Math 181 Miniproject 2: Population and Dosage.md
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Math 181 Miniproject 2: Population and Dosage
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**Overview:** In this miniproject you will use technological tools to turn data and into models of real-world quantitative phenomena, then apply the principles of the derivative to them to extract information about how the quantitative relationship changes.

**Prerequisites:** Sections 1.1--1.6 in *Active Calculus*, specifically the concept of the derivative and how to construct estimates of the derivative using forward, backward and central differences. Also basic knowledge of how to use Desmos.

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1\. A settlement starts out with a population of 1000. Each year the population increases by $10\%$. Let $P(t)$ be the function that gives the population in the settlement after $t$ years.
(a) Find the missing values in the table below.
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(a)

| $t$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|--------|------|---|---|---|---|---|---|---|
| $P(t)$ | 1000 | 1100 |1210 |1331 |1464 | 1610 |1771 | 1948 |

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(b) Find a formula for $P(t)$. You can reason it out directly or you can have Desmos find it for you by creating the table of values above (using $x_1$ and $y_1$ as the column labels) and noting that the exponential growth of the data should be modeled using an exponential model of the form
\\[
y_1\sim a\cdot b^{x_1}+c
\\]
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(b) Entering the values I got in the table above into Desmos, then using the exponential model, the formula I got was:

$P(t)=1002.29*1.09976^t -2.26115$

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(c\) What will the population be after 100 years under this model?
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(c\)
$P(100)=1002.29*1.09976^{100} -2.26115$
$P(100)=13,514,042$

After 100 years, the population of the settlement will have $13,514,042$ people.

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(d) Use a central difference to estimate the values of $P'(t)$ in the table below. What is the interpretation of the value $P'(5)$?
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(d)
| $t$ | 1 | 2 | 3 | 4 | 5 | 6 |
|--- |---|---|---|---|---|---|
| $P'(t)$ |105|115.5|127|139.5|153.5|169|

Central Difference:

$P'(t)=$ $\frac{f(2)-f(0)}{2-0}$ $=$ $\frac{1210-1000}{2}$ $=$ $\frac{210}{2}$ $=$ $105$
$P'(t)=$ $\frac{f(3)-f(1)}{3-1}$ $=$ $\frac{1331-1100}{2}$ $=$ $\frac{231}{2}$ $=$ $115.5$
$P'(t)=$ $\frac{f(4)-f(2)}{4-2}$ $=$ $\frac{1464-1210}{2}$ $=$ $\frac{254}{2}$ $=$ $127$
$P'(t)=$ $\frac{f(5)-f(3)}{5-3}$ $=$ $\frac{1610-1331}{2}$ $=$ $\frac{279}{2}$ $=$ $139.5$
$P'(t)=$ $\frac{f(6)-f(4)}{6-4}$ $=$ $\frac{1771-1464}{2}$ $=$ $\frac{307}{2}$ $=$ $153.5$
$P'(t)=$ $\frac{f(7)-f(5)}{7-5}$ $=$ $\frac{1948-1610}{2}$ $=$ $\frac{338}{2}$ $=$ $169$

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(e) Use a central difference to estimate the values of $P''(3)$. What is the interpretation of this value?
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(e)

$P''(t)=$ $\frac{f'(4)-f'(2)}{4-2}$ $=$ $\frac{139.5-115.5}{2}$ $=$ $\frac{24}{2}$ $=$ $12$

If $P(t)$ is the population of a settlement in terms of time in years, then $P'(t)$ is the average increase in the population over the year. So the interpretation of $P''(t)$ is that it is the average increase of the population of the settlement per month in the the third year.

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(f) **Cool Fact:** There is a constant $k$ such that $P'(t)=k\cdot P(t)$. In other words, $P$ and $P'$ are multiples of each other.

What is the value of $k$? (You could try creating a slider and playing with the graphs or you can try an algebraic approach.)
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(f)
$P'(1)=k\cdot P(1)$
$105=k\cdot 1100$
$\frac{105}{1100}$ $=k$
$k = 0.095$

$P'(2)=k\cdot P(2)$
$115.5=k\cdot 1210$
$\frac{115.5}{1210}$ $=k$
$k= 0.095$

$\frac{127}{1331}$ $=0.095$
$\frac{139.5}{1464}$ $=0.095$

The value of $k$ is 0.095.

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2\. The dosage recommendations for a certain drug are based on weight.
| Weight (lbs)| 20 | 40 | 60 | 80 | 100 | 120 | 140 | 160 | 180 |
|--- |--- |--- |--- |--- |--- |--- |--- |--- |--- |
| Dosage (mg) | 10 | 30 | 70 | 130 | 210 | 310 | 430 | 570 | 730 |
(a) Find a function D(x) that approximates the dosage when you input the weight of the individual. (Make a table in Desmos using $x_1$ and $y_1$ as the column labels and you will see that the points seem to form a parabola. Use Desmos to find a model of the form
\\[
y_1\sim ax_1^2+bx_1+c
\\]
and define $D(x)=ax^2+bx+c$.)
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(a)

$D(x) = 0.025x^2-0.5x+10$
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(b) Find the proper dosage for a 128 lb individual.
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(b)
$D(128)=0.025(128)^2-0.5(128)+10$
$D(128)=355.6$

The proper dosage of the medication for an individual that is $128$ $lbs$ is $355.6$ $mg$

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(c\) What is the interpretation of the value $D'(128)$.
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(c\)
First, finding the expression for the derivative using the variables $x$ and $h$ so that I have an expression I could enter any value from the function into if I need it:
$D'(x)$ $=$ $\lim_{h \to 0}\frac{D(x+h)-D(x)}{h}$
$D'(x)$ $=$ $\lim_{h \to 0}\frac{0.025(x+h)^2-0.5(x+h)+10-(0.025x^2-0.5x+10)}{h}$
$D'(x)$ $=$ $\lim_{h \to 0}\frac{0.025x^2+0.05xh+0.025h^2-0.5x-0.5h +10-0.025x^2+0.5x-10)}{h}$
$D'(x)$ $=$ $\lim_{h \to 0}\frac{0.05xh+0.025h^2-0.5h}{h}$
$D'(x)$ $=$ $\lim_{h \to 0}\frac{0.05xh+0.025h^2-0.5h}{h}$
$D'(x)$ $=$ $\lim_{h \to 0}\frac{~~h~~(0.05x+0.025h-0.5)}{~~h~~}$
$=$ $0.05x-0.5$

$D'(128)$ $=$ $0.05(128)-0.5$
$D'(128)$ $=$ $5.9$$\frac{lbs}{mg}$

For an individual that is $128$ lbs, the dosage of the drug at that weight increases by $1$ mg for every $5.9$ lbs the person weighs.
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(d) Estimate the value of $D'(128)$ using viable techniques from our calculus class. Be sure to explain how you came up with your estimate.
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(d)
$D'(130)$ $=$ $\frac{D(140)-D(120)}{140-120}$
$D'(130)$ $=$ $\frac{430-310}{20}$
$D'(130)$ $=$ $\frac{120}{20}$
$D'(130)$ $=$ $6$

I came up with my estimate by doing a central difference for $D'(130)$. It is the closest "nice" function value to $D'(128)$.

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(e) Given the value $D'(130)=6$, find an equation of the tangent line to the curve $y=D(x)$ at the point where $x=130$ lbs.
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(e)
First I have to find the $D(a)$ at $a= 130$

$D(130)= 0.025(130)^2-0.5(130)+10$
$D(130)= 367.5$

Now I can find the equation of the tangent line to the curve $y=D(x)$ at $x=130$ lbs.

$D(a)= L(x)$$=D(a)+D'(a)\cdot(x-a)$
$D(a)= L(x)$$=367.5+6\cdot(x-130)$
$D(a)= L(x)$$=367.5+6x-780$
$D(a)= L(x)$$=6x-412.5$

The equation for the tangent line to the curve $y=D(x)$ at $x=130$ lbs is
$L(x)$ $=6x-412.5$

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(f) Find the point on the tangent line in the previous part that has $x$-coordinate $x=128$. Does the output value on the tangent line for $x=128$ lbs give a good estimate for the dosage for a 128 lb individual?
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(f)
$L(128)$$=6(128)-412.5$
$L(128)$$=355.5$

The dosage for a 128 lb individual is $355.6$ mg. (Found in section (b).) The estimate I got using the equation of the tangent line in $355.5$ mg. I believe that, yes the tangent line does give a good approximation of the dosage, but I am not a doctor and don't know the margin of error for this drug.

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