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Don T’s Range of Motion

Posted by Denny in Hmmm..., Math Teachers Derivate, Mathematics and Life's Details on 02 21st, 2010 | no responses

A guy I know, Don Thomas (known to his family as Donut Head), told me that just recently he hurt his wrist and that its range of movement was only about 20% its normal range of movement. He went on to say that he thought his wrist function was improving at a rate of 1% a day.

So how long, Don T asked me, will it take until my range is back to 100%?

Assuming that he, that would be Don T, was down to 20% his full range of motion, we calculated as follows.

We let $F$ represent Don’s full range of motion.

Then, 20% of full range can be represented by $0.20F$ . Its true. Think about it this way and just write what you say.

20% of current range.
The word of translates into mathematics as times $\cdot$ .
Since the current range is $F$ , 20% of $F$ is represented as $0.20\cdot F$ , or just $0.20F$ .

You know, I think I will use the star symbol $\star$ rather than the multiplication dot $\cdot$ to indicate multiplication. With decimal points and multiplication dots, there are too many dots.

We called the day of injury Day 0.
The amount of motion on Day 0 is $0.20F$ .

Now, we will let $A_n$ represent the total amount of motion at the end of Day $n$ .

Then, on Day 0, the total amount of motion is

$A_0 = 0.20F$

We called the first day following Day 0, Day 1.
A 1% increase in range is represented by $0.01\star \{\hbox{the current amount of range}\}$ .

Since the current range is $0.20F$ , a 1% increase is represented as $0.01 \star 0.20F$ .

Then, the amount of motion at the end of Day 1 is the range of motion on Day 0 plus the 1% increase in that range. That is, the total amount of motion at the end of Day 1 is

$A_1 = 0.20F + 0.01 \star 0.20F$ .

We can simplify a little by factoring out the $0.20F$ that is common to the two terms on the right side of the formula. Doing so gives us

$\begin{aligned} A_1 &=& 0.20F + 0.01 \star 0.20F \hfill \\ A_1 &=& 0.20F \bigl(1 + 0.01\bigr) \hfill \\ A_1 &=& 0.20F \bigl(1.01\bigr)\hfill \end{aligned}$

We can use exponents to related the day number and the total amount of motion at the end of that day.

At the end of Day 0, the total amount of motion $A_0 = \bigl(0.20F\bigr)^0$ .

At the end of Day 1, the total amount of motion $A_1 = 0.20F \bigl(1.01\bigr)^1$ .

We called the first day following Day 1, Day 2.
A 1% increase in range is represented by $0.01\star \{\hbox{the current amount of range at the end of Day 1}\}$ .

Since the amount of range at the end of Day 1 is $0.20F \bigl(1.01\bigr)^1$ , a 1% increase is represented as $0.01 \star 0.20F \bigl(1.01\bigr)$ .

Then, the amount of motion at the end of Day 2 is the range of motion at the end of Day 1 plus the 1% increase in that range. That is, the total amount of motion at the end of Day 2 is

$A_2 = 0.20F \bigl(1.01\bigr)^1 + 0.01 \star 0.20F \bigl(1.01\bigr)$ .

We can simplify a little by factoring out the $0.20F \bigl(1.01\bigr)$ that is common to the two terms on the right side of the formula. Doing so gives us

$\begin{aligned} A_2 &=& 0.20F \bigl(1.01\bigr)^1 + 0.01 \star 0.20F \bigl(1.01\bigr) \\ A_2 &=& 0.20F \bigl(1.01\bigr) \bigl(1 + 0.01\bigr) \\ A_2 &=& 0.20F \bigl(1.01\bigr)^2 \end{aligned}$

You may be able to see the general form, but if not, we’ll make one more iteration.

At the end of Day 0, the total amount of motion $A_0 = \bigl(0.20F\bigr)^0$ .

At the end of Day 1, the total amount of motion $A_1 = 0.20F \bigl(1.01\bigr)^1$ .

At the end of Day 2, the total amount of motion $A_2 = 0.20F \bigl(1.01\bigr)^2$ .

We called the first day following Day 2, Day 3.
A 1% increase in range is represented by $0.01\star \{\hbox{the current amount of range at the end of Day 2}\}$ .

Since the amount of range at the end of Day 2 is $0.20F \bigl(1.01\bigr)^2$ , a 1% increase is represented as $0.01 \star 0.20F \bigl(1.01\bigr)^2$ .

Then, the amount of motion at the end of Day 3 is the range of motion at the end of Day 2 plus the 1% increase in that range. That is, the total amount of motion at the end of Day 3 is

$A_1 = 0.20F \bigl(1.01\bigr)^2 + 0.01 \star 0.20F \bigl(1.01\bigr)$ .

We can simplify a little by factoring out the $0.20F \bigl(1.01\bigr)$ that is common to the two terms on the right side of the formula. Doing so gives us

$\begin{aligned} A_3 &=& 0.20F \bigl(1.01\bigr)^2 + 0.01 \star 0.20F \bigl(1.01\bigr) \\ A_3 &=& 0.20F \bigl(1.01\bigr)^2 \bigl(1 + 0.01\bigr) \\ A_3 &=& 0.20F \bigl(1.01\bigr)^3 \end{aligned}$

Now we have it. We can see the general form.

The total amount of motion at the end of Day $n$ is given by
$A_n = 0.20F \bigl(1.01\bigr)^n$

We can use this formula to determine how many days it will take to reach 100% range.

100% of the total range is represented by $1.00F$ , or $1F$ .

We want to find the number of days, $n$ , it takes to get to $1F$ .

Set $A_n$ equal to $1F$ in the formula $A_n = 0.20F \bigl(1.01\bigr)^n$ and solve for $n$ .

$1F = 0.20F \bigl(1.01\bigr)^n$

Divide both sides by $F$ .

$\displaystyle{\frac{1F}{F}} = \displaystyle{\frac{0.20F \bigl(1.01\bigr)^n}{F}}$

The $F$ ’s divide out (cancel), and we are left with

$1 = 0.20 \bigl(1.01\bigr)^n$

Divide both sides by 0.20.

$\frac{1}{0.20} = \frac{0.20 \bigl(1.01\bigr)^n}{0.20}$

Which gives us

$5 = \bigl(1.01\bigr)^n$

To solve for $n$ we will take the natural log of each side.

$\begin{aligned} \ln(5) &=& \ln\bigl(1.01\bigr)^n \\ \ln(5) &=& n\cdot \ln(1.01) \\ n &=& \frac{\ln(5)}{\ln(1.01)} \end{aligned}$

In decimal form, $n=~162$ days.

Now, Donut Head’s doctor told him his wrist would take about 60 days to get back to 100%. So, that constant 1% increase in range of motion must be incorrect or he started with more than 20% his original range. Let’s assume he really did start with 20% his original range.

Then the problem is, given 100% range of motion in 60 days, that is, given $n = 60$ , find the rate at which the range of motion increases. Can you do it?