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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  |1464.1  	| 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) The formula for P(x) is P(x)=$1000*1.1^x+0$
![](https://i.imgur.com/4h2yZ4r.png)





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(c\) What will the population be after 100 years under this model?
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(c\) After 100 years the population will be 13,780,612.3398 




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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 |
|---      |---|---|---|---|---|---|
| $P'(t)$ |110 | 121  |  133.1 |  146.41 | 161.05100  |  177.1561 |
The interpretation of P'(5) is an increase of 161 people a year after 4 years of settlement  




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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) The estimated value of P''(3) using a central difference is 13.31

The avergae rate of change using the derivative values P'(3) and P'(4) , f''(a)= f'(b)-f'(a)/b-a was used to find this estimate 
$$ P''(3)=146.41-133.1/(4-3)$$
$$P''(3)=13.31$$



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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 0.1
This value was solved algebraically simply inserting a corresponding values from the table of P(t) and P'(t). This value was observed using different corresponding values for t 

$$P'(t)=k*P(t)$$
$$P'(1)=k*P(1)$$
$$100=k* 1100$$
$$k*1100/1100=100/1100$$
$$k=0.1$$



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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) defined is D(x)=$0.025x^2+(-0.5)x+10$
![](https://i.imgur.com/WTfdCND.png)






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(b) Find the proper dosage for a 128 lb individual.
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(b) The proper doasage for a 128 lb individual is 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 the value D'(128) is that the dosage for a 128 lb person has a specific increase from a smaller weight value in mg/lb, this is the slope. 


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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) The estimated value of D'(128) is 6.2 mg/lb 
The way I found an estimated value of D'(128) is using central difference with an interval of the closest values to D(128)from the table values of D(X), the interval I used was [128,140]. These values were input into the formula for the average rate of change 
$$D'(128)= 430-355.6 /140-128$$
$$D'(128)=6.2$$



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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 to the curve applying D'(130)=6  is y=6x-412.5
 This was found using the point slope forumla for the tangent line. Knowing the point involved is (130,367.5) and the slope is 6 
 
 $$y-y1=m(x-x1)$$
 $$y-367.5=6(x-130)$$
 $$y-367.5=6x-780$$
 $$y=6x-412.5$$
 

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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) Using the equation from part(d) and inputting x=128 the estimated dosage for a 128 lb individual is 355.5 mg. This is a good estimate compared to the actual value of D(128) being 355.6 mg
$$y=6x-412.5$$
$$y=6(128)-412.5$$
$$y=768-412.5$$
$$y=355.5$$




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