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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 |1610|1664|1771 |1948|





:::info
(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)
$p(t)=(1002.29)(1.09976)^t+(-2.26115)$.
$p(t)=(1002.29)(1.09976)-(-2.26115)$.



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(c\) What will the population be after 100 years under this model?
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(c\)Replace t=100 in the equation found in b
$p(100)=(1002.29)(1.09976)^100(-2.26115)$
=13,516,062.82077.




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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 |
|---      |---|---|---|---|---|---|
| $P'(t)$ |   |   |   |   |   |   |
$p'(t)=(1002.29)(1.09976)^5log_e(1.09976)-2.26115$
$p'(5)=(1002.29)(1.09976)^5log(1.09976)-2.26115$
=151.06




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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)$p''(3)=(1002.29)(1.09976)^3ln(1.09976)^2-2.26115$
=979606

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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


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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)https://www.desmos.com/calculator/
$D(x)=0.025x^2+-0.5x+C$





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(b) Find the proper dosage for a 128 lb individual.
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(b)$D(x)=.025(128)^2+-0.5(128)+10$
=356 dosages.





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(c\) What is the interpretation of the value $D'(128)$. 
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(c\)$D'(x)=0.05x-0.5$
$D'(128)=0.05(128)-0.5$
$D'(128)=5.9$.
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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)By using linear approximation formula $y=f(a)+f'(a)(x-a)$ use value x=120 from the table use the corresponding value of y=310 and a=128
$y=D(a)+D'(a)(x-a)$
$310=D(128)+D'(128)(120-128)$
$310=356+D'(128)(-8)$
$D'(128)=\frac{310-356}{-8}$
=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)$y_130=D(130)=0.025(130)^2-0.5(130)+10$
$y_130=D(130)=367.5$
The equation of tangent line
$y-y_1=m(x-x_1)$
$y-367.5=6(x-130)$
$y-367.5=6x-780$
$6x-y=412.5$

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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)$6x-y=412.5$ at x=128lbs
$y=6x-412.5$
$y=6(128)-412.5$
$y=355.5$
Which is estimated to 356
Yes it is a good estimate for the dosage.



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