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 |1948 |

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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) $P(t)$= $1002.3(1.1)^x+(-2.3)$

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(c\) What will the population be after 100 years under this model?
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substitute x for 100

(c\) $P(t)$=$1002.3(1.1)^{100}+(-2.3)$

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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 | 139.5 | 153.5 | 169 |
(d) p'(5) is telling us how much the population will increasng for that 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)
$$f''(3)=Av[2,4]$$
$$=\frac{f'\left(139.5\right)-f'\left(115.5\right)}{4-2}$$
$$f''(3)=\frac{24}{2}$$
$$=12$$
Population increasing 12 people 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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$$p'(t)k*(P(t))$$
$$105=k*(1000)$$
$$(f) k=0.105$$

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

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(c\) What is the interpretation of the value $D'(128)$.
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(c\) $D'(x)=0.05x-0.5$
$D'(128)=0.05(128)-0.5$
$D'(128)=5.9mg/lb$

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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)$$f(x)=D(a)+D'(a)(x-a)$$
$$310=D(128)+D'(128)(120-128)$$
$$310=356+D'(128)(-8)$$
$$D'(128)=\frac{\left(310\right)-\left(356\right)}{-8}
=5.75mg/lb$$

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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) $$L(x)=D(130)+D'(130)(x-130)$$
$$=130+6(x-130)$$

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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)$$y=6x-412.5$$
$$y=6(128)-412.5$$
$$y+768-412.5$$
$$y=355.6mg$$

$355.5=356$ This estimate for dosage at x=128 is a good estimate.

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