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 |1464.1|1610.51|1771.561|1948.7171|
According to the information intitially provided in the table, at t=0 years the population was 1000. In order to find the missing values for the population between years 1 and 7, the first population was multiplied by 0.10. This calculated the 10% increase to find current population.

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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)The link provided below is where you may find the table that was created in Desmos using the values from the table in part a, as well as the formula P(t) found after inputting the exponential model \\[
y_1\sim a\cdot b^{x_1}+c
\\]
The values found on Desmos were as followsL: a=1000, b=1.1 and c=-1.6399x10^-12 and thus the formula P(t)=1000*1.1^x+(-1.6399*10^-12). The units in this case are years for the x-axis, and population for the y-axis.
https://www.desmos.com/calculator/ongx3r8wz8

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(c\) What will the population be after 100 years under this model?
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(c\) Using Desmos again, I calculated P(100). The link to this graph is provided below. P(100) is equal to 13780612.3398.

https://www.desmos.com/calculator/5cgs6raqfb

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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)The table above provided values from t 1 through 6. All P'(t) values were worked out by using the central difference of every point in the table from part a. The central difference formula is as follows:
f'(a)=(f(b)-f(a))/(b-a).

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

The calculations for this work are provided in Desmos in the link below. The value for P'(5) shows that the average rate of change in the population at year t=5. Thus at year 5, the population increased at a rate of 153.7305.

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(f) Use a central difference to estimate the values of $P''(3)$. What is the interpretation of this value?
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(f)The formula used for the central difference in order to calculate P'(3) is as follows:
(f'(b)-f'(a))/(b-a)=
f'(4)-f'(2)/4-2=
139.755-115.5/2=
12.1275 people per year^2. The second derivate demonstrates the rate that f' changes. Thus this calculation shows an increase in the population at a rate of 12.1275, at the time t=3.

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(g) Plot $P'(t)$ and $k\cdot P(t)$, creating a variable $k$. Play with the slider until you find the value of $k$ so that $P'(t)=k\cdot P(t)$. What is the value of $k$? (Note this shows that for this function $P$ and $P'$ are just multiples of each other.)
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(g)The table from above was used for this question. The link for the graph and following values can be found below:
https://www.desmos.com/calculator/nvgsob6qg2

Using Desmos, I found that the value of k=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)
The chart linked below was used to find the approximate dosage after inputting the weigh of an individual. The following formula was used:

\\[
y_1\sim ax_1^2+bx_1+c
\\]

The following values were calculated using Desmos:
a=0.025, b=-0.5, and c=10. Thus the formual would be D(x)=0.025x^2-0.5+10.The figure below shows this info. The x-axis is symbolic of weight (in pounds) and the y-axis represents dosage (mg).
and
\\[
y_1\sim ax_1^2+bx_1+c
\\]
and

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(b) Find the proper dosage for a 128 lb individual.
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(b)
Desmos was used to calculate this, and the work done is in the graph linked above. d(128) = 355.6 mg per drug.

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(c\) What is the interpretation of the value $D'(128)$.
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(c\)
The derivative is what is used to calculate the average rate of change. When I calculated D'(128), the answer tells me the average rate of change in dosage in mg is for an indidual with the exact weight of 128 pounds. Units: dosage in mg.

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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)
For the purpose of this question, I used the h method we learned in class. The limit was taken as h approaches 0. The formula was D'(128)= lim h->0 (f(x+h)-f(x)/h. After doing some algebra this means:

lim h->0 (5.9+0.025h)/h

=5.9mg/lbs.

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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)
The equation of the tangent line is equal to L(x)=f(a)(x-a). D'(x) was substitued to f'(x) for consistency. Therefore,
y1=f(a)_f'(a)(a-a)
F'(130)=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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(f)
F'(128)=367.5+6(128-130)
=367.5-12
359.5mg.

This is a good estimate for a person weighing 128 pounds because the output is near to what I calulated in Desmos. The point was very close to 130, meaning it was a good estimate in order to keep the tangent line accurate. If it was further from the point, it would be less accuate. The larger the space, the more inaccurate the results.

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