Understanding Slope Better

Question

Recently, I have realized how much I have taken for granted in understanding the slope of a line and the slope of a curve in general. With that said, I wanted to clear up my understanding to make sure I am on the right track.

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After reading this question: "Understanding the slope of a line as a rate of change", I see now how the slope of the line $y = f(x): = mx + b$ (where $f: \mathbb{R} \to \mathbb{R}$) is the rate of change of $y$ with respect to $x$. However, where I get a little confused is this: since the value $y$ of the function changes by an amount $\Delta y$ for every $\Delta x$ change in $x$, therefore $y$ changes by an amount $ m =\frac{\Delta y}{\Delta x}$. So, if we were to study the line $y = \frac{2}{3}x$, does that mean the slope of the line can be interpreted as "two units of $y$ every three units of $x$"? With more relatable units, if $y$ is measured in meters and $x$ is measured in seconds, would the rate of change of a particle traveling along this line be read as "two meters every three seconds"?

This seems right, however in general for any differentiable curve $g: \mathbb{R} \to \mathbb{R}$ that may not be a line i've seen for example that $g'(3) = \frac{2}{5} $ (given in meters/seconds) is read as "two-fifths meters per second"; simply because the derivative is the rate of change at a point (or at an instant).

Is my understanding correct at all?

Answer 1

So, if we were to study the line $y = \frac{2}{3}x$, does that mean the rate of change of the line can be interpreted as "two units of every three units of "? With more relatable units, if is measured in meters and is measured in seconds, would the rate of change of a particle traveling along this line would be read as "two meters every three seconds"?

Yes, you are exactly right. And if it helps you build a more physical intuition, the units of the slope are the units of $y$ divided by the units of $x$. So the number of hot-dogs Takeru Kobayashi eats in a hot-dog eating competition can be modeled with the line $y = m x$, where $y$ has units of hot dogs, $x$ has units of minutes, and $m \simeq 6.6$ hot dogs / minute. Very impressive.

As for more general differential curves on any domain: a slope is a slope. The fact that a derivative exists guarantees that $g(x)$ can be approximated as a line in the neighborhood centered at $x = 3$. It is possible to overthink this. Why not eat a hot dog?

Answer 2

The derivative at a point $x_0\in\mathbb{R}$ of a differentiable function $f:\mathbb{R}\to\mathbb{R}$ is a local linearization of the function. The derivative $f'(x_0)$ is associated with a linear transformation: $$ h\mapsto f'(x_0)h\tag{1} $$

Depending on whether one is talking about the derivative $f'(x_0)$, for the linear transformation (1), one has all sorts of "interpretations".

Geometrically, the graph of the linear function (1) is the tangent line to the graph of the function $f$. The "slope" $m=f'(x_0)$ of the tangent line $y=f'(x_0)(x-x_0)$ says that whenever $x$ increases $h$ units at $x_0$, i.e., from $x_0$ to $x_0+h$, the dependent variable $y$ increases $mh$ units.

If one interprets the function $f$ as a one-dimension motion of a particle (so that the variable $x$ is interpreted as time), then the sign of $m$ tells which direction ($+1$ for right and $-1$ for left) the particle is traveling, and the absolute value $|m|$ says how fast the particle travels at the moment of $x=x_0$.