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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| $t$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|--------|------|------|-----|---|---|---|---|---|
| $P(t)$ | 1000 | 1100 |1210 |1331| 1464.1|1610.51|1771.561 |1948.7171 |

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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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y=1000 * 1.1^x + (1.2976x10^-12)

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(c\) What will the population be after 100 years under this model?
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y=1000 * (1.1^100 + (1.2976x10^-12)
y=1000 * (13780612.34) + (1.2976x10^-12)
$y=13780612.32$

At 100 years, the population will be approximately 13,780,612.

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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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| $t$ | 1 | 2 | 3 | 4 | 5 | 6 |
|--- |---|---|---|---|---|---|
| $P'(t)$ |105|115.5|127.05|139.755|153.7305|169.10355|

After 5 years, the rate at which the population will be growing will be by approximately 153.7305 members.

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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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$P''(3)= \frac{P'(4)-P'(2)}{4-2}$
$P''(3)= \frac{139.755-115.5}{2}$
$P''(3)= \frac{24.255}{2}$
$P''(3)= 12.1275$

After 3 years, the rate at which the population will be growing will increase by approximately 12.1275 members/year per 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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$P'(t)=k*P(t)$
$P'(1)=k*P(1)$
$105=k*1100$
$k=0.095454545454$

$P'(t)=k*P(t)$
$P'(2)=k*P(2)$
$115.5=k*1210$
$k=0.095454545454$

$P'(t)=k*P(t)$
$P'(3)=k*P(3)$
$127.05=k*1331$
$k=0.095454545454$

$P'(t)=k*P(t)$
$P'(4)=k*P(4)$
$139.755=k*1464.1$
$k=0.095454545454$

$P'(t)=k*P(t)$
$P'(5)=k*P(5)$
$153.7305=k*1610.51$
$k=0.095454545454$

$P'(t)=k*P(t)$
$P'(6)=k*P(6)$
$169.10355=k*1771.561$
$k=0.095454545454$

The value for $k$ is $0.095454545454$.

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

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(b) Find the proper dosage for a 128 lb individual.
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$D(x)=0.025x^2-0.5x+10$
$D(128)=0.025(128)^2-0.5(128)+10$
$D(128)=409.6-64+10$
$D(128)=355.6$mg

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(c\) What is the interpretation of the value $D'(128)$.
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$D'(x)=\lim_{h \to 0}\frac{D(x+h)-D(x)}{h}$
$D'(x)=\lim_{h \to 0}\frac{[0.025(x+h)^2-0.5(x+h)+10]-[0.025x^2-0.5x+10]}{h}$
$D'(x)=\lim_{h \to 0}\frac{[0.025(x^2+2hx+h^2)-0.5x-0.5h+10]-[0.025x^2-0.5x+10]}{h}$
$D'(x)=\lim_{h \to 0}\frac{0.025x^2+0.05hx+0.025h^2-0.5x-0.5h+10-0.025x^2+0.5x-10}{h}$
$D'(x)=\lim_{h \to 0}\frac{0.025h^2+0.05hx-0.5h}{h}$
$D'(x)=\lim_{h \to 0} 0.025h+0.05x-0.5$

$0.025(0)+0.05x-0.5$

$D'(x)=0.05x-0.5$
$D'(128)=0.05(128)-0.5$
$D'(128)=5.9$mg/lb

At 128 lbs, the required dosage (mg) for the drug based on their weight (lbs) is rising by 5.9 mg/lb.

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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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Using the central difference formula:
$D'(128)= \frac{D(b)-D(a)}{b-a}$
$D'(128)= \frac{D(140)-D(120)}{140-120}$
$D'(128)= \frac{430-310}{140-120}$
$D'(128)= \frac{120}{20}$
$D'(128)= 6$mg/lb

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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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Linear approximation:
$L(x)=D(x)+D'(x)(x-a)$
$L(x)=D(130)+D'(130)(x-130)$
$L(x)=367.5+6(x-130)$

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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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$L(x)=367.5+6(x-130)$
$L(128)=367.5+6(128-130)$
$L(128)=367.5+6(-2)$
$L(128)=367.5-12$
$L(128)=355.5$mg

Yes, the output value on the tangent line for $x=128$ does give a good estimate (the value is just off by 0.1) for the dosage for a 128 lb individual.

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