Temperature Gradient and Fourier's Law I'm having trouble understanding temperature gradient in the context of fourier's law of heat conduction. 
Namely, I do not understand how the change in temperature is approximately equal to what appears to be the formula for a tangent plane in 3 dimensions. 
Furthermore, I do not understand the first equality, where they say "the rate of change of temperature in the direction alpha-hat is the directional derivative."
I apologize for submitting a picture and not using latex. I'm in an area with limited mobility and am posting this from my phone. Thanks in advance for your help. 
 A: If you have a temperature function $T$, $T$ defines an infinite set of surfaces in the tridimensional space, on each of which the temperature is the same. The equation of the surface with temperature $T_{0}$ is
$$T(x,y,z)=T_{0}$$
From the equation you can get, let's say, an explicit value of $z$, so that the surface is
$$(x,y,z(T_{0}))$$
If you have a single-valued function $f$, its derivative at point $x$ gives you the direction of the tangent line at point $(x,f(x))$. So the derivative of $T$ with respect to $x$ and $y$ gives you the "direction" of the tangent plane at point $(x,y,z(T_{0}))$. If you add to it the derivative with respect to $z$, thus forming the gradient, you get the direction of the vector normal to the tangent plane at the same point.
If you follow this (normal) vector, you go to the other surfaces where the temperature is constant, increasing it if you go in its direction, decreasing it if you follow the opposite direction. Therefore, the gradient is the change in the direction of maximum change of the temperature. If you want to know the change in any other direction, it is sufficient to project your vector onto the gradient vector. The proof for this comes from explicitly differentiating (I underline differentiating, not deriving) the temperature field, as was implicitly done in you textbook. 
A: Let $\vec r(s)$ parameterize a space curve, where $s$ is the arc length.  We note that $\frac{d\vec r}{ds}$ is a unit tangent to the curve since $|\frac{d\vec r}{ds}|=1$.  Thus, using the chain rule reveals that the derivative in the direction tangent to the space curve is 
$$\frac{dT}{ds}=\nabla T\cdot \frac{d\vec r}{ds}$$
In the picture posted of the text, we have $\hat \alpha =\frac{d\vec r}{ds}$

ADDITIONAL NOTE:
Now, suppose that $\vec r(s)$ parameterizes a curve on a constant temperature surface (i.e. $T(\vec r(s))=\text{constant}$).  Then, we have 
$$\frac{dT}{ds}=\nabla T \cdot \frac{d\vec r}{ds}=0$$
Since $\frac{d\vec r}{ds}$ is tangent to the constant temperature surface, and since $\nabla T \cdot \frac{d\vec r}{ds}=0$ for any tangent vector, then $\nabla T$ is normal to the constant temperature surface.
