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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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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|1611|1772|1949|





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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) To find this formula we simply use Desmos, plugging in our values from our table from a and using the template fromula I got: 
$P\left(t\right)=994.072+1.10044^{t}+6.08172$




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(c\) What will the population be after 100 years under this model?
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(c\) Using our equation we just created above, just plug in the number 100 for t: 
$P\left(t\right)=994.072+1.10044^{100}+6.08172=15343$  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) To get the values for the table below we use the formula:
$P'(t)=\frac{\left(f\left(t+1\right)-f\left(t-1\right)\right)}{2}$ 
So for the first one we just input our t values so 1 then 2, then 3 and so on:
$P'(t)=\frac{f\left(2\right)-f\left(0\right)}{2}$ 
we can look at our values from the table and input those into our equation to solve:
$P'(t)=\frac{1210-1000}{2}= 105$
And we follow this formula for each remaining number for t in the chart.

| $t$     | 1 | 2 | 3 | 4 | 5 | 6 |
|---      |---|---|---|---|---|---|
| $P'(t)$ |105|115.5|127.05|139.7|153.7|169.1|

The interpretation of $P'(5)$ is the rate of increase in the population by the end of the 5th year. 


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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) So for this question we will use the formula:
$P''\left(x\right)=\frac{P\left(x+h\right)-2P\left(x\right)+P\left(x-h\right)}{h^{2}}$
So we know that x=3 so we have that variable. And we also know that h is 1 so we just plug in all those values to determine what $P''(3)$
$P''\left(3\right)=\frac{P\left(3+1\right)-2P\left(3\right)+P\left(3-1\right)}{1^{2}}=12.1$ 
To interpret this, it means that the population is changing at the rate of 12.1 per unit of time this case our unit is years. So overall the population is increasing and it is still increasing at this point. 

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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 value of K is 1.1, I found this value while playing with the slider on Desmos. Also, the one constant variable in our equation is 1.1 as we multiplied our population by it to increase 10% so for me that made sense.




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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) So using desmos, I made a table for all the points above. In doing this it gave me a value for a, b, and c. And the following is the equation I was able to get: 
$D\left(x\right)=0.025x^{2}+\left(-0.5\right)x+10$





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(b) Find the proper dosage for a 128 lb individual.
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(b) To determine the dosage or a 128 lb individual would be to simply plug in 128 for x in our equation in the last question.
$f\left(128\right)=0.025(128)^{2}+\left(-0.5\right)(128)+10= 355.6$ mg.





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(c\) What is the interpretation of the value $D'(128)$. 
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(c\) The interpretation of $D'(128)$ would be:
$D'(x)= 0.025x-0.5$
$D'(128)=0.025(128)-0.5$
$D'(128)=2.7$ mg. 
This means basically that after the 128 lb individual dosage, the dosage increases by 2.7 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) To do this we can use the linear approximation formula: 
$y= f(a)+f'(a)(x-a)$
From the table we have above, we can take a point so x=120 and then the y coordinate beside it y=310 and  a=128. So using the correct letters given in the problem we can rewrite the formula as:
$y= D(a)+D'(a)(x-a)$ and from this we plug in our values we just figured out:
$310= D(128)+D'(128)(120-128)$
$D'(128)=310-356/(-8)=5.75.$




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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) So we are given $D'(130)=6$ so we know that our x here will be 130, to find y we just use our formula from earlier in the question and insert 130 for x:
$D(130)= 0.025(130)^{2}+\left(-0.5\right)(130)+10= 367.5$
To find the tangent line we can use the formula: 
$y-y_{0}\ =\ m\left(x-x_{0}\right)$ we know what our $y_{0}$ and our $x_{0}$ are, they are 367.5 and 130 respectively. So we plug this into our new formula: 
$y-367.5\ =\ 6\left(x-130\right)$
$y-367.5\ =\ 6x-780$
$y=6x-412$ and this is our final tangent equation. 


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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) For this problem we can just use the equation we created from the last problem and plug in 128 for x: 
$y=6(128)-412.5=355.5$ mg. 
With this we can say that this is a good estimate for the 128 lb individual and we found this same value a different way in part b. 




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