Is it correct to say that the total derivative is the maximum directional derivative? I am studying electromagnetism this semester, and this equation appeared: $$dT = \nabla T \cdot dr$$ where $dr$ is the infinitesimal displacement vector. The right hand side is the equation for the directional derivative. Before we get this equation, however, we have this equation: $$dT = (\frac{\partial T}{\partial x}) dx + (\frac{\partial T}{\partial y}) dy + (\frac{\partial T}{\partial z}) dz$$ I bring this second equation up because it is the result of my ultimate confusion. It has been years since I have taken vector calculus, and even when I did, I did not feel comfortable with gradients and such. I know that $\frac{\partial T}{\partial x}$ is the partial derivative of T with respect to x, while keeping y and z constant -- and I can easily calculate these, I just need an intuitive approach. My main confusion comes from the meaning of adding the $dx, dy,$ and $dz$ after the partials. Is this just notation? Is it just a way to make the gradient (a vector) into the scalar function $dT$?
Thank you! :)
This is a good question and it's actually pretty complicated. Mostly, the symbol $d$ is difficult to deal with. There is always a very precise mathematical meaning to it but it is hard to describe, especially in higher dimensions. It involves some multivariable calculus and linear algebra. Below, I'll let $d$ denote the total derivative rather than an infinitesimal displacement vector. For $dx$, $dy$, and $dz$, they can be regarded as both as I'll show below. I'll introduce the general idea first but it's much simpler in your case.
We can regard $T$ as a differentiable function $T: \mathbb{R}^3 \to \mathbb{R}$. If $f: \mathbb{R}^n \to \mathbb{R}^m$ is differentiable, then we usually regard the Total Derivative $df$ to be the linear map associated with the matrix $$df = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \dots &\frac{\partial f_1}{\partial x_n}\\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \dots &\frac{\partial f_2}{\partial x_n}\\ \vdots & & \ddots & \vdots\\ \frac{\partial f_m}{\partial x_1} & \frac{\partial f_m}{\partial x_2} & \dots &\frac{\partial f_m}{\partial x_n} \end{bmatrix}$$ called the Jacobian of $f$. Here we take the $f_i$ to be defined by $$f(x_1, \dots, x_n) = (f_1(x_1, \dots, x_n), f_2(x_1, \dots, x_n), \dots, f_m(x_1, \dots, x_n))$$ which is just $f$ expanded into the coordinates of $\mathbb{R^m}$. This is found by taking the best linear approximation of $f$ at any point and is then, a streightforward generalization of the usual derivative.
Then, we can do some calcuations. Let $x: \mathbb{R}^3 \to \mathbb{R}$ be the function which takes $(x_1,x_2,x_3) \mapsto x_1$ and we define $y$ and $z$ similarly. $y$ takes $(x_1,x_2,x_3) \mapsto x_2$ and $z$ takes $(x_1,x_2,x_3) \mapsto x_3$. Then, we find that the total derivative of $x$ is $$dx = \begin{bmatrix} \frac{\partial x}{\partial x_1} & \frac{\partial x}{\partial x_2} & \frac{\partial x}{\partial x_3} \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}$$ by the definition of $x$. Similarly, we have $$dy = \begin{bmatrix} 0 & 1 & 0 \end{bmatrix} \;\text{and}\;dz = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}.$$
In the case of $T$, the Jacobian is given by $$dT = \begin{bmatrix} \frac{\partial T}{\partial x} & \frac{\partial T}{\partial y} &\frac{\partial T}{\partial z} \end{bmatrix}.$$ Then, we note that given this $dT(x,y,z)$ is not a real number: it's a linear map from $\mathbb{R^3}$ ro $\mathbb{R}.$ Because of that, it would not make sense to write $dT$ as the sum of partial derivatives. It wouldn't take into account the different coordinates of the input. Using the linear properties of matrices, though, we can obtain the next best thing. \begin{align*} dT &= \begin{bmatrix} \frac{\partial T}{\partial x} & \frac{\partial T}{\partial y} &\frac{\partial T}{\partial z} \end{bmatrix} \\ &= \frac{\partial T}{\partial x} \begin{bmatrix} 1 & 0 & 0 \end{bmatrix} + \frac{\partial T}{\partial y} \begin{bmatrix} 0 & 1 & 0 \end{bmatrix} +\frac{\partial T}{\partial z} \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} \\&= \frac{\partial T}{\partial x_1} dx + \frac{\partial T}{\partial x_2} dy +\frac{\partial T}{\partial x_3}dz. \end{align*}
This is the second formula you mentioned.
Essentially, the total derivative is a little different from partial derivatives in the sense it gives a linear map rather than a scalar function. In multivariable calculus and differential geometry, there are numerous "derivatives" and they always have some intersection with each other. In the above case, the Total Derivative and its relation to the Partial Derivative is brought to light. I use these two derivatives much more often than the gradient, but that may be due to my lack of experience in fields that don't. A lot of this notation can be converted to gradients but I find it more intuitive to work with just partial derivatives.