Blog 2

Brieanna Bookter

Blog Post 2

February 9, 2015

Part A:

a)      Wind Speed  of Hurricane Irene

b)      Hurricane Irene first made her debut as a tropical storm that hit Puerto Rico on August 22, 2011. Shortly after the storm hit land, Irene strengthened and was categorized as a hurricane. By August 24^th, Irene reached Category 3 strength. Once hitting US land on August 27^th, the hurricane weakened to Category 1 strength and remained at the level as it traveled up the east coast. The average maximum wind speed recorded on August 27^th was 21.53 m/sec. While traveling up the United States’ east coast, did the storm slow up or slow down?

c)       Recorded Maximum Wind Speeds

Time of Day	Maximum Wind Speed (m/sec)
18:06	10
19:00	14.3
02:36	25
08:06	17.8
09:42	25.5
20:42	26.3
21:11	31.9
16:00	26.2
16:42	20.1
22:30	19.2
22:06	21.1

  E) The math is worked out on the attachment with the graph. Once you determine the slopes, you see that all of the values are the same. Each calculation is getting smaller and smaller but resulted in a value that rounded to .047.

G) The slope of the line is -2.43. Mathmetically, this value means that the graph begins to fall at this point. In terms of the experiment, this mathematical value proves that the storm is slowing down at this time of day because it is a negative number.

H) Through the math in part e, we narrowed in on a point that we could draw the tangent line as exact as possible. It was important to narrow in the results because we wanted to find out the value at 21:06 to determine whether or not the storm was slowing down. For this particular value, the slope was negative, which was a sure way to know the storm was slowing down. 

Amy ( Sin Yan) Lau - Blog Post 2

Instantaneous Rate of Change: Severe Acute Respiratory Syndrome Death in Hong Kong 

Introduction

On 12 March 2003, the World Health Organization issued a global alert on atypical pneumonia, called severe acute respiratory syndrome, after numerous reports of outbreaks from China, Singapore, Vietnam, Thailand, Indonesia, Taiwan, Philippines, Canada, Germany, and the United States (Severe). On 17 March 2003, your hometown Hong Kong also became a city under siege.

Severe acute respiratory syndrome, also known as SARS, is a respiratory illness caused by SARS-associated coronavirus, According to the World Health Organization, a total of 8,098 people worldwide became sick with SARS during the 2003 outbreak. Of these, a total of 744 people resulted in death with the majority of cases and deaths found in Hong Kong.

As a student who is concerned about the rate this outbreak will be happening, you have decided to use your knowledge in calculus in order to answer the question of “ Assuming that March 17^th is t=0, what is the instantaneous rate of change at t=5?”

Data

Date	Day	Cumulative number of Death(s)
March 17	Day 1	1
March 18	Day 2	2
March 19	Day 3	7
March 20	Day 4	9
March 21	Day 5	19
March 22	Day 6	26
March 23	Day 7	30

Slopes of the Secant Lines

This can be found using the slope formula. Since I wanted to find what happens when t=5, I used the ordered pair (5,19) and another pair that was nearby. In this case, I will use t=4, t=6, and t=7.

t=3 to t=5

(3,7) to (5,19)

(19 – 7) / (5-3) = 6

t = 4 to t =5

(4, 9) to (5,19)

(19-9) / (5-4) = 10

t = 5 to t = 6

(5,19) to (6, 26)

(26-19)/(6-5)= 7

t= 5to t=7

(5,19) to (7,30)

(30-19) / (7-5)= 5.5

From the result, it can be seen that the secant lines of the left side of t=5 were getting bigger (from 6 to 13).  The secant lines of the right side of t=5 were getting larger as well (from 5.5 to 7). Hence, it can be seen that the average number of deaths is increasing. Interestingly, the rate of change is slowing down after the fifth day. This may be caused by the fact that it takes longer for people to die than to catch the disease. Moreover, as death rates are increasing, hospitals probably began to employ different medications in order to slow down the death rates. Hence, people lives continue to be taken away, but in a much slower rate than the first four days since the outbreak. Moreover, the calculation shows that interval of IRC should be within the interval of (7, 10), and the average of the two values is 8.5. So the IRC that I should get later on should be around 8.5.

Tangent Line

IRC

Based on the tangent line, I chose the point (5,19) and (6,27) to calculate the slope of the tangent line

(27 – 19) / (6-5) = 8

The slope of the line represents the instantaneous rate of change, which is the rate of change at the specific point. Hence, the IRC is also the derivative at t=5. In this case, the rate of change in deaths caused by SARS is approximately 8 deaths per day. Or in other words, 8 means that by the fifth days since the outbreak, the death caused by SARS is increasing at a rate of 8 deaths per day.

Conclusion

I know 8, or 8 deaths per day, is my IRC because the rate of change determined by the ARC using secant lines were approaching 8.  Moreover, my IRC that I calculated (8) falls within the range (7,10) as predicted by my ARC.