Calculus

Before You Can Begin

• Know Algebra. You can't even begin to understand the concepts with this.
• Know Geometry and Trigonometry. Calculus assumes you know these topics and uses concepts from them frequently.

The specific things you must know:

• Understand that variables can take on any value, up until the point you put constraints on them in the form of inequalities and equations.
• Understand what a function is, what <math>f(x) = x^2</math> means.
• Be familiar with how polynomial functions behave. Can you draw the graph for <math>y = 3x^2 + 5?</math> If not, you'll have a hard time even considering the discussions below.

As for the rest, you can probably pick it up along the way. Don't know what something is? Just look it up.

Limits

Definition

The formal definition of a limit is

<math>\lim_{x \rightarrow c} f(x) = L\,</math>

means that for all real values of <math>\epsilon > 0 \,</math>, there exists <math>\delta \,</math> for all <math>x \,</math> such that <math>0 < |x - c| < \delta \,</math>, we have <math>|f(x) - L| < \epsilon \,</math>.

Although this won't mean much to the junior mathematician, it is important to understand since limits are used a lot more than you think.

In My Words

1. Assume that the limit of <math>\lim_{x \rightarrow c} f(x) \,</math> is L.
2. Choose an arbitrarily small number for <math>\epsilon</math>, something very, very close to 0 but not 0. (Hint: Choose <math>10^{-10}</math> or some other large, negative exponent.)
3. Find the value of <math>x</math> in <math>|f(x)-L|</math> that will get you something smaller than <math>\epsilon</math>.
4. Make sure that <math>|x-c|</math> gives you a real number bigger than 0. That is, you can't use <math>x=c</math>.

If you can do the above for every imaginable <math>\epsilon</math>, then you have found the limit.

In Practice

Finding the limit can be tricky, especially if you don't follow some careful rules.

1. Even if the function is not defined at <math>c</math>, the limit may still exist. Just put your finger over the graph at that point and ignore it because you can't even think about it. (If you're dealing with infinity, this is easy. You can never, ever reach infinity so you can't even consider it.)
2. Think about what happens to the function as you approach <math>c</math> from both sides.
   • If it is a different value from one side than another, the limit doesn't exist. (Unless you're doing one-sided limits, which are rare.)
   • If it switches between two values without stabilizing, the limit doesn't exist.
   • If it shoots off to infinity, the limit doesn't exist.
3. If the function smooths out before reaching <math>c</math>, and tends to a single value, that's the limit.

Notice that the definition of the limit doesn't even enter into the discussion of how to find the limit. We always work backwards, starting with the limit, and see if the definition holds up precisely, rather than the other way around.

Example

As a simple example, let's find:

<math>\lim_{x \rightarrow 0}\frac{2xy+y^2}{x}</math>

(This is what you get when you try to differentiate <math>y^2</math>, and substitute <math>\delta y = x</math>.)

Given the above, the values from the definition are: <math> \begin{align} f(x) &\equiv \frac{2xy+x^2}{x} \\ c &\equiv 0 \\ \end{align} </math>

Keep in mind that we can never, ever plug in <math>x = c</math>, because <math>0 < |x - c| < \delta \,</math>. However, I know, already, that this is going to end up <math>L = 2y</math>.

<math> \begin{array}{l | l | l | l} x & \delta & f(x) & \epsilon \\ \hline 1 & > 1 & 2y+1 & > 1 \\ 0.1 & > 0.1 & 2y+0.1 & > 0.1 \\ -0.1 & > 0.1 & 2y-0.1 & > 0.1 \\ n & > n & 2y+n & > n \\ \end{array} </math>

It is clear that for any number what the limit will be as we approach 0.

Table of Limits

These are some common operations you can perform on limits:

<math> \begin{align}

\lim_{x \rightarrow n}(f(x) \pm g(x)) &= \lim_{x \rightarrow n}f(x) \pm \lim_{x \rightarrow n}g(x) \\

\lim_{x \rightarrow n}(f(x) g(x)) &= \left (\lim_{x \rightarrow n}f(x) \right ) \left ( \lim_{x \rightarrow n}g(x) \right ) \\

\lim_{x \rightarrow n}\frac{f(x)}{g(x)} &= \frac{\lim_{x \rightarrow n} f(x)}{\lim_{x \rightarrow n} g(x)} \\

\lim_{x \rightarrow n}\frac{f(x)}{g(x)} &= \lim_{x \rightarrow n}\frac{f'(x)}{g'(x)}\text{ (L'Hôpital's rule)} \\ \end{align} \,</math>

Otherwise, limits are pretty much common sense. I could write down a few equations for you, but they're really not that useful. You should be able to figure them out just by thinking a little bit.

Derivatives

The derivative is simply the slope of a curve at a given point.

Recall that the slope between two points on a curve is simply:

<math> \frac{f(x + \delta x) - f(x)}{\delta x} </math>

If you want to find the slope at a point, you simply have to bring <math>\delta x\,</math> all the way down to 0. But you obviously can't divide by 0, so you have to find the limit of the slope as it approaches zero.

<math>\frac{d}{dx} f(x) \equiv f'(x) \equiv \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} \,</math>

Forms

There are a few forms of derivatives you might see.

Liebniz: This is the oldest, and arguable, the least confusing, although it is cumbersome. Physicists don't use this much beyond introductory courses.

<math> {d \over dx} f(x) </math>

Oftentimes, the parameters to the function is dropped. You have to know that the function varies in x, otherwise the derivative is just 0.

<math> {d \over dx} f </math>

You can take the derivative twice:

<math> {d \over dx}{d \over dx} f \equiv {d^2 \over dx^2} f </math>

It should be obvious what the 3rd and 4th derivatives look like, although they are rarely, if ever, used in Physics.

Newton: The newton form only applies to time derivatives, and involves putting dots on top. This is extraordinarily useful for basic physics.

<math> \begin{align} {d \over dt} f &= \dot f \\ {d^2 \over dt^2} f & = {d \over dt} \dot f = \ddot f \\ \end{align} </math>

Total derivative: Take Liebniz notation apart you can deal with dt, dx, dy, etc... separately. It's a bit easier to make mistake with this notation, but if you know what you are doing, it's very valuable.

Prime Notation: If you know what you are deriving against, you can use hash marks to note how many derivatives you need. For functions of one variable, the derivative is obvious.

Subscript Notation: If a function is multi-variable, you'll need to use subscripts that describe the derivative. (This is multi-variable calculus, which isn't described here.)

Physical Interpretation

The derivative gives you rates. This could the rate of the change in position of a particle over time, commonly known as the speed or velocity. It could also be the rate of the rate of the change of the position over time, or the acceleration. It could also represent the rate of change in the temperature as you vary the pressure and keep the volume of a gas constant.

Derivatives, as you might imagine, are used extensively in physics. In fact, modern physics would not exist without this basic concept.

List

Finding the derivative of a particular function is not hard, but it's best to memorize as many derivatives as you can. (When we do integrals, we'll have to work backwards.)

I don't recommend memorizing the proofs for each of the following. Just put the tools in your toolbox, and remember how to use them.

<math>\text{Note: }a,\ b\ \text{are constants, and}\ f,\ g\ \text{are functions of }x.\,</math>

<math> \begin{align}

          \frac{d}{dx}a &= 0 \\
        \frac{d}{dx}x^a &= ax^{a-1} \\
      \frac{d}{dx}(f+g) &= \frac{d}{dx}f + \frac{d}{dx}g \\
       \frac{d}{dx}(fg) &= \frac{d}{dx}(f)g + f\frac{d}{dx}(g) \\
\frac{d}{dx}\frac{f}{g} &= \frac{\frac{d}{dx}(f)g - f\frac{d}{dx}(g)}{g^2} \\
       \frac{d}{dx}f(g) &= \frac{d}{dx}g \frac{d}{dg}f \text{ (Chain Rule)} \\
        \frac{d}{dx}e^x &= e^x \\
   \frac{d}{dx}\log_n x &= {\log_n e \over x} \\
     \frac{d}{dx}\sin x &=  \cos x \\
     \frac{d}{dx}\cos x &= -\sin x \\
     \frac{d}{dx}\csc x &= -\cot x \csc x \\
     \frac{d}{dx}\sec x &=  \tan x \sec x \\
     \frac{d}{dx}\tan x &=  \sec^2 x \\
     \frac{d}{dx}\cot x &= -\csc^2 x \\
  \frac{d}{dx}\arcsin x &=   {1 \over \sqrt{1 - x^2}} \\
  \frac{d}{dx}\arccos x &= - {1 \over \sqrt{1 - x^2}} \\
  \frac{d}{dx}\arctan x &=   {1 \over 1 + x^2} \\
  \frac{d}{dx}\arccsc x &= - {1 \over x \sqrt{x^2 - 1}} \\
  \frac{d}{dx}\arcsec x &=   {1 \over x \sqrt{x^2 - 1}} \\
  \frac{d}{dx}\arccot x &= - {1 \over 1 + x^2} \\

\end{align} </math>

Practice

It takes a fair bit of practice to learn how to apply the derivatives above. The ones you must be good at:

• Addition rule.
• Multiplication rule.
• Chain rule.
• Power rule.
• Logs and e.
• Trigonometric rules.

The rest you can look up in a book when you get stumped. The above should get you 99% of the way there.

Note on the Chain Rule

The Chain Rule is the one most people get wrong, and is the hardest one to see in the wild. When you do spot it, and properly apply it, give yourself a pat on the back because that puts you in the top 1% of the 1%.

I won't go over how to use the chain rule here, because your calculus textbook should devote a large portion of time to it, and there should be plenty of good problems and examples to work out on your own.

Many derivative tables include derivatives that are found simply by applying the chain rule. I wouldn't memorize those that are obviously chain rule derivatives, such as <math>{d \over dx}e^{ax} = ae^{ax}</math>.

Memorization

From here on out, you will need to know the derivatives fluently. You should be able to take the derivative of almost any formula in a matter of seconds. Think of how well you can add and multiply, and aim for the same proficiency of differentiation.

If you do not choose to memorize the differentials, when you approach integrals, you will be lost in the math and unable to grasp the core concept which is the fundamental theorem of calculus. You'll miss the forest for the trees.

Integrals

Integrals are a measure of the area under a curve.

Indefinite integrals do not have bounds attach. You can think of these as formulas for how to calculate definite integrals. You solve these simply by finding the anti-derivative, or the thing that gives you the derivative of what's in the integral.

Definite integrals have bounds attached. They may be infinity.

Evaluating definite integrals is really easy. Plug the high value into the indefinite integral, and subtract the low value. Don't forget that you have a constant to add at the end.

Multi-Variable, Path, etc...

I won't touch advanced integrals here. Note that all you need is a few tricks and you can do any kind of integral you can imagine. I'll cover these in advanced math.

List of Integrals

Because it's sometimes not as obvious how to find the anti-derivative, common integrals are listed here. Note that the number of integrals in books is much larger because there is no simple chain rule like there is for derivatives.

NOTE WELL: These are simply anti-derivatives! And because the derivative of a constant is 0, then you ALWAYS need to add the constant term!

Fundamental Theorem of Calculus

This is simply that the anti-derivative is the integral, as we've already said. This is one of those things that should make you go, "Wow!"

In plainer language: Take any function. Now find the function that describes how the function changes at every point in its domain. Now take the area under that secondary function. You'll get the original function back.

Alternatively, take the area under the first function's curve. Now find how the area changes at every point. That will be the same as the function itself.

In definite terms:

Let's let <math>f(x) = x^2</math>.

What is the area under the curve from 0 to a specific point? <math>f(x) = x^3 / 3.</math> (This is the partial integral from 0 to x: <math>\int_0^x x^2\, dx</math>.) What is the slope of the tangent (ie, derivative)of that function at every point? <math>x^2</math>, the original function.

The other way: What is the slope of the tangent at every point along the curve? <math>f(x) = 2x</math>. (This is the derivative). What is the area under the curve from 0 to any point? <math>x^2 = \int_0^x 2x\,dx</math>.