and insulated boundaries

In summary, an isobar is a line of constant temperature perpendicular to any part of an insulated boundary.
  • #1
end3r7
171
0
Need help checking if my reasoning is sound for this.

Homework Statement


Isobars are lines of constant temperature. Show that isobars are perpendicular to any part of the boundary that is insulated.


Homework Equations


[tex]u(t,\underline{X})[/tex] is the temperature at time [tex]t[/tex] and spatial location [tex]\underline{X}[/tex]

Let V be the region of the interest, then the at the points in the boundary of V, we have [tex]\nabla{u}\cdot\underline{n} = 0[/tex] (insulation), where [tex]\underline{n}[/tex] is the outward normal at that point.


The Attempt at a Solution


Consider, for a fixed time [tex]t = \alpha[/tex], the isobar [tex]u(\alpha,\underline{X}(s)) = C[/tex], where [tex]C[/tex] is an arbitrary constant and [tex]s[/tex] parametrizes a curve [tex]\partial{X}[/tex]in n-dimensional space. Its derivative for a fixed time is [tex]\displaystyle \displaystyle \frac{du}{ds} = \nabla{u}\cdot\frac{d\underline{X}(s)}{ds} = 0[/tex]. Therefore, the gradient [tex]\nabla{u}[/tex] is perpendicular to the tangent line.

Next, consider a point [tex]X = X_{0}[/tex] in an insulated boundary [tex]\nabla{u(t,X_{0})}\cdot\underline{n} = 0[/tex] [tex]\forall t[/tex], where [tex]\underline{n}[/tex] is understood as the outward normal to the boundary at [tex]X = X_{0}[/tex], which means that the gradient [tex]\nabla{u}[/tex] at that point lies in the tangent n-1 hyperplane.

This means that at [tex]X = X_{0}[/tex], the tangent of [tex]\partial{X}[/tex] lies normal to the n-1 tangent hyperplane, and they are thus perpendicular.
 
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  • #2
Actually isobars are curves of constant pressure not temperature. ("bar" is from "barometric"). I'm not sure if there is a special name for constant temperature.

If T(x,y,z) is the temperature at the point (x,y,z) then [itex]\nabla f\cdot\vec{v}[/itex] is the rate of change of T in the direction of the unit vector [itex]\vec{v}[/itex]. In particular, if [itex]\vec{v}[/itex] is tangent to a curve of constant temperature then [itex]\nabla T\cdot\vec{v}= 0[/itex] because T does not change in that direction. Combine that with the condition on an insulated boundary.
 
  • #3
Then, errr, isn't that what I wrote?
 

Related to Vector Calculus: Level curves and insulated boundaries

1. What is vector calculus?

Vector calculus is a branch of mathematics that deals with the study of vector fields, which are functions that assign a vector to each point in space. It is used to model and analyze phenomena such as fluid flow, electromagnetism, and heat transfer.

2. What are level curves in vector calculus?

Level curves, also known as contour curves, are curves on a two-dimensional surface that connect points with the same value of a given function. In vector calculus, level curves are often used to visualize scalar fields, which are functions that assign a scalar value to each point in space.

3. How are level curves related to insulated boundaries?

In vector calculus, level curves can be used to represent insulated boundaries in two-dimensional space. Insulated boundaries are surfaces that do not allow the transfer of heat or other forms of energy. Therefore, the level curves of temperature in a two-dimensional space can be used to represent insulated boundaries.

4. What is the significance of level curves in vector calculus?

Level curves have several important applications in vector calculus. They can be used to visualize scalar fields and understand their behavior in different regions of space. They also help in solving optimization problems and finding critical points of a function.

5. How can I calculate level curves in vector calculus?

To calculate level curves, you need to first find the contour equation, which is a relationship between the coordinates of the points on the curve and the value of the function at those points. You can then use this equation to plot the curve on a graph or create a contour map to visualize the level curves in a two-dimensional space.

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