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 | 1464.1 | 1610.51 | 1771.561 | 1948.7171 |

:::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)
In this case I will be using the growth rate formula.
$P(t)=P_0(1+r)^t$
Where $P_0$ equals the initial amount, r equals the rate of change (the percentage), and t is time interval.

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

(c\)
$P(t)=P_0(1+r)^t$
$P(100)=1000(1+.10)^100$
$P(100)=1000(1.10)^100$
$P(100)=13,780,612.34$

:::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 | 127.05 | 139.755 | 153.7305 | 169.10355 |
The value of $P'(5)$ is just the slope (or rate of change) of of $P(t)$ at t=5.

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

(e)
$P''(3)=\frac{139.755-115.5}{4-2}$
$P''(3)=\frac{24.255}{2}$
$P''(3)=12.1275 \frac{people}{year^2}$
After 3 years, the rate at which the population is increasing is increasing at a rate of $12.1275 \frac{people}{year^2}$.

:::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)
I believe that the constant k is 10%, because if I take 10 percent of 105 (or $P'(t)$) I get 10.5. If I add that 10.5 to the original 105 I get 115.5, which is the value of $P''(2)$. This goes for on for every number too.

:::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)
$D(x)=0.025x^2-0.5x+10$
$D(128)=0.025(128)^2-0.5(128)+10$
$D(x)=409.6-64+10$
$D(x)=355.6 mg$

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

(c\)
$D'(128)$ gives us the value of the rate of change of the dosage at the exact point of x=128 in mg/lb.

:::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)
To estimate the value of $D'(128)$ I will use the central difference.
$D'(x)=\frac{P(b)-P(a)}{b-a}$
$D'(130)=\frac{P(140)-P(120)}{140-120}$
$D'(130)=\frac{430-310}{20}$
$D'(130)=\frac{120}{20}$
$D'(130)=6 mg/lb$

:::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)
$L(x)=d(a)+d'(a)(x-a)$
$L(x)=d(130)+d'(130)(x-130)$
$L(x)=[0.025(130)^2-0.5(130)+10] +6(x-130)$
$L(x)=367.5 +6(x-130)$

:::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)
$L(x)=367.5 +6(x-130)$
$L(128)=367.5 +6(128-130)$
$L(128)=355.5 mg$
I belive that this output value of $355.5 mg$ is a good estimate for the dosage for a 128 lb individual because it is almost spot on to the actual value of $355.6 mg$.

---

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