Vector Calculus: Level curves and insulated boundaries

[end3r7]

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

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

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

The Attempt at a Solution

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

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

This means that at X = X_{0}, the tangent of \partial{X} lies normal to the n-1 tangent hyperplane, and they are thus perpendicular.

 

[HallsofIvy]

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 \nabla f\cdot\vec{v} is the rate of change of T in the direction of the unit vector \vec{v}. In particular, if \vec{v} is tangent to a curve of constant temperature then \nabla T\cdot\vec{v}= 0 because T does not change in that direction. Combine that with the condition on an insulated boundary.

 

[end3r7]

Then, errr, isn't that what I wrote?