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) $y_1\sim 1000\cdot 1.1^{t}+0$

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

(c\) $y_1\sim 1000\cdot 1.1^{100}+0$

~13,780,612.339

:::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 |155.2305 |169.10355 |

$\frac{f(2)-f(0)}{2-0}$ ~ F'(1)

$\frac{(1210)-(1000)}{2}$ ~ 105

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

(e) $\frac{f'(4)-f'(2)}{4-2}$ ~ F''(3)

$\frac{(139.755)-(115.5)}{2}$ ~ 12.1275 popultaion per year
After 3 years, the rate of change will be increasing at the rate of 12.1275 population per additional 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) $\frac{P'(t)}{P(t)}$ = k

$\frac{105}{1000}$ = 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)

$D(128)=0.025(128)^2-0.5(128)+10$

= 355mg

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

(c\) The dosage will increase at 5.6mg per additional pound at the weight of 128 pounds.

:::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)=$\lim_{h \to 0}\frac{f(0.047619(45+h)^2+0.0119048(45+h)+-0.07145286)-F(0.047619(45)^2+0.0119048(45)+-0.07145286)}{h}$
$\lim_{h \to 0}=0.095238x+0.0119048$
=0.095238x+0.0119048

=$\lim_{h \to 0}\frac{0.025(x^2) +0.05hx+0.025(h^2) -0.5x -0.5h+10-0.025(x^2) +0.5x-10}{h}$

=$\lim_{h \to 0}\frac{(h)(0.025h+0.05x-0.5)}{h}$

=$\lim_{h \to 0}(0.025h+0.05x-0.5)$

= 0.025(0)+0.05x-0.5

=0.05x-0.5

Input: 128

=0.05(128)-0.5

D'(128)=5.9

First I found the derivitive formula by solving for X in original equation, then I input value we were looking for (128) and calculated.

:::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)

$D(130)=0.025(130)^2-0.5(130)+10$

= 367.5mg

L(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)

L(128)= 367.5+ 6(128-130)

~367+(-12)
~355.5 mg

Yes it provides a good estimate, as the actual is 355mg and the estimate is 355.5mg. The difference is only 0.5, a closer estimate can be obtained using 129lbs or 127lbs.

---

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