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title: 'Math 181 Miniproject 2: Population and Dosage'

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Math 181 Miniproject 2: Population and Dosage.md
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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. 

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:::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|





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(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
\\]
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(b)
$P(t)=1002.29\cdot1.09976^{x}-2.26115$



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(c\) What will the population be after 100 years under this model?
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(c\)
$P(100)=1002.29\cdot1.09976^{100}-2.26115=13514042\ people$



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(d) Use a central difference to estimate the values of $P'(t)$ in the table below. What is the interpretation of the value $P'(5)$?
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(d)
| $t$     | 1 | 2   | 3 | 4   | 5   | 6 |
|---      |---|-----|---|-----|-----|---|
| $P'(t)$ |105|115.5|127|139.5|153.5|169|
$P'(1)=\frac{P(2)-P(0)}{2-0}=\frac{1210-1000}{2}= 105$
$P'(2)=\frac{P(3)-P(1)}{3-1}=\frac{1331-1100}{2}= 115.5$
$P'(3)=\frac{P(4)-P(2)}{4-2}=\frac{1464-1210}{2}= 127$
$P'(4)=\frac{P(5)-P(3)}{5-3}=\frac{1610-1331}{2}= 139.5$
$P'(5)=\frac{P(6)-P(4)}{6-4}=\frac{1771-1464}{2}= 153.5$
$P'(6)=\frac{P(7)-P(5)}{7-5}=\frac{1948-1610}{2}= 169$
$P'(5)=153.5$ can be interpreted as is people per year



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(e) Use a central difference to estimate the values of $P''(3)$. What is the interpretation of this value? 
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(e)
$P''(3)=\frac{P'(4)-P'(2)}{4-2}=\frac{139.5-115.5}{2}=\frac{24}{2}=12$
$P''(3)$ can be interpreted as the rate of population increasing per year per year
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(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)
$P'(t)=k\cdot P(t)$
$P'(1)=k\cdot P(1)$
$105=k\cdot 1100$
$\frac{105}{1100}=k$
$k=0.095$


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




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(b) Find the proper dosage for a 128 lb individual.
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(b)
$D(128)=345.18mg$ for a 128 lb individual




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(c\) What is the interpretation of the value $D'(128)$. 
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(c\)
$D'(128)=6.02\frac{mg}{lbs}$
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(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.
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(d)
$D(x)=0.0275035x^2-1.01556x+24.5551$
$D'(128)=\lim_{h\to 0}\frac{(0.0275035(128+h)^2-1.01556(128+h)+24.5551)-(0.0275035(128)^2-1.01556(128)+24.5551}{h}$
$D'(128)=\lim_{h\to 0}\frac{7.04h +h^2-1.01556h+345.18082-352.18082}{h}$
$D'(128)=\lim_{h\to 0}\frac{h^2+6.0244h}{h}$
$D'(128)=\lim_{h\to 0}\frac{h(h+6.0244)}{h}$
$D'(128)=\lim_{h\to 0}h+6.02$
$D'(128)=0+6.02$
$D'(128)=6.02\frac{mg}{lb}$
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(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. 
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(e)
$L(x)=f(a)+f'(a)(x-a)$
$L(x)=f(130)+f'(130)(x-130)$
$L(x)=357.24+6(x-130)$

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(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? 
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(f)
$L(x)=357.24+6(x-130)$
$L(128)=357.24+6(128-130)$
$L(128)=357.24-12$
$L(128)=345.24mg$
The output value does give a good estimate of the dosage needed for a 128 lb individual.

