# Homework Help: Vector Calculus: Level curves and insulated boundaries

1. Aug 29, 2008

### end3r7

Need help checking if my reasoning is sound for this.

1. The problem statement, all variables and given/known data
Isobars are lines of constant temperature. Show that isobars are perpendicular to any part of the boundary that is insulated.

2. Relevant 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.

3. 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.

2. Aug 29, 2008

### 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.

3. Aug 29, 2008

### end3r7

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