Algebra 2

Alzheimer’s and Exponential Growth

The incidence of Alzheimer’s doubles every five years after the age of 65 until, by 85 years, nearly one in three adults is afflicted.

Are you wondering what percent are afflicted at age 65?  Can you work backards from 33% being affected at age 85?

Let’s build up the equation.  When something doubles, you multiply by 2.  When it doubles a bunch of times, you will be multiplying by a bunch of 2’s… or simply \(2^n\)  Since it is doubling every five years, you only want to multiply by 2 when the year hits a multiple of 5.  So, change it to \(2^\frac{n}{5}\)  You also need some sort of initial amount that is getting doubled in the first place.  Our initial equation to represent the percent of people with Alzheimer’s at  65 is \(y=a(2^\frac{n}{5})\)  Next, we use the given (n,y) pair to solve for that constant (a).  At 85 years (ie: 20 years later), the percent is .33  So, plug it in.  \(.33=a(2^\frac{20}{5})\)     (Also, notice how the division by 5 comes into play?  20/5 = 4 doubles in that 20 years.  ie:  Doubles every 5 years)   Solve and get a = .02  Plug that back into a, and you have \(y=.02(2^\frac{n}{5})\)   Well, let’s answer the question.  What percentage of people have Alzheimer’s at age 65 (n=0)?  \(y = .02(2^\frac{0}{5}) = .02(2^0) = .02(1) = .02\)    So, 2% of people at age 65.  A quick calculation verifies this:  2, 4, 8, 16, 32 at ages 65, 70, 75, 80, 85, respectively.

Seeds of Dementia: What Do Alzheimer’s, Parkinson’s and Lou Gehrig’s Have in Common?
Algebra 2

Power Steering Fluid

I was replacing the power steering fluid in my car when I stumbled upon some exponential decay math.  First, some background:  There is no drain plug on a power steering system.  You need to siphon out fluid from the reservoir and replace it with new fluid.  This new fluid then mixes into the rest of the system to create a slightly cleaner mixture.  This idea is that if you repeat this a few times, you’ll replace most of the old fluid with new fluid.

So, as you can see, there exists some sort of formula that can determine exactly how many times you need to extract and replace to reach X% replacement.

My car’s entire power steering system holds about 1 liter, and the reservoir itself holds .4 liters of that. So, 40% of the steering fluid is replaced each time I drain and refill the reservoir, and 60% of the old fluid remains elsewhere in the system.  I can do this repeatedly, each time replacing 40% of the “mixed” fluid with brand new fluid.

Let p = Percentage of the system replaced each time you empty the reservoir.
Let n = number of times you empty/fill the reservoir (“flush”)
Let FD = Percentage of dirty fluid in the system.
Let FN = Percentage of new fluid in the system.

If p = percentage of new fluid introduced by a flush, then (1-p) is percentage of old fluid remaining.  (eg: if \(p = .40, then (1-.40) = .60)\)

\(FD_0 = 1\)  (initial proportion that is dirty)
\(FD_1 = (1-p)\)
\(FD_2 = (FD_1)(1-p) = (1-p)(1-p)\)
\(FD_3 = (FD_2)(1-p) = (1-p)(1-p)(1-p)\)

\(FD_n = (FD_{n-1})(1-p) = (1-p)^n\)


\(FD = (1-p)^n\)
\(FN = 1 – FD\)


For my car, p=.40 so \(FD = (1-.40)^n\)

How many times do I need to empty and fill the reservoir to get to 80% clean?
Just set \(FN = .8\) and solve for n:

\(.80 = 1- (1-.40)^n\)
\(.80 = 1- (.60)^n\)
\(.6^n = .2\)
\(log(.6)^n = log(.2)\)
\(n*log(.6) = log(.2)\)
\(n = \frac{log(.2)}{log(.6)}\)
\(n = 3.15\)


With a reservoir that holds 40% of capacity, I need to empty and replace it about 3 times to get 80% of the old fluid replaced.

You can also use the formula to figure out what percentage of the system contains old vs. new fluid, based on the number of flushes you’ve done.  For this, you just plug in n and calculate \(FD\)  eg:  If you’ve done 5 flushes, \(FD = (1-.4)^5 = .08\)   So, after 7 refills, 8% is dirty, and 92% is new.

Algebra 2

The Beatles meet Mathematics & Physics

About this sound Listen to the opening chord


Mathematics, Physics and A Hard Day’s Night

In this article we shall use mathematics and the physics of sound to unravel one ofthe mysteries of rock ’n’ roll – how did the Beatles play the opening chord of A Hard Day’s Night? The song may never sound the same to you again.


I just love this paper.  Professor Jason Brown sampled the famous opening chord of this song and (using Math) separated out all the distinct frequencies in the clip.  From this, he was able to determine each individual note that was played.  From there, he determined exactly what each member of the band played.  He even discovered a surprising element relating to George Martin’s 5th Beatle status.

Professor Brown took each frequency and converted it to a musical note on the Western scale.  Here is the function he used to do this:

\(f(x) = 12 log_2(\frac{x}{220})\)    (…where 220 hertz is the frequency for A natural.)

I thought I’d try his calculation myself, because it’s a good opportunity to use the change of base formula for logarithms.  If you look in the original white paper, the first frequency in the table is 110.34   Let’s plug this into the function:

\(f(110.34) = 12 log_2(\frac{110.34}{220}) = 12 log_2(.5015)\)

How do you evaluate the above expression?  There is no \(log_2()\) button on most calculators.  Well, here’s where the change of base formula for logs comes in:  \(log_b(x) = \frac{log_d(x)}{log_d(b)}\)   So, let’s choose \(log_{10}\) (since calculators do have this) and continue:

\(12 log_2(.5015)=12(\frac{log_{10}.5015}{log_{10}2}) = 12(-.9957) = -11.9466\)


In other words, 110.34hz is -11.9466 semi-tones below the note of A.  It should really be 12, but as Professor Brown noted, the Beatles’ instruments weren’t in perfect tune, so the values are not whole numbers!

So, which note is 12 semi-tones below A?  Actually, 12 semi-tones makes an octave, so the answer an A note.  Using this method, he determined every note that was played.  The rest of the paper describes how he deduced which groups of notes were played by which instrument/band member.  Fascinating.