Instantaneous Rate of Change: Life or Death Situation (Blog Post 2)

Introduction 

You are in a life and death situation. You have been kidnapped, and you don’t know where you’ve been taken. Your kidnapper approaches you and hands you a paper. Displayed on the right side of the paper is data for the average monthly temperature for China from 1990-2009 provided by The World Bank (www.worldbank.org). There is a question at the very bottom of the page. You squint at the small text. It reads: “Assuming that January is t=0, what is the instantaneous rate of change at t=5?” 

Your kidnapper gives you twenty minutes to solve the problem. If you solve it within the time constraint, they will grant you a chance to leave safely. 

If you do not solve it within the time constraint, well... needless to say, there will be consequences. 

Average Rate of Change (Secant Lines/ARC) 

(5, 63.86) to (6, 67.82)

(67.82 – 63.86) / (6 – 5) = 3.96

(5, 63.86) to (7, 65.84) 

(65.84 – 63.86) / (7 – 5 ) = .99 

(5, 63.86) to (4, 56.58)

(56.58 – 63.86) / ( 4 – 5 ) = 7.28

The slopes of the secant lines were found using the slope formula. Because I wanted to know what happened at t=5, I used the ordered pair (5, 63.86) and another corresponding pair on the line that was nearby. I calculated the slope (ARC) of the secant line between these points. I noticed that the secant lines on the right side of t=5 were getting larger (.99 to 3.96). However, the secant lines on the left side were getting smaller (8.37 to 7.28). If I had values that were smaller than t=5 but were getting closer, I could’ve had a tighter range of where the instantaneous rate of change (IRC) would fall. However, in this case, I can determine that the IRC is between 3.96 and 7.28. Assuming that it’s somewhere in the middle, the average of the two values is 5.62. Therefore, the IRC should be approximately around 5.62.  These calculations represent the average rate of change (i.e., the rate of change of the temperature over a period of time) happening over an interval, rather than what the rate of change is at a specific point. 

Instantaneous Rate of Change (IRC)

Based on the tangent line, I chose the points (5, 63.86) and (7, 75) to calculate the slope of the tangent. 


(5, 63.86) to (7, 75) 

(75 – 63.86) / ( 7 – 5 ) = 5.57

The instantaneous rate of change of any function represents the rate of change at that specific point—i.e. it’s a specific slope at a given point. In other words, the IRC is the derivative at a given point. Another way to view this is that it’s also the slope of the tangent line at a given point. 

In this context, this means that the rate of change at t=5, i.e. in June, the rate of change in temperature is 5.57 degrees Fahrenheit. This means that there was a 5.57 degrees Fahrenheit increase in June.

Conclusions

I know that the value that I calculated as the IRC is indeed the instantaneous rate of change/derivative at that point as the ARC calculations I made were getting smaller and smaller and closer and closer to each other. The slopes of the secant lines were getting smaller—and this could have been better displayed if I had smaller intervals to work with. Furthermore, the IRC found is within the interval of (3.96, 7.28). To provide even more evidence that 5.57 is the IRC, the calculations were also very close to the average between the interval, suggesting that the IRC was definitely somewhere in the middle of the two numbers.