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)

$$P\left(t\right)=\left(1000\right)\left(1.1\right)^{x}+1.2838\cdot10^{-12}$$

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

(c\)

$$P\left(100\right)=13,780,612.3398$$

So the population will be about 13,780,612 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) $P'(5)$ can be interpreted as the average rate of change in population on year 5. In the chart below I determined the rate of change to be 153.5.
| $t$ | 1 | 2 | 3 | 4 | 5 | 6 |
|--- |---|---|---|---|---|---|
| $P'(t)$ |104.841|115.3|126.9|139.5|153.5|168.85|

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

(e)
$$p''\left(3\right)=\frac{p'\left(4\right)-p'\left(2\right)}{4-2}$$
$$p''\left(3\right)=\frac{\left(139.5\right)-\left(115.3\right)}{2}$$
$$P''\left(3\right)=12.09$$
$P"(3)$ can be interpreted as the rate of which $P"$ is changing. In this case, the calculation shows at year 3 the population is increasing at a rate of 12.09 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)
$$k=\frac{P'\left(t\right)}{P\left(t\right)}$$

$$k=0.095$$

:::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) $$D\left(128\right)=0.025(128)^2-0.5(128)+10=355.6mg$$

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

(c\)
$$D'(128)=5.9\frac{\frac{mg}{lb}}{lb}$$
$D'(128)$ can be interpreted as 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)
$$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$
$$=\lim_{h \to 0}\frac{f(128+h)-f(128)}{h}$$
$$=\lim_{h \to 0}\frac{(0.025(128+h)^2-0.5(128+h)+10)-(0.025(128)^2-0.5(128)+10)}{h}$$
$$=\lim_{h \to 0}\frac{409.6+6.4h+0.025h^2-64-0.5h+10-355.6}{h}$$
$$=\lim_{h \to 0}\frac{6.4h+0.025h^2-0.5h}{h}$$
$$=\lim_{h \to 0}\frac{h(6.4-0.5+0.025h)}{h}$$
$$=5.9+0.025(0)$$
$$D(128)=5.9 \frac{\frac{mg}{lb}}{lb}$$

:::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)
$$L\left(x\right)=f\left(a\right)+f'\left(a\right)\left(x-a\right)$$
$$L\left(x\right)=f\left(130\right)+f'\left(130\right)\left(x-130\right)$$
$$L\left(x\right)=367.5+6\left(x-130\right)$$

:::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)
$$L\left(x\right)=367.5+6\left(128-130\right)$$
$$L(x)=355.5$$
I believe the output value from the tangent line equation does give a good estimate for the dosage for a 128lb individual. Desmos calculated the value to be $355.6$. That is only a $0.0003$% error from the actual dosage. The reason for this output value being so close to the actual value is because 128 and 130 are not that far apart. The reason for this is because the further away from the point 128, the less accurate the output value will be.

---

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