Grand Finale (Part III): A Crash Course in Calculus with an Important Outcome

“I found a discarded textbook on calculus in a wastebasket and read it from cover to cover.”  — John Pople


The first fundamental teaching from calculus discusses the effect of changing some independent variable on a given dependent variable. In other words, if I change a variable x and x is related to another variable y (i.e. y is a function of x), how much does y change?

Believe it or not, but you have seen this concept before in your algebra class. Recall that the equation for a line in the xy-plane is expressed as y = mx + b, where m is the slope of the line and b is the y-intercept of the line. Now, say I change x a little bit in this equation such that it takes on the value x + ∆x, where ∆x represents the amount I changed x. If I asked for the change in y as a result of changing x from x to x + ∆x, one could simply evaluate the y values at both x and x + ∆x using the equation for a line and, from there, subtract the two values for y to find the change in y. (For our purposes, we will represent a change in y with the symbol ∆y.)

Evaluating the equation for the line at x yields y = mx +b. Evaluating the equation for the line at x + ∆x yields y + ∆y = m(x + ∆x) + b. Subtracting the two equations (and doing a little pinch of algebra) yields the following relationship ∆y = m∆x. Therefore, we recognize that lines have the amazing property that, if I want to find a change in y as a result of a change in x, I need only multiply the change in x by the slope of the line. (You may recognize this relationship from the definition of slope, which is m = ∆y/∆x.)

Calculus extends this theory to graphs in the xy-plane that are not lines.

In general, we can consider a curve in the xy-plane to be comprised of very tiny (infinitesimal) line segments. We can calculate how much y changes when x changes over these very tiny lines. In other words, we can use the relationship ∆y = m∆x, where m is the slope of the small line segment. To indicate that we are over a very small line segment, we consider our changes in x and y to be infinitesimal, and so we use the notation dx and dy to represent infinitesimal changes in x and y, making our formula dy = mTdx.

Notice I changed m to mT. There is a definite reason for this change. Recall m is the slope of a tiny line segment. Recall also that these tiny line segments are essentially describing some generic curve in the xy-plane. Thus, from the curve’s perspective, m appears to be the slope of the tangent line at one of its points. (Hence, I included the subscript T for tangent.) This may explain why you hear discussion of tangent lines from students taking calculus classes.

As it turns out, mT becomes a function of x that gives you the value of the slope of the tangent line at some point on a curve y = f(x). mT can be tricky to calculate. Here, I will only go over how to find mT for simple curves.

Consider the function f(x) = a*xn plotted on the xy-plane, where a and n are constants. The slope of the tangent line for some generic x on the graph of f is then given by mT = a*n*xn-1. (I will not attempt to prove this rule, for it requires some extensive mathematics beyond the scope of this blog.) To make sure you are comfortable with this notion, however, let us do a quick example. Let us say that f is the graph of the standard parabola in the xy-plane, y = x2. Following from our formula, the slope of the tangent line to f must be equal to mT = 2x. If we consider the point where x = 4 (i.e. (4,16)) on our graph, we should then expect that the slope of the tangent line to our curve at that point equals 8 because mT = 2*4.

There is another very special curve in the xy-plane that we can very easily calculate mT. That is the curve whose y values are equal to its mT values for all values of x. (In other words, it is the graph of a function y = f(x) such that y = f(x) = mT for all values of x.) We can actually use the previous formula to help us discover this curve.

Recall y and mT are both dependent on x. Let us assume x begins at zero but can vary over the real numbers after starting at zero. Therefore, we have the following set of equations when x starts.

y = 0

mT = 0

At this point, you would say that the two values are equal and so we have found our answer, but you would be incorrect. Look at the previous formula mT = a*n*xn-1. If mT = 0, then either a, n, or xn-1 must be zero. In this case, n = 0. Now, if we take a look at the other part of the formula, y = f(x) = a*xn, we can conclude n = 0 and, assuming a = 1, y = 1. Therefore, for this set of equations to hold true, we need to add one to y.

y = 0 + 1

mT = 0

To keep the equality true, however, we must also add one to mT.

y = 0 + 1

mT = 0 + 1

Now, we have mT = 1 = a*n*xn – 1. Notice, there are no x terms in the expression for mT; we just have the value one. This implies that the exponent over x must now be zero. Thus, n – 1 = 0, or n = 1. If n = 1, then we can use the other part of the formula, y = f(x) = a*xn, to discover y = x, assuming a is one. Therefore, our equations become updated once again.

y = 0 + 1 + x

mT = 0 + 1

Again, to keep the equality true, we must also add x to mT.

y = 0 + 1 + x

mT = 0 + 1 + x

Hopefully, you can see that this process is never ending: We keep adding terms to the first equation only to add another term in the second equation that will go back and change the first equation. If you continue this process a few more times, however, you may start to recognize a pattern, which saves us a lot of time. (You can feel free to keep expanding the expressions for y and mT, but it becomes more and more complicated as you expand outwards, especially if you think about it the way I have taught you, i.e. the “non-Calculus” version.) The final result is a curve with the following equation.

y = 1 + x + x2/2 + x3/6 + x4/24 + … + xn/n! + …

The curve of this equation has tangent slope values equal to the y value along its length.

Believe it or not, but this curve can be simply expressed as the more familiar curve y = ex. (I will not show the proof here, for it requires mathematics beyond the scope of this blog.)

Therefore, I can claim the following.

ex = 1 + x + x2/2 + x3/6 + x4/24 + … + xn/n! + …

I can do a similar analysis to the one I have done here for other special curves in the xy-plane like y = sin(x) and y = cos(x). At the end of the day, the following two equations can be obtained.

sin(x) = x – x3/6 + x5/120 – x7/5040 + x9/362880 + … + (-1)nx2n + 1/(2n+1)! + …

cos(x) = 1 – x2/2 + x4/24 – x6/720 + x8/40320 + … + (-1)nx2n/(2n)! + …

We will use the subsequent three equations to transform our understanding of complex numbers next week, wrapping it up nicely with the most amazing formula in all of mathematics. One more blog to go!

Summary Exercise

Approximate f(1) = sin(1) to within a maximum possible error of 0.01.


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3 Responses to Grand Finale (Part III): A Crash Course in Calculus with an Important Outcome

  1. Caitlin Edinger says:

    I actually had a fantastic calculus teacher who, like you was very passionate about teaching mathematics. I find that your explanations do an excellent job at supplementing what I learned in high school. However, in contrast to my current math class – you outshine the professor by a long shot. I enjoyed the lessons, Ryan!

  2. Lewis Esposito says:

    My high school calc teacher also wasn’t the best. He truly wanted us to learn, and he really enjoyed teaching the material, but he was rather old and at times he would forget how to do certain problems…Needless to say, my previous calc experience was underwhelming. Your math explanations always make more sense than those of any of my high school teachers!

  3. eje5073 says:

    I took a Calculus class in high school, but it was poorly run and the teacher really couldn’t care less if we retained the material (I was fortunate enough to get the only Calc teacher in my school that was like this). This exlanation makes a lot of sense, and I’m genuinely glad that I can read and actually learn the fundamentals of Calculus, rather than simply doing derivatives and integrals without a base knoweldge for how they actually apply.

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