Math 181 Miniproject 2: Population and Dosage.md
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Math 181 Miniproject 2: Population and Dosage
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**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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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.
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(a)
P(1)=1000 + 1000 x0.1= 1100
| $t$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|--------|------|---|---|---|---|---|---|---|
| $P(t)$ | 1000 | 1100 |1210 | 1331 | 1464 | 1611 | 1772 | 1949 |

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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 *1.09976^(t) + -2.26115

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(c\) What will the population be after 100 years under this model?
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(c\) P(t)= 1002.29 *1.09976^(100) + -2.26115= 13,516,062.82077 million

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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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f'(x)=(f(x+h) - f(x-h))/2h
P(1)= (P(1+1)-P(1-1))/2x1 = P(2)-P(0)/2
1210-1000/2 = 105

(d)
| $t$ | 1 | 2 | 3 | 4 | 5 | 6 |
|--- |---|---|---|---|---|---|
| $P'(t)$ | 105 | 115.5 |127 | 139.5 | 153.5 |169 |

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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''(t)= (P'(t+1) -P'(t-1))2*1 =
P''(3) = (P'(4)-P'(2))/2
=12

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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.)
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(f) P'(t)=1000 (1.1)^t*ln1.1
P'(t)=1.1P(t)
k=1.1

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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 $D(x)=ax^2+bx+c$.)
D(x) = 0.025x^2 + -0.5x + c
where c= 10
* 
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(a)

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(b) Find the proper dosage for a 128 lb individual.
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(b) D(x) = .025(128)^2 + -0.5(128)+10 = 356mg dosage

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(c\) What is the interpretation of the value $D'(128)$.
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(c\) D'(x) = 0.05x -0.5
D'(128) = 0.05(128)-0.5
D'(128)= 5.9

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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) use linear apporximation
y=f(a)+f'(a)(x-a)
x=120,y=310. a=128
y= D(a)+D'(a)(x-a)
310=D(128)+D'(128)(120-128)
310=356+D'(128)(-8)
D'(128) = (310-356)/-8
=5.75

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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)D'(136)= 6
Y at x=130 will be
Y130 = D(130)= 0.025(130)^2 -0.5(130)+10
=367.5

Y-367.5=6(x-130)
6x - y=412.5

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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)
y=6x - 412.5
=6(128)-412.5
y=355.5
Yes this is a good estimate since at 120 lbs the required dosage is 310mg and at 140 lbs the recommended is 430 mg, 355.5 lies directly in the middle

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