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) The formula for *P(t)* is *P(t)*=$1002.29*1.09976^x+(-2.26115)$.

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

(c\) The population after 100 years would be about $13514043$.

:::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 |
|--- |---|---|---|---|---|---|
| $P'(t)$ |43 |48 | 53 |58 |64 |71 |
The interpretation of *P'(5)* is $$P'(5)=1002.29*1.09976^5log(1.09976)-2.26115=64.33.$$

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

(e)The interpretation of *P''(3)* is $$P''(3)=1002.29*1.09976^3log(1.09976)^2-2.26115=107.85.$$

:::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) The value of *k*=$1.1$.

:::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)The function D(x) that approximates the dosage of an individual is $$(Dx)=0.025x^2+(-0.5)x+c$$

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

(b)The proper dosage for an individual who is 128 lbs is 356mg.

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

(c\)The interpretation of the value $D'(128)$ is:
$=0.025(128)^2+(-0.5)(128)+10$
$=410-64+10$
$=356$

:::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)If we use the linear approximation formula: $D(x)=D(a)+D'(a)(x-a)$
We would just need to plug in the given information from the table and the value we are trying to solve. After plugging in the information, begin to solve as any other function. After step 2, we are solving for $D'(128)$ so we need to set the equation to solve it by subtracting 356 from the right side to bring it over to the left then dividing that difference by -8 to isolate the $D'(128)$.

$310=D(128)+D'(128)(120-128)$
$310=356+D'(128)(-8)$
$D'(128)=5.75$

:::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)$y-367.5=6(x-310)$
$y-367.5=6x-780$
$6x-y=412.5$

:::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)$6x-y=412.5$ at $x=128$
$y=6x-412.5$
$y=6(128)-412.5$
$y=355.5$
Yes, the output value on the tangent line does give a good estimate for the value because they were nearly identicala had we not rounded the values in 2c to whole numbers so they would be easier to deal with.

---

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