Isobars are perpendicular to any part of insulated boundary > Isobars are lines of constant temperature. Show that isobars are perpendicular to any part of the boundary that is insulated. Consider the heat equation $$u_t=k \nabla^2 u$$ Lines of constant temperature should satisfy $u_t(\vec{x},t)=0$ or $\nabla^2 u =\nabla \cdot \nabla u= 0$. A part of the boundary that is insulated must satisfy $\nabla u = 0$. And that's where I'm stuck. In my understanding, we need to express isobars as vector, as well as the boundary, and then show that the dot product is $0$. I'd appreciate a hint.

Insulated (adiabatic) boundary implies no heat is getting out, thus $$\boldsymbol\
abla u \cdot\boldsymbol n=0$$ where $\boldsymbol n$ is a vector normal to the boundary which is another way of saying it is normal to the tangent plane. Isobars are parametrized curves of constant value. Use $\boldsymbol y(r)$ as the parametrized curve. Then $$u(\boldsymbol y(r),t)=c$$ where $c$ is a constant. Take the derivative of this equation to obtain $$\boldsymbol\
abla u\cdot\frac{d\boldsymbol y(r)}{dr}=0$$ This can be read as $\boldsymbol\
abla u$ is perpendicular to the tangent of the isobars. This makes it perpendicular to the tangent plane at the boundary, which is what we wanted to prove.

**Aside:** Wherever this is from needs to fix its definition of isobar. Iso- means same while -bar is a measurement of pressure. Isobars are actually the lines of constant pressure, _not_ temperature. Isothermal means constant temperature. Just my two cents.