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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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:::info
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.
:::

(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)
$P(t)=P_0(1+r)^2$



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(c\) What will the population be after 100 years under this model?
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(c\)
$P(t)=P_{1000}\left(1.1\right)^{100}$
$P(t)=13780612.34$




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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   |127.05   |139.755   |153.7305   |169.10355   |




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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)
$P''(3)=\frac{2-4}{2-4}$

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

$P''(3)=12.1275 \frac{people}{years^{2}}$

The value of $P''(3)$ tells you about the rate of change of the rate of change from $P'(t)$. After 3 years the population is increasing at a rate of $12.1275 \frac{people}{years^{2}}$


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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) 
 The constant value of $k$ is 10%. This is becasue when you take a value from the table of $P'(t)$ and multiply 10% It is possible to get the next value simply by adding. For example, take the value of $105$ and multiply this by the constant of $0.1$.
 $\left(105\right)\left(0.1\right)$=$10.5$
 Now add it to $105$
 $105+10.5=115.5$.




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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)

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



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





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(c\) What is the interpretation of the value $D'(128)$. 
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(c\) 
 $D'(128)$ Tells you about the rate of change of the dosage at the point of x=128 in 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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(d)
Use the central difference to find the value of $D'(128)$

$D'\left(x\right)=\frac{P\left(b\right)-P\left(a\right)}{b-a}$

$D'\left(128\right)=\frac{P\left(140\right)-P\left(120\right)}{140-120}$

$D'\left(128\right)=\frac{430-310}{140-120}$

$D'(128)=6$




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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)
$L(x)= D(a)+D'(a)(x-a)$

$L(x)=367.5+6 (x-130$)


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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)
$128= D(128)$
$\approx L (128)$
$=367.5+6(128-130)$
$=355.5$ $mg/lbs$

 It gives a good estimate. Because when I tried  to calculate in desmos the accurate Dosage for a $128 lb$ indivudual, it would be $355.6$ The result that I got is super close to the actual value.





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