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- 1. THE CALCULUS CRUSADERS Accumulation Functions: The Beetles Question
- 2. .:. THE QUESTION .:. E(t) is the rate of which the beetles rush into the chamber, whereas L(t) is the rate of which the beetles rush out of the chamber. Both E(t) and L(t) are measured in beetles per minute.
- 3. .:. THE QUESTION .:. At t = 0, Jamie hears the noise of the beetles. The beetles start rushing in and out of the chamber at t = 1.84. However, Bench estimates that everyone must escape the chamber until t = 15.17. (After t = 15.17, the beetles would have filled up the chamber completely.)
- 4. .:. THE QUESTION .:. a) How many beetles have entered the chamber at t = 10? b) With all these beetles filling up the chamber, Bench, Jamie, and Zeph have limited space. 3 m3 of the chamber is filled up for every beetle that enters the chamber until t = 10. After t = 10, 5 m3 of the chamber is filled up for every beetle that enters the chamber. How many cubic metres of the chamber would be filled up with beetles at t = 15.17?
- 5. .:. THE QUESTION .:. c) Let H(t) for 1.84 ≤ t ≤ 15.17. Determine H’(10) and explain the meaning of H’(10). d) At what time, during 1.84 ≤ t ≤ 15.17, will H(t) reach a maximum?
- 6. PART A How many beetles have entered the chamber at t = 10?
- 7. .:. THE SOLUTION .:. E(t) is measured in beetles per minute. To obtain an answer in beetles, we multiply beetles per minute by a change in time. This is the definition of an integral. This way of thinking is called a unit analysis.
- 8. .:. THE SOLUTION .:. We know the domain is 1.84 ≤ t ≤ 15.17. We know the upper limit of what we are integrating is t = 10.
- 9. PART B With all these beetles filling up the chamber, Bench, Jamie, and Zeph have limited space. 3 m3 of the chamber is filled up for every beetle that enters the chamber until t = 10. After t = 10, 5 m3 of the chamber is filled up for every beetle that enters the chamber. How many cubic metres of the chamber would be filled up with beetles at t = 15.17?
- 10. .:. THE SOLUTION .:. NOTE THAT: “3 m3 of the chamber is filled up for every beetle that enters the chamber until t = 10. After t = 10, 5 m3 of the chamber is filled up for every beetle that enters the chamber.”
- 11. .:. THE SOLUTION .:. Because of the statement written in the question, the rates of the chamber filling up with beetles are two different rates. Therefore, we integrate the function E(t) from the intervals where beetles would accumulate at the rate of 3m3 and 5m3.
- 12. PART C Let H(t) for 1.84 ≤ t ≤ 15.17. Determine H’(10) and explain the meaning of H’(10).
- 13. .:. THE SOLUTION .:. H(t) is defined as the integral of the difference of L(x) and E(x). (Integrating beetles per minutes gives us beetles as discussed in Part A.)
- 14. .:. THE SOLUTION .:. We are to determine H’(t). In this case, we are differentiating an anti derivative. (Note the quot;∫quot;.) Differentiation and anti differentiation can be seen as inverse processes of each other; The derivative of x2 is 2x; an antiderivative of 2x is x2.
- 15. .:. THE SOLUTION .:. Not only are we differentiating an antiderivative, we’re differentiating an accumulation function, a function that measures the accumulating area under a graph.
- 16. .:. THE SOLUTION .:. The derivative of an accumulation function is the original function, by The Second Fundamental Theorem of Calculus. The Second Fundamental Theorem of Calculus: If f is continuous in a closed interval, A’(x) = f(x), where A(x) is the accumulation function and f(x) is the original function.
- 17. .:. THE SOLUTION .:. We can see there is a function within a function in s(t). (Note there are two variables, t and x.) To differentiate a function within a function, we use The Chain Rule.
- 18. .:. THE SOLUTION .:. The Chain Rule: [fg]’(x) = f’(g(x))g’(x)
- 19. .:. THE SOLUTION .:. Since H’(t) is a transcendental function, a function that contains an exponential function and a trigonometric function, we cannot apply the algebra we know to solve for the roots of v’(t), so we have to use our calculator and solve numerically.
- 20. .:. THE SOLUTION .:. H’(10) is the rate at which the number of beetles in the chamber is changing. The number of beetles in the chamber is increasing at approximately beetles per minute.
- 21. PART D At what time, during 1.84 ≤ t ≤ 15.17, will H(t) reach a maximum?
- 22. .:. THE SOLUTION .:. By The First Derivative Test, the critical point of a derivative indicates the original function has a local extrema.
- 23. .:. THE SOLUTION .:. The First Derivative Test If c is a critical number and if f’ changes sign at x = c, then f has a local minimum at x = c if f’ is negative to the left of c and positive to the right of c; f has a local maximum at c if f’ is positive to the left of c and negative to the right of c.
- 24. .:. THE SOLUTION .:. Using our calculator’s features to determine roots and intersections, we find that t = 1.8400082 minutes.
- 25. .:. THE SOLUTION .:. By The Extreme Value Theorem, the endpoints are considered local extrema too. (In this case, the critical number found previously is also an endpoint.) The Extreme Value Theorem If the function f is continuous on the interval [a, b], then there exist numbers c and d in [a,b] such that for all x in [a, b], f(c) ≤ f(x) and f(d) ≥ f(x).
- 26. .:. THE SOLUTION .:. Looking at all the local extrema, we find that H(15.17) yields the largest number, the absolute maximum.
- 27. AAAAH!!!! BEETLES!

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