Math 181 Miniproject 2: Population and Dosage.md
---
Math 181 Miniproject 2: Population and Dosage
===

**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.

---

:::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|1464|1610.51|1771.56|1948.72

:::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
\\]
:::

(b)$y_1~a*b^x_1+c$
$a=908.405$
$b=1.10005$
$c=.717479$

$P(t)~908.405*1.10005^x_1+.717479$

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

(c\)$P(100)=142944.932856$

:::info
(d) Use a central difference to estimate the values of $P'(t)$ in the table below. What is the interpretation of the value $P'(5)$?
:::

(d)
| $t$ | 1 | 2 | 3 | 4 | 5 | 6 |
|--- |---|---|---|---|---|---|
| $P'(t)$ |105|115.5|127|139.755|153.78|169.105 |

:::info
(e) Use a central difference to estimate the values of $P''(3)$. What is the interpretation of this value?
:::

(e)
12.1375 means the rate in which the population of the people increases per year per year.

:::info
(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.)
:::

(f)$k=\frac{p'(1)}{p(t)}$
$p'(1)+k*p(1)$
$p'(1)=k*p(1)$
$k=\frac{105}{115.5}$
$k=.909$

:::success
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 $D(x)=ax^2+bx+c$.)
:::

(a)$y_1~ax^2_1+bx_1+c$
$a=.025$
$b=-.5$
$c=10$

$D(x)~.025x^2+-.5x+10$

:::success
(b) Find the proper dosage for a 128 lb individual.
:::

(b)$D(x)=.025(128)^2+-.5(128)+10$
$D(128)=355.6$

:::success
(c\)$What is the interpretation of the value D'(128).
:::

(c\)The interpretation of D'(128) is the rate of change as the individual's weight increaseswith respect to the proper dosage in mg.

:::success
(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.
:::

(d)$D(x)~.025x^2+-.5x+10$
$\lim_{h \to 0}\frac{f(128+h)^2-f(128)}{h}$
$\lim_{h \to 0}\frac{(.025(128+h)^2-.5(128+h)+10-(.025(128^2-.5(128)-10))}{h}$
$5.75 mg/lb.$

:::success
(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.
:::

(e)$f(a)+f'(a)(x-a)$
$f(130)+f'(130)(x-130)$
$x=130,y=367.5$

:::success
(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?
:::

(f)x=128 then y or dosage is 355.5 milligrams.

---

To submit this assignment click on the Publish button . Then copy the url of the final document and submit it in Canvas.