1. MAT 213 Brief Calculus Section 5.1 Results of Change and Area Approximation

2. Recall… Distance = Velocity x Time

3. You are speeding down a very perilous mountain road in your brand new fancy yellow Porche when you spot a fuzzy white rabbit sitting in the middle of the road 400 feet directly ahead of you. You immediately apply the brakes. At the time you apply the brakes you are going 100 ft/sec (approximately 68 mph) Your velocity decreases throughout the 10 seconds it takes you to stop. Do you hit the rabbit?

4. Here is some information that may help you. v(t) gives the speed of the car (in ft/sec) as a function of time (in sec) since the brakes were applied Time since brakes applied (sec) 0246810 Velocity (ft/sec) 10064361640 v(t)

5. Do you hit the rabbit? How far has the Porche traveled? Recall… distance = rate x time It goes at most 100·2=200 feet in the first 2 seconds And at most 64·2=128 feet in the next 2 seconds And so on… So during that 10-second period, it goes at most 100·2 + 64·2 + 36·2 + 16·2 + 4·2 = 440 feet Time since brakes applied (sec) 0246810 Velocity (ft/sec) 10064361640

6. Do you hit the rabbit? How far has the Porche traveled? You could also say… It goes at least 64·2=128 feet in the first 2 seconds And at least 36·2=72 feet in the next 2 seconds And so on… So during that 10-second period, it goes at least 64·2 + 36·2 + 16·2 + 4·2 + 0·2 = 240 feet Time since brakes applied (sec) 0246810 Velocity (ft/sec) 10064361640

7. Do you hit the rabbit? Therefore, 240 feet ≤ Total distance traveled ≤ 440 feet There is a difference of 200 feet between the upper and lower estimates!!!!! Upper Estimate Lower Estimate Time since brakes applied (sec) 0246810 Velocity (ft/sec) 10064361640

8. Do you hit the rabbit? How could we get a more accurate estimate? Time since brakes applied (sec) 0246810 Velocity (ft/sec) 10064361640

9. Do you hit the rabbit? Upper Estimate 100  1 + 81  1 + 64  1 + 49  1 + 36  1 + 25  1 + 16  1 + 9  1 + 4  1 + 1  1 = 385 Lower Estimate: 81  1 + 64  1 + 49  1 + 36  1 + 25  1 + 16  1 + 9  1 + 4  1 + 1  1 + 0  1 = 285 t (sec) 012345678910 V(t) (ft/sec) 1008164493625169410 ft Therefore, 285 feet ≤ Total distance traveled ≤ 385 feet

10. Time since brakes applied (seconds) Speed (ft/sec) Consider the graph of v(t)

11. Left-hand sums Time since brakes applied (seconds) Speed (ft/sec) Consider the graph of v(t) The area of the first purple rectangle is 100·2=200, the upper estimate of the distance traveled during the first two seconds. The area for the second purple rectangle is 64·2=128, the upper estimate during the next two seconds. The total area of the purple rectangles represents the upper estimate for the total distance moved during the ten seconds.

12. Right-hand sums Time since brakes applied (seconds) Speed (ft/sec) Consider the graph of v(t) The area of the first green rectangle is 64·2=128, the lower estimate of the distance traveled during the first two seconds. The area for the second green rectangle is 36·2=72, the lower estimate during the next two seconds. The total area of the green rectangles represents the lower estimate for the total distance moved during the ten seconds.

13. Time since brakes applied (seconds) Speed (ft/sec) Based on these rectangles, how could you calculate the difference between the two estimates? Consider the graph of v(t)

14. Time since brakes applied (seconds) Speed (ft/sec) Based on these rectangles, how could you calculate the difference between the two estimates? Consider the graph of v(t) 100 2

15. The difference in the upper and lower estimates for the two-second data is 200 feet. What is the difference in the upper and lower estimates for the one-second data?

16. Time since brakes applied (seconds) Speed (ft/sec) Based on these rectangles, how could you calculate the difference between the two estimates? Consider the graph of v(t) 100 2

17. Time since brakes applied (seconds) Speed (ft/sec) One-Second Data Consider the graph of v(t) 100 1

18. The difference in the upper and lower estimates for the two- second data is 200 feet. The difference in the upper and lower estimates for the one- second data is 100 feet. What would the difference be if the velocity were given every tenth of a second? Every hundredth of a second? Every thousandth?

19. So…the more measurements we take (the more rectangles we use), the more accurate our estimate will be. What would give us the exact distance? The total distance traveled is the area between the velocity curve and the x-axis

20. So…the more measurements we take (the more rectangles we use), the more accurate our estimate will be. It won’t matter if we are using left- or right- hand rectangles because the sums will eventually be so close we can consider them equal.

21. Left-Hand Sums n rectangles The top LEFT corner of the rectangles touch the graph of f(t) Let’s see how this works with our velocity function

22. Right-Hand Sums n rectangles The top RIGHT corner of the rectangles touch the graph of f(t) Let’s see how this works with our velocity function

23. Now we take the limit of these sums as n goes to infinity. If f is continuous for a ≤ t ≤ b, the limits of the left-and right-hand sums exist and are equal.

24. The DEFINITE INTEGRAL is the limit of these sums. f is called the integrand, and a and b are called the limits of integration.

25. The DEFINITE INTEGRAL as AREA When f(x) ≥ 0 and a < b: The area under the graph of f and above the x-axis between a and b.

26. Examples

27. The DEFINITE INTEGRAL as AREA When a < b, and f(x) is positive for some x-values and negative for others: The sum of the areas above the x-axis (+) and those below the x-axis (-) Example

28. Time since brakes applied (seconds) Speed (ft/sec) Time since brakes applied (seconds) Speed (ft/sec) Units!!!

29. Example Suppose C(t) represents the daily cost of heating your house, measured in dollars per day, where t is measured in days since January 1, 2007. Interpret Give units.