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)$P(t)=119.4\cdot1.15^x+957.259$ :::info (c\) What will the population be after 100 years under this model? ::: (c\) There will be appoximately 1401383 people. The actual number of the popultion is 140213983.2726. I rounded it because you can not have tenths of a person. :::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|231/2|127.05|139.755|153.7305|169.10355 :::info (e) Use a central difference to estimate the values of $P''(3)$. What is the interpretation of this value? ::: (e)Using the values of (2,231/2) and (3,127.05), the central difference is 12.1275. The value could mean the amount of people per week since it can be assumed P(t)=year, P'(t)=monthly and P''(t)=weekly. :::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.) ::: I took the central difference of the intersections of 10 and 20 on each function of P(t) and P'(t). The k appromimately equals 159.5518. :::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)=0.025$x^2$-0.5(x)+10 :::success (b) Find the proper dosage for a 128 lb individual. ::: (b)For a 128 pound individual the dosage needed is 355.6 milligrams. :::success (c\) What is the interpretation of the value $D'(128)$. ::: (c\)The interpretation for D'(128) is the amount 5.9 milligrams for each grouping of appormimately 50 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) I used the values of (127,349.725) and (129, 361.525). To find those values I plugged in 127 and 129 into the formula $d(x)=0.025x^2^-0.5x+10$ then found the central difference. I chose this method because it is easiest for me. The value is appoximately 5.9 milligrams :::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)D'(130)=6, the tangent line at that point is d(x)=6x-825/2 :::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) D(128)=355.5 milligrams according to the tangent line of D9x0+6x-825/2. This is a good estimate since it was a tenth off from the actual dosage is 355.6 milligrams. --- To submit this assignment click on the Publish button ![Publish button icon](https://i.imgur.com/Qk7vi9V.png). Then copy the url of the final document and submit it in Canvas.