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Understanding Slope Better

Mathematics Asked by Taylor Rendon on February 22, 2021

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.


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?

2 Answers

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?

Correct answer by true on February 22, 2021

The derivative at a point $x_0inmathbb{R}$ of a differentiable function $f:mathbb{R}tomathbb{R}$ is a local linearization of the function. The derivative $f'(x_0)$ is associated with a linear transformation: $$ hmapsto f'(x_0)htag{1} $$

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 $x=x_0$.

Answered by mrsamy on February 22, 2021

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