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It's easy to accept that some concepts were immaculately concepted. There's just too much math out there to understand all of it. However, there were always a few bits of math that I felt like had been unjustly left behind. That I just used, but never really understood. I call those math crumbs, because they fell off the edges, and their explanations got left behind. Here are a small collection of concepts that are easy to explain, but I never took the time to seek them out. Until now.

The Power Rule

If you want to know how quickly a function is changing - maybe you have a position function and you want velocity - you can take a derivative. To find this, we want to take a small run and see what the resulting rise is. In other words, you take a step along the function and see how much your position changed.

Let's take a function, say $f(x) = x^2$. If we take a small step along that function of size $\Delta x$, that starts at position $x$, our final x position is $x + \Delta x$. If we evaluate the function at that point, $f(x + \Delta x)$, and subtract the value at the original position, $f(x + \Delta x) - f(x)$, we are given the change in position. You may recognize that ass the numerator of the limit definition of the derivative.

But what I always missed was what that expression actually looks like. Let's evaluate it for $f(x)$ from earlier.

So for any step $\Delta x$, we will increase our height by $2x\Delta x + \Delta x^2$. This represents the value that the function will change by. The rate at which the function changes comes when we consider the slope of that change. That is the rise that resulted from the run $\Delta x$, so if we divide by our run, we get our slope, which is $2x + \Delta x$.

If we make increasingly smaller steps, we can say that the $\Delta x$ term becomes insignificant. This leaves us with the rate of change of the function, which is $2x$.

If you say that the power rule is a direct result of the limit definition of a derivative, it's only satisfying if you understand the limit definition. And in general, the power rule is a result of FOILing. Multiplying out $(x + \Delta x)^3$ or $5(x + \Delta x)^3 - 2(x + \Delta x)^2$ will only leave you with the original, which you will subtract off, and a most significant term.

That most significant term has the next highest $x$ power, which means it has only 1 instance of $\Delta x$, which gets infinitely small.

Let's just foil this out.

If we divide this by $\Delta x$ again, to get a slope And as $\Delta x$ gets infinitely small, everything but $3x^2$ goes away. This is a direct result of FOILing. Only one term of the fully FOIL'd out expression will have exactly 1 $\Delta x$. After being divided by $\Delta x$, to get a slope, that term will have no $\Delta x$, so it will be all that is left.

Fractional Powers

This one is short, but nobody ever explained this one to me either. Maybe because it was so blindingly obvious to them, but I was always confused. How you have a fraction as an exponent?

If we have a number, say $2^3$, and multiply it by another exponent of $2$, say $2^5$, we can write out all of those $2$s.

and then convert them to a new exponent, which is $2^8$, which you get just by counting the $2$s. You have 3 copies of 2. You add 5 copies of 2, you get 8.

So in general, $2^a \cdot 2^b = 2^{a + b}$.

But sometimes we put fractions in the power, like $2^\frac{1}{2}$. If $2^8$ means writing a 2 out 8 times, how do you write 2 a half a time?

You can't. It's completely right to think that writing $2^\frac{1}{2}$ is nonsense. However, consider this little puzzle. Is there a number $a$ that satisfies $2^a \cdot 2^a = 2$? In other words, $2^{a + a} = 2?$

Well we know that $2^1 = 2$, so $a$ must be $\frac{1}{2}$

And if this number, $(2^a)^2 = 2$, then $2^a = \sqrt{2}$ And if $a = \frac{1}{2}$, then $2^\frac{1}{2} = \sqrt{2}$

We write square roots as fractional powers becuase we want $2^\frac{1}{2} \cdot 2^\frac{1}{2}$ to equal $2^{\frac{1}{2} + \frac{1}{2}}$, which equals 1.

(Also, did you catch me cheating? I snuck in the fact that $2^{2a} = (2^a)^2$. That was because we had $2^a \cdot 2^a$, and multiplying a number times itself is the same as squaring. That's one motivation/explanation for why $(2^a)^b = 2^{ab}$. The other is just to write $2^a$ down $b$ times, but that's no fun)

Re-Inventing Trig

All right triangles are circles. Let me explain.

If you want to make a right triangle, all you need is a hypotenuse.

And then you can drop the hypotenuse onto each axis, to create the other two legs of the triangle.

If you want to make a right triangle with hypotenuse length $r$, you create a circle of radius $r$.

And pick some angle

And drop the x and y components down.

And you've created a right triangle. What are the lengths of those $x$ and $y$ components (highlighted in red)? How can we figure that out?

Well, we could just measure them. If you put a ruler to the x axis and y axis, you could just read off the lengths of those components. That's a little inconvenient to do for every right triangle you want to get the components of.

Instead, let's only measure once. If you take a circle of radius 5 and make a right triangle out of it, you have to make the circle and then pick an angle for your hypotenuse. If you made a right triangle from a circle of radius 1, but with the exact same angle, its components will be exactly one fifth the size of the components of the radius 5 right triangle. (To quickly justify this, imagine two triangles, one with angle 0, and the other with angle $\frac{\pi}{2}$. Those are a perfectly horizontal line, and a perfectly vertical line. They have lengths equal to the radius of the circle, becuase they are themselves radii. Any other triangle from that circle has components who are built from some factor of those components, so that combination will also be scaled by the radius of the circle)

This picture makes it clearer. The left circle has radius $r$, the right has radius $2r$, because $r + r = 2r$. You can visually see that the components of a triangle created from the hypotenuse of the right circle will have components that are twice the size of the small triangle's, because we made the radius 2 triangle from stacking 2 radius 1 triangles next to each other, and that new triangle has an x-component with length $2x$ and a y-component with length $2y$

So if we know the measurements of every triangle of hypotenuse length 1, we can calculate the measurements of every other right triangle. We figure out what angle the hypotenuse has, find the measurements of the length 1 right triangle with that angle, and multiply it by the length of the original triangle's hypotenuse.

To find those measurements of the length 1 hypotenuse, you can just look them up in a table. We have named those tables $\sin(\theta)$ and $\cos(\theta)$. When you plug in the angle of your right triangle, $\theta$, it spits out the measurement of the component of the hypotenuse length 1 triangle with the same angle $\theta$.

Let's go ahead take all of those measurements. We'll use $\sin(\theta)$ - the size of the y component.

At each point of the graph on the right, the function has the same height as the tip of the radius on the left. In other words, each point's height is the length of the y component of the triangle formed from the radius on the left.

I hope this has illuminated how trig functions can be relevant to both circles and triangles - because triangles are circles! And that the $\sin$ and $\cos$ functions are just a table of values, precalculated based on the measurements we took of a right triangle with hypotenuse length 1. When we do all of those calculations and plot them, we see the classic sine wave, as the animation drew.