Tag Archives: industry

Exponential Growth IRA Application – Real Enough World

This application isn’t real world but it’s real enough world.  It’s not real world in the sense that nobody needed to figure out this exact question at their job.  But it’s real enough world because a financial advisor at a well known wealth management firm told me he does calculations like this a lot.  He simply invented the following scenario because he thought young kids could relate to it.  I’m not sure if they can – but I’m damn sure they can learn from it.  Since he’s the expert I’ll just leave it all in his words:

Additionally, here are some well- known abbreviations that I’ll reference in two scenarios:
PV = present value
FV = future value
N = years to goal
i = assumed annual growth rate
PMT = annual payment
Scenario 1 – When his son is age 18, a dad opens a Roth IRA for the boy with a $1000 investment (PV).  The dad tells the boy “I’m giving you this money under one condition…and that is, you must contribute  $600 per year (PMT) and leave it alone until you turn age 65, which is 47 years from now (N).  We’re going to invest the money in an aggressive growth stock mutual fund that over time, I expect, should grow 9% per year on average (i).  At age 65, I expect your account value will be in the neighborhood of $433,535 (FV).”   Pretty amazing what time and compounding will do, huh?
Scenario 2 – A 13 year-old girl wants to purchase a used car at age 18, 5 years out (N).  She expects the car to cost $8000 (FV).   So far, she has saved $3000 (PV) and wants to know how much she must save annually (PMT) if her money is invested at a 4% annual rate (i).  Solving for PMT:
PV = $3000
FV = $8000
i = 4%
N = 5 years
PMT = $803.13
 I only ended up using scenario 1 and my teacher move was to block out the $433,535 and have students go through the estimation process about that account value after 47 years.  What’s it going to be?  Give me a couple dollar amounts it definitely won’t be because they are too high or low.  Brave guesses only.  (I heard Dan use the term “brave” after prompting for an estimation and it works well).



Well Flow Rate – “Real World” Math

Keeping with the easy definition of “real world” being “math someone needed to do at their job” – I actually prefer the term “industry math”, but regardless here’s the question that needed to be answered:

My wife’s working on a project where they are building a house on a property without a well or access to city water.  So they dug a 425ft well that was 8 inch in diameter.  When they finished the dig at 3:00 pm it was completely dry.  The next day, at 7:00 am, the well had filled with water up to 45ft below the surface.

How fast is the well filling up in gallons per minute?

They wanted 1.25 gallons per minute.  Do they need to dig another well?  Wells are about $50 a foot – so yeah, they would rather not dig another one.

(digging the well)

(well finished! That’s the cap they left on it)

They used a falling rock to determine how much of the well had filled with water.  Well (not the noun), they also used a cell phone connected to a string but isn’t that just too obvious?

Here’s the video of the rock drop:

Well Rock Drop



“Real-Life” Annulus Problem!

I just can’t type “real-life” without quotes because I’m yet to resolve what “real-life” means in a math class.  But for this post it means – “Math someone needed to do at their job”.  And in this post that someone is my wife – who much to her distain and my joy – does a lot of geometry as a project manager at a construction firm.

So here’s what we need to know:  How much cement is needed to make that border?  We need the answer in cubic yards because you buy cement in cubic yards.


The image below provides some of the context for the problem – The cement border is being used to circle an existing tree.


I was actually surprised how open the middle was on this problem.  Yet students used two main strategies.  The first was the standard subtract the areas and multiply by the height, or they subtracted the volumes of both cylinders directly.  The second was the find the perimeter, think of the wall as a rectangle (dimensions of 2*pi*r X 1′), find the area of that rectangle and then multiply by the height.

One student using the second strategy used a radius of 14ft and got a solution of 16.2 cubic yards.  He told me he knew the answer would be a little too small because of only using the inner radius.  Another students used 14.5 and got the same solution as the area subtracting syndicate.

Converting from cubic feet to cubic yards is a great time to practice your perplexed I wonder why you divide by 27? face.

By the way the wall ended up costing around $35000.  I can’t believe how long I would have to work to put a wall around a tree 🙁


– B


Cable Design for Satellites

The Description

This is an activity that was created by a harness and cable engineer in the aerospace industry. He is in charge of designing all interconnects between all the various systems / components on a satellite. This is an actual design problem that he made during the course of his job. The only thing he changed was the length of the wire from the ICB300 to the LAE ,because he wanted the 24AWG wire to result in a voltage drop greater than one, in order to test whether the students would see that and move to a larger wire.

The solutions are on the second page of the pdf. These solutions were authored by the engineer who wrote the problem.

I have a simplified version of this problem in my Satellite Design Teams activity.

The Advice

I would make sure that the students know this is an actual design problem for a satellite.  It is not a simplified representation of a problem someone might do, rather it is a real problem that someone must do, in order for the satellite to achieve mission success.

Looking at the equation again:  V = L * R * A.  V is the voltage drop, which is what they are solving for.  L is the lenght of the wire, which is given in the problem.  R is resistance, which they get from the table (they first choose a wire size, then look at the table for its resistance).  A is amps, which is given to be 0.8 in the first bullet point of the problem.

So basically, A is constant, and the students are inputing some value of L and R, in order to find V.  Then they are adding up all their V’s, and seeing if the sum is less than 1.

The design tradeoff between weight / voltage drop is key here.  Large wires have very low voltage drop, and since we cannot have a voltage drop greater than 1V, we are tempted to just use very large wires.  But large wires are also heavy, and satellites need to be as light as possible.  Thus the tradeoff between weight and voltage drop.  We must select the smallest possible wire, that still has a voltage drop less than 1V.

The Goods

Wire harness exercise


These are a listing of hastags that I use to catagorize my lessons plans.  Each catagory represents a different style lesson plan.  My instructional goal is typically to make sure that I use each hashtag at least once a month.  The goal of this blog is to share all the lesson plans that I use under each hashtag.

My detailed lesson plans are my Keynote slides.  But along with those, I make a quick, calendar-style overview to me a general idea of what I am doing.  It’s on this calender where I place the hashtags at the bottom of each day.  This allows me  to quickly look back at what I have been doing, and know whether of not I am differentiating.  For example, here is two weeks worth of my lesson plans in geometry.  Notice that I can quickly see whether or not I have differentiated my instruction, without having to analyze each specific lesson plan.  The hashtags allow me to get a quick sense of what I have been doing, and what I have not been doing.


*Notes –

-The term “perplexity” is being used as described by Dan Meyer here