---
# System prepended metadata

title: 'Math 181 Miniproject 2: Population and Dosage'

---

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|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
\\]
:::

(b)
$P(t)$=$1002.29$*$1.09976^x$-2.26115



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


(c\)
$P'(t)$=$1002.29\cdot1.09976^{100}-2.26115$=$13514042.8208$




:::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|139.5|153.5|169|

$P'(5)$=153.5 indicates the rate in which the population will increase by the end of the 5th year. So P(5)=1610 at the end of the 5th year the predicted population is 1610+153.5=1763.5


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


(e)
$P''(3)$=$139.5-115.5/12$=12
indicates that at 3years that rate at which the population is increasing is by 12 people 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)
$k$=$P'(t)$/$P(t)$=0.095



:::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)
$D(x)=0.025x^2+-0.5x+10$





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

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





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


(c\)$D'(128)$=0.05(128)-0.5=5.9 mg/lb. per lb. 
is the dosage change per lbs.

:::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)
By using central difference we can compute D'(128)
$AV$[$120,140$] D($140$)-D($120$)/($140-120$)

= 430-310/140-120=6mg per lb per 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)
$D(130)=0.025(130)^2-0.5(130)+10$=367.5
$L(130)=D(130)+D'(130)(x-130)$
            =367.5+6(x-130)




:::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)
128lb=355.6mg dosage
367.5+6(128-130)
367.5+-12=355.5
YES, the output value on the tangent gives a good estimate.


---

To submit this assignment click on the Publish button ![Publish button icon](https://i.imgur.com/Qk7vi9V.png). Then copy the url of the final document and submit it in Canvas.

