Simple physical intuition for differential 1-form of a scalar function acting on a tangent vector? I'm trying to gain some very basic physical intuition as to how the differential 1-form $df$ of a smooth function $f$ acts on a tangent vector $\mathbf{v}$. Say the temperature of a hot plate is given by $f\left(x,y\right)=100-\left(x^{2}+y^{2}\right)$. I'm walking along a curve on the hot plate defined by some parametric equation $\Phi\left(t\right)$. When I get to the point $p\left(3,7\right)$ the tangent vector $\mathbf{v}$ is given by $$\mathbf{v}=\frac{d\Phi}{dt}=\left(\frac{dx}{dt},\frac{dy}{dt}\right)=\left(1,3\right).$$
At that point am I entitled to say that the rate the temperature is changing in the direction of $\mathbf{v}$ is given by$$df\left(\mathbf{v}\right)=\left(-2x\,dx-2y\,dy\right)\left(1,3\right)=-\left(2\times3dx+2\times7dy\right)\left(1,3\right)=-48.$$So if $t$ was time in minutes, and $f\left(x,y\right)$ was in Celsius, that would mean the temperature was changing at the rate of -48 degrees Celsius per minute in the direction of $\mathbf{v}$. Have I got that right?
EDIT 10 Jan 2021
Apologies, I messed up the arithmetic by computing for the wrong point (hopefully now fixed). Also, as pointed out by @Semiclassical the units of temperature are unknown. Having said that, I would like to ask again if my interpretation is correct that at point $p\left(3,7\right)$ the instantaneous rate of change of temperature is equal to -48 units of temperature (whatever they are) per minute in the direction of $\mathbf{v}$?
 A: Based on your calculation that
$\frac{d \Phi}{dt} = (\frac{d \Phi}{dx}, \frac{d \Phi}{dy}) * (\frac{dx}{dt}, \frac{dy}{dt})^T $, then it is interpreted as that the change of temperature at a specific point is equal to the inner product of gradient of $f$ and its mobility at this point.
A: Setup

that would mean the temperature was changing at the rate of -48 degrees Celsius per minute in the direction of $\mathbf v$. Have I got that right?

I would not use that exact phrasing to refer to any relevant quantity, but it's quite possible that what you had in mind was correct. There are three different calculations you might be interested in, that I might phrase as follows:

*

*The rate of temperature change in the direction of $\mathbf v$.

*The rate of temperature change along my path/the curve (taking into account my/its speed).

*The thing you get when you calculate $\mathrm df(\mathbf v)$.

Calculation 1

The rate of temperature change in the direction of $\mathbf v$.

Without units
If $\mathbf v=\langle1,3\rangle$, then I would say "the direction of $\mathbf v$" is essentially $\mathbf w=\dfrac{\mathbf v}{\Vert\mathbf v\Vert}=\left\langle\dfrac1{\sqrt{10}},\dfrac{3}{\sqrt{10}}\right\rangle$. And then we can take a directional derivative of $f$ at $(3,7)$ in the direction $\mathbf w$. Since $f$ is differentiable, this is just a dot product: $\left\langle\left.\dfrac{\partial f}{\partial x}\right|_{(x,y)=(3,7)},\left.\dfrac{\partial f}{\partial y}\right|_{(x,y)=(3,7)}\right\rangle\bullet\mathbf w$ $=\left\langle-2(3),-2(7)\right\rangle\bullet\left\langle\dfrac1{\sqrt{10}},\dfrac{3}{\sqrt{10}}\right\rangle$ $=\dfrac{-48}{\sqrt{10}}$.
With units
The meaning of this is difficult to understand properly without units.
Let's say $f(x,y)=100^{\circ}\mathrm{C}-\dfrac{1^{\circ}\mathrm{C}}{\mathrm{m}^2}(x^2+y^2)$. Then we could write $\left.\dfrac{\partial f}{\partial y}\right|_{(x,y)=(3\,\mathrm m,7\,\mathrm m)}$ as $-14^{\circ}\mathrm{C}/\mathrm{m}$. A geometrically relevant direction vector would be unitless, since dividing a vector* by its norm would cancel any units. Therefore, the final answer here is like $\dfrac{-48^{\circ}\mathrm{C}}{\sqrt{10}\,\mathrm m}$; the "rate" is with respect to meters on the hot plate, not time.
*For clarity, $\mathbf v$ might be $\langle1\,\mathrm m/\mathrm{min},3\,\mathrm m/\mathrm{min}\rangle$ and then $\Vert\mathbf v\Vert$ would be $\sqrt{\left(1\,\mathrm m/\mathrm{min}\right)^2+\left(3\,\mathrm m/\mathrm{min}\right)^2}=\sqrt{10}\,\mathrm m/\mathrm{min}$.
Calculation 2

The rate of temperature change along my path/the curve (taking into account my/its speed).

Without units
In the calculation above, we found $\dfrac{-48}{\sqrt{10}}$ as the rate of temperature change in a direction. And the speed of the path at $(3,7)$ is defined to be $\Vert\mathbf v\Vert$, so the rate of change of temperature for you as you're traveling along the curve at a speed of $\sqrt{10}$ would be $\sqrt{10}*\dfrac{-48}{\sqrt{10}}=-48$.
With units
In the calculation above, we found a geometric rate of $\dfrac{-48^{\circ}\mathrm{C}}{\sqrt{10}\,\mathrm m}$. And if your velocity at $(3,7)$ is $\langle1\,\mathrm m/\mathrm{min},3\,\mathrm m/\mathrm{min}\rangle$, then your speed is $\Vert\mathbf v\Vert=\sqrt{10}\,\mathrm m/\mathrm{min}$, so the rate of change of temperature at your location with respect to time (not position) is $\sqrt{10}\dfrac{\mathrm m}{\mathrm{min}}*\dfrac{-48^{\circ}\mathrm{C}}{\sqrt{10}\,\mathrm m}=-48^{\circ}\mathrm{C}/\mathrm{min}$.
Calculation 3

The thing you get when you calculate $\mathrm df(\mathbf v)$.

Without units
As you found in the question, this comes out to be $\left\langle\left.\dfrac{\partial f}{\partial x}\right|_{(x,y)=(3,7)},\left.\dfrac{\partial f}{\partial y}\right|_{(x,y)=(3,7)}\right\rangle\bullet\mathbf v=-48$.
With units
$\left\langle\left.\dfrac{\partial f}{\partial x}\right|_{(x,y)=(3,7)},\left.\dfrac{\partial f}{\partial y}\right|_{(x,y)=(3,7)}\right\rangle\bullet\mathbf v$ $=\left\langle-6^{\circ}\mathrm C/\mathrm m,-14^{\circ}\mathrm C/\mathrm m\right\rangle\bullet\langle1\,\mathrm m/\mathrm{min},3\,\mathrm m/\mathrm{min}\rangle$ $=-48^{\circ}\mathrm{C}/\mathrm{min}$
Conclusion
The calculation of $\mathrm df(\mathbf v)$ gives you the rate of temperature change along the curve (taking into account its parametrization/speed) at the given point. This is not the same thing as what some would call "how the temperature was changing in the direction of $\mathbf v$" (that's calculation 1), even though the direction you're traveling at that point is indeed "the direction of $\mathbf v$". The reason is that your speed matters for $\mathrm df(\mathbf v)$.
