Taxes as a calculus problem

Around this time of year, no one wants to think any more about taxes, but it makes for a very excellent calculus problem.

Secants vs. tangents

First, let us say that we have a function t(x), where t is the total tax owed and x is the pre-tax income. What is t '(x), the derivative of t(x)? It is the instantaneous tax rate: at a given income x, t ' is the additional tax paid on earning one additional cent. This is distinct from the effective tax rate, which is t(x) / x. To make this distinction clear, I have my students first work with the current U.S. income tax bracket system:

This is a graph of the current U.S. function for t(x). The x-axis shows pre-tax income; the y-axis shows total tax owed (note: the scales are logarithmic). I ask my students questions about the effective tax rate for someone earning $93,100 -- a rate represented by the secant line from (93.1, 17.03) to the origin. The effective rate is 17.03/93.1 = .183 = 18.3%. I ask my students to compare that to the instantaneous rate a person earning $93,099.99 pays on one additional cent -- a rate represented by the tangent line at x = 93.09999. From incomes in the $44,000 to $93,100 range, each additional cent is taxed at the same rate. This is the 25% bracket: (17.03 - 4.75)/(93.1 - 44) = .25. If a taxpayer's income falls in this range, she pays a top marginal rate of 25%, and a raise of $.01 results in .25 cents of additional tax. However, no taxpayer whose income falls in this range pays an effective rate of 25%. The top marginal rate a taxpayer pays never equals her effective rate! In fact, it is the person who earns $228,100 who pays an effective rate of 25%; she is in the 33% top marginal rate, but a lot of her income is taxed at lower rates.

Here is the graph for t '(x), the derivative of the t(x) above:

Note that the x-axis is still logarithmically scaled pre-tax income; the y-axis is instantaneous tax rate and is normally scaled.

Modeling, first derivatives, second derivatives, and c

Here is a graph of the effective tax rates, t(x) / x, at the endpoints of each bracket:

The x-axis is logarithmically scaled pre-tax income in thousands of dollars (note: all subsequent formulas assume x is in thousands of dollars). The effective tax rate (total tax paid per $1,000 pre-tax income) is shown on the y-axis. Intriguingly, the above graph looks linear (but remember, the x-axis is ln(x)). The equation T(x) / x = .076 ln(x) - .17 models the key points fairly well. Thus, the real t(x) can be modeled closely by T(x) = .076 x ln(x) - .17x. And T '(x) would be .076 ln(x) - .094. Here is this T '(x) versus the actual tax brackets:

I also ask them to think about T "(x), which is 0.076 / x. What conclusion can they draw from the fact that, when x > 0, T " > 0? This shows that T '(x) is always increasing, and T (x) is concave up. Of course, they can see from the graph above that T '(x) continues increasing; however, the real tax brackets top out at 35% marginal rate. This function T '(x) eventually equals 1 at about x = 1.8 million, meaning it would hit a 100% instantaneous rate and this high-earner could keep no additional money. At about 5 million, T(x) = x, so this person would have $0 income after paying taxes! Obviously, this makes no sense; T '(x) should have an asymptote at no greater 1, a conclusion my students reach with little prompting. An asymptote means that T '(x) will always be positive and increasing (progressive) but never reach confiscatory levels.

Furthermore, I ask my students to think about what increasing or decreasing the c in T(x) will do. It will not change T '(x) or T "(x), but it will affect the x- and y-intercepts. Why does it make sense for the y-intercept to be negative? This might correspond to the earned income tax credit: cash payments, in the form of a tax rebate, to people with very low incomes.

Setting up an ideal function

So, we need to set up an ideal tax function, i(x), such that its derivative will have an asymptote (unlike the linear T '(x)). Generally speaking, the easiest way to accomplish this is by setting up its derivative in the form:

where a and b can be adjusted to change the x- and y-intercepts, and c can alter how quickly the instantaneous tax rate increases (where c > 0). Furthermore, it makes sense to set a = b so that i '(x) > 0 where x > 0, otherwise one would create the unusual situation where additional income pushed the total tax down (i.e., a negative slope on the integral i (x)). The simplest example of i '(x) is

and thus its integral, i(x), would be x - ln(x+1) + c. At this point, I set my students loose to play with the function i '(x). Here is one example that matches up well against the current brackets:

This i '(x) is

I ask my students to compute its antiderivative -- good practice for them -- and compare it to T(x). They look quite similar below $600,000. After $600,000, T(x) becomes much bigger, eventually crossing the line y = x. The function i(x), however, looks like it starts to plateau, but appearances are deceptive: i(x) continues to increase, its slope continues to increase, and it is always concave up -- yet it never crosses y = x. Therefore, i(x) could create a progressive tax system without brackets.

Taxes Project

Here's the GeoGebra mentioned.

A neat application for derivatives

Whoa! May was crazy. But I survived. Here's a fun problem.

The function gives a Bell curve. The function
gives a downward-opening parabola that fits exactly under the Bell curve, like so:


In other words, f(x) is a lower boundary to g(x):
How to go about proving this? Algebraically, simplifying the above doesn't do much. It turns out, though, that looking at the derivatives does prove this.
Using the Chain Rule for compounded functions on g(x) reveals the similarity between the two functions. Here's a graph of the two derivatives:


On the lower right-hand side, quadrant IV, f(x) has a more negative slope than g(x), which is easy enough to prove:
On the upper left-hand side, quadrant II, f(x) has a more positive slope than g(x), which has virtually the same proof:
(Just remember that the xs in these inequalities are negative, so the inequality signs are flipped twice, back to their original position.)

What does this prove? One final fact: at x = 0, f(0) = g(0).

As x increases, the slope of f(x) is always more negative than the slope of g(x) -- they start equal at x = 0, but f(x) decreases more rapidly, so f(x) must always be less than g(x) as x increases. On the other side, the logic is simply reversed.