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 | 1610 | 1771 | 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)

$P\left(t\right)=1002.29\cdot1.09976^{t}-2.26115$

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

$P\left(100\right)$= 13514042.8208 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)
| $t$ | 1 | 2 | 3 | 4 | 5 | 6 |
|--- |---|---|---|---|---|---|
| $P'(t)$ | 104.975874306 | 115.448267527 |126.965386695 | 139.631453672 | 153.56108749 |168.880341578 |

$\frac{\left(P\left(b\right)-P\left(a\right)\right)}{b-a}$ = x People per year

$\frac{\left(P\left(2\right)-P\left(0\right)\right)}{2}$= 104.975874306
$\frac{\left(P\left(3\right)-P\left(1\right)\right)}{3-1}$=115.448267527
$\frac{\left(P\left(4\right)-P\left(2\right)\right)}{4-2}$=126.965386695
$\frac{\left(P\left(5\right)-P\left(3\right)\right)}{5-3}$=139.631453672
$\frac{\left(P\left(6\right)-P\left(4\right)\right)}{6-4}$=153.56108749
$\frac{\left(P\left(7\right)-P\left(5\right)\right)}{7-5}$=168.880341578

The interpretation of P'(5) is that at 5 years, the population is increasing by 153.56108749 people per 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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$\frac{\left(P'\left(b\right)-P'\left(a\right)\right)}{b-a}$= people/ $years^2$
P''(3)= $\frac{\left(P'\left(4\right)-P'\left(2\right)\right)}{4-2}$=12.0733892735 people/$years^2$

At 3 years the rate at which the population is rising, is increasing at a rate of 12.0733892735 people 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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(f)

$k\ =\ \frac{P'\left(t\right)}{P\left(t\right)}$

k~0.095

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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)
$D\left(128\right)$= 355.6 mg

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(c\) What is the interpretation of the value $D'(128)$.
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D'(128)= 5.9
This means at 128 pounds the dosage is 5.9 mg per pound

(c\)

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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) using the central difference formula of $D'(x) =\frac{\left(D\left(b\right)-D\left(a\right)\right)}{b-a}$
$\frac{\left(D\left(140\right)-D\left(120\right)\right)}{140-120}$=6 the derivative of 128 should be slightly smaller than 6, because the central difference of D(140) and D(120) is a close approximation of D'(130) and since 128 is a little less than 130 the answer would be slightly smaller.

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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(130)=D(130)+D'(130)(x-130)
L(130)= 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)
x=128= D(128)=L(128)
L(130)= 367.5 + 6(x-130)
L(128)= 367.5 +6(128-130)
L(128)=355.5 mg

The answer gives a very close estimate for the dosage for a 128 pound individual. The answer found for part b was 355.6 mg which is only .1 mg greater than the amount found using the tangent line formula

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