Blog 2

Stacey Schwartz
Blog 2


Part a:A.     Find a real world application OR design your own experimental application relating to rates of change. (in the blog folder you will find plenty of examples to get you started)
How far does a ball travel in a certain amount of seconds
B.     Write a narrative or synopsis explaining your application/experiment and include a question. (for example, what is the velocity of the snowball at exactly 2 seconds? Or how can I find the velocity of the baseball at exactly 3 seconds?)What is the rate of change of the distance the ball has traveled at 8 seconds?C.     Create a table of values for the data that you have recorded from your application/experiment.

D. 

E. The slopes at the three points calculated in the first picture show that the line gets flatter as time     goes on meaning the ball is slowing down.

F. 

G.  Finding the slope of line (PQ): 

(160-120) / (8 - 4.8) = 12.5

The ball is moving about 12.5 feet per second.

H. As the secant line approaches x=8 the slope of the lines gets smaller meaning its approaching the tangent line.

Assignment 2

1. Recorded data of the prevalence of HIV in females (%, ages 15-24) in Botswana. 
2. Africa has the most serious HIV and AIDS epidemic globally, with Southern Africa particularly afflicted. This experiment tracks the prevalence of HIV in females (in percentages) ages 15-24, leading to the experimental question: What is the velocity of HIV’s prevalence in females of Botswana in 2009?
3. 
   

1. above
2. above
3.  above
4. above
5. On graph. As the secant lines get closer together, the ARC approaches the instantaneous rate of change. In terms on my experiment, as the secant lines near (2000,18), the calculated slope is closer to the instantaneous rate of change in year 2000.
6. on graph
7. Answer is -1.  In 2000, the instantaneous rate of change for the prevalence if HIV in females 15-226 in Botswana was -1%.
8. The slope of the secant lines was getting more and more precise. They continuously neared 1. They were approaching the slope value of the tangent line.The tangent line touches the line drawn to define the slope of that point on the curve. 

Blog Post 2

A) Find a real world application or design your own experimental application relating to rates of change:

A pond was stocked with a type of fish call “walleye” and over a 25-year period, the population of the fish has increased.  Initially, the population of the fish was 3000.   

B) Write a narrative or synopsis explaining your application/experiment and include a question:

What happens to the to the population of fish at exactly 5 years?

C)

Year	Walleye Population
0	3000
1	3400
2	3720
3	3976
4	4181
5	4345
6	4476
7	4581
8	4665
9	4732
10	4786
11	4829
12	4863
13	4890
14	4912
15	4930
16	4944
17	4955
18	4964
19	4971
20	4977
21	4982
22	4986
23	4989
24	4991
25	4993

D)

  

E) find the population at exactly x=5 (5,4345).  I will draw three secant lines with the points: (3,3976), (4,4181), (6,4476), (7,4581), (8,4665)  

- y2-y1/x2-x1= 4345-3976/5-3= 184.5

- y2-y1/x2-x1= 4345-4181/5-4 = 164

- y2-y1/x2-x1= 4476-4345/6-5 = 131

- y2-y1/x2-x1= 4581-4345/7-5 = 118

- y2-y1/x2-x1= 4665-4345/8-5= 106.67

the thing I am noticing about the rate of change of the secant lines is that the rate of change is decreasing as we move towards year 5 and away from year 5.  Also, the difference between the rates of change fluctuates.  However, since the difference in population is less as the number of years increases then the rate of change will continue to decrease. 

F)

G) (7,4600)

IRC= y2-y1/x2-x1= 4600-4343/7-5= 128.5

This means that at exactly point 5, the rate of change is 128.5

H) According to part d of the experiment, the population is beginning to increase at a slower rate as the number of years is growing.  The IRC proves that the population of the fish is increasing per/year but the population change is according to the year itself.  And in year 5, the population has increased 

Blog post #2

Math Blogpost #2

Ana Maria Lopez

Car depreciation

a.         Car depreciation

b.         In the real world, when you buy a car at price X, each year that goes by the price of that same car decreases. In this example, the owner bought the car for $22,000 and then ten years later ends up with a car that its worth $4,050. Even though the car looses value of $17,950 over a period of 10 years, the price did not depreciate by the same amount each year. So, what is the depreciation rate in year 5?

c.           

Time (years)	Value (in dollars)
0	22,000
1	16,200
2	14,350
3	11,760
4	8,980
5	7,820
6	6,950
7	6,270
8	5,060
9	4,380
10	4,050

d.          

Explanation

Question #	Explanation
e.	For this portion, I calculated three different secant lines. I wanted to find the IRC for point a (x=5). Therefore I labeled points a1  (4,8980), a2 (3,11760), and a3  (2,14350) . I drew a secant line from point a to all three points (a1, a2, a3 ). In this case a3  is the point that is furthest away from point a. I found that the slope between these two points was – 3/6530. I found that as the point got closer to point a ( in this case point a1 is the closest), the slope became larger ( -1/1160). This connects back to my question since the closest to the actual depreciation rate in year 5 or x=5 is point a1 where the rate is -1/1160.
g.	For this section, I drew a tangent line that passes through point x= 5. In this case I took two points A and B and found the slope between them. This gave me the IRC or the slope/ derivative at point x=5. The INC I got was -3/6530. I found it interesting that this number is the same slope I got for the secant line a3 . In terms of my experiment it means that my depreciation rate at x=5 is approximately -3/6530 according to the tangent line experiment.
h.	In this case, as the secant lines approached the value x=5, the numbers got bigger.  Therefore the slope of the secant is becoming closer to the tangent line. By definition, any points in the tangent line will have the same slope or derivative than the point being analyzed (in this case x=5). Therefore, when finding the slope

Bouncy ball

Blog #2

Ricardo Zapata

a.Find a real world application OR design your own experimental application relating to rates of change. (in the blog folder you will find plenty of examples to get you started

I chose the experiment of the ball bounce. 


b.Write a narrative or synopsis explaining your application/experiment and include a question. (for example, what is the velocity of the snowball at exactly 2 seconds? Or how can I find the velocity of the baseball at exactly 3 seconds?)

This is an experiment in which the ball for time(s) goes up to some height(cm). The results 
of this experiment can be seen in the table below.
What is the velocity of the ball at 0.53 seconds?

c.Create a table of values for the data that you have recorded from your application/experiment.

d.Graph the points using the data from your table of values (connect the dots).

e.Calculate the slope (ARC) of at least three secant lines originating from the same point on your graph to three different points on your graph (i.e. maybe you want to know what happens exactly at x = 20, so your points might be (20, 62), (20, 56), (20, 50)).  Explain what you notice about the ARC of these secant lines and what the calculations mean/represent in terms of your experiment/application.

This means that at points like (0.23,40) and (0.30,141) the ball is in a lower height and is in process to get to the height of the point (0.53,146), this is the reason why the slope is positive, because the point has to increase. And in point (0.46,178) the ball has a higher height, and this is why the slope is negative, because it has to get to an inferior height to get to the point (0.53,146).

f.Sketch an approximation of a tangent line that passes though the same point (P) from part e to which you connected your secant lines (i.e. you would draw a tangent line through the point 20, since that is the same point that you used to calculate your three different secant lines)

g.Choose a second point (Q) on the tangent line, and calculate the slope of the line (PQ). This calculation will be the instantaneous rate of change ((IRC or derivative at a point)…be sure to identify the units correctly).  Explain what this calculation means mathematically and in terms of your experiment/application.

This calculation means that the ball is going to decrease 800 centimeters for every second that passes.

h.Explain in detail how you know that the value from part g is the IRC. (i.e. since the values of calculations from part d are getting smaller and smaller, this shows that the slope of the secant is getting closer and closer to the tangent line … or some explanation similar to this). BE DETAILED!!!

The values of the maximum height keeps getting smaller and smaller, as time increases,  so the speed will decrease. This is why the slope of the point in the tangent is negative. So the ball will decrease 800cm/sec until it comes to a complete stop.