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 |1610.5|1771.5|1948.7|
The values given in the table were at t=0 years and the population was 1000. To find the missing population values for years 1 through 7, the previous population was multiplied by .10 to find out the 10% increase and added to the previous years' population number to find the current population value.

:::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 values are a= 1000, b=1.1 and c=1.2658x10-12. Therefore; the formula P(t)= 1000 * 1.1x +(1.2658*10-12).

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

(c\) 

The population will be about 13,780,613 people

:::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) The only values given in the table was t 1 through 6. All P'(t) values were estimated through calculating the central difference of every point from the values of table 1 from part a. Each was done as shown for 1 - 6

| $t$ | 1 | 2 | 3 | 4 | 5 | 6 |
|--- |-----|-----|------|------|------|-----|
| $P'(t)$ | 105 |115.5|127.05|139.75|153.73|169.1

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

(e) Similar to part d. we will use the central difference formula to comput f"(3).
$f"(3)=(f'(4)-f'(2))/(4-2)$
$f"(3)=((139.75-115.5)/2)=12.1275$
The second derivative shows the rate at which f' is changing. This calculation is showing an increase in population at an increasing rate of 12.1275 people per year at the time t=3, which is at 3 years.
:::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 that matches the graph is 10 . https://www.desmos.com/calculator/6gjwcnwt1q

:::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 $d(x)$

:::

(a) The chart about was used to enter in the values into a table and find the values for the model form
provided. After entering in all the provided information, Desmos gave values for the variables to be : a=0.025, b=-0.5, and c=10. Therefore the formula is $d\left(x\right)=0.025x^2+\left(-0.5x\right)+10$

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

(b) $D\left(128\right)=0.025\left(128\right)^2+\left(0.5\left(128\right)\right)+10$, so this means $D(128)=355.6mg$

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

(c\) The derivative calculates the average rate of change. When calculating D'(128) the answer would tell you the average rate of change the dosage in mg would be for an individual weighing exactly 128 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)! To calculate the derivative we can use $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$ to find D'(128) with $D\left(x\right)=0.025x^2+\left(-0.5x\right)+10$

$=\lim_{h \to 0}\frac{f(128+h)-f(128)}{h}$
$=\lim_{h \to 0}\frac{(0.025\left(128+h\right)^2+\left(0.5\left(128+h\right)\right)+10-(0.025\left(128\right)^2+\left(0.5\left(128\right)\right)+10))}{h}$
$=\lim_{h \to 0}\frac{(5.9)-(0.025h)}{h}$
$=5.9+0.025(0)$
$D'(128)=5.9 mg/lb$

:::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) $L(x)=f(a)+f'(a)(x-a)$
$L(x)=f(130)+f'(130)(x-130)$
From looking at my expoential graph from earlier plotted coordinates, I located where x=130 to find the f(a) value for x=130lbs. The f(130)=367.5, then we were given f'(130) so plug in for the tangent line equation
$L(x)=367.5+6(x-130)$

:::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) Since 128 is close to 130 we can use the previous tangent line equation to make an estimation.
$L(128)=367.5+6(128-130)$
$=367.5-12$
$=359.5$
Yes, this would be a good estimate for a 128lb individual because the output is close to what Desmos calculated. The point used was close enough to 130 to give a good estimate and keep the equation of the tangent line accurate. The further away from the point, the less accurate the results are due to the space that is created between the tangent line and the original function. The larger the space the more
inaccurate the results tend to be in estimating a result.

---

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