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# System prepended metadata

title: 'Math 181 Miniproject 2: Population and Dosage'

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* 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




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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)\\[y\sim 100\cdot 1.1^{x}+(1.2838\cdot10^{-12})\\]




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(c\) What will the population be after 100 years under this model?
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(c\)\\[y\sim 100\cdot 1.1^{100}+(1.2838\cdot10^{-12})\sim 13780612.34\space people\space after\space 100\space years   \\]




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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)\\[P'(5)=\frac{P(4)-P(6)}{4-6}=\frac{1464.1-1771.561}{-2}=153.7305\space people\space per\space yeaer \\]
| $t$     | 1 | 2 | 3 | 4 | 5 | 6 |
|---      |---|---|---|---|---|---|
| $P'(t)$ |105   |115.5   |127.05   |139.755   |153.7305   |169.10355   |
After five years the population will be increasing by 153.7305 people. 




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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''(3)=\frac{P'(2)-P'(4)}{2-4}=\frac{115.5-139.755}{-2}=6.3525\space people\space per\space year^2 \\]
At double prime of 3 it means that at 3 years the population is increasing at 6.3525 per 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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(f)\\[\frac{P'(t)}{P(t)}=k\\]
\\[\frac{P'(105)}{P(1000)}=k\\]
 The value of k would equal 0.105 as your constant. 





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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)$D(x)=0.025(128)^2+-0.5(128)+10=355.6 mg/lb$





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(c\) What is the interpretation of the value $D'(128)$. 
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(c\)The value of D'(128) explains the increase of the dosage in miligrams of the drug for 128lbs. 

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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'(128)=.05(128)-.5=5.9mg/lb\\]
 This formula comes from getting the derivative and plugging in 128lbs for x therefore the dosage increasment is of 5.9 mg per pound.



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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)\\[L(x)=D(x)+D'(x)(x-a)\\]
\\[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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(f)\\[L(x)=367.5+6(x-130)\\]
\\[L(x)=367.5+6(128-130)=355.5 mg/lb\\]
According to the previous dosage chart, having 355.5 mg/lb for an individual weighing 128 lbs is accurate because its in between the values of 120 and 140 pounds.




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