caroline cheung pd 2&3 what to do?. a zoo sponsored a one-day contest to name a new baby...
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AP Calculus Free Response Question
Caroline Cheung Pd 2&3
What to do?
2010 - Question #2
A zoo sponsored a one-day contest to name a new baby elephant. Zoo visitors deposited entries in a special boxbetween noon (t = 0) and 8 P.M. (t = 8). The number of entries in the box t hours after noon is modeled by adifferentiable function E for 0≤t≤8. Values of E(t ), in hundreds of entries, at various times t are shown inthe table above. The lesson
today is…
t (hours) 0 2 5 7 8
E(t) (hundreds of entries)
0 4 13 21 23
Part A :Use the data in the table to approximate the rate, in hundreds of entries per hour, at which entries were beingdeposited at time t = 6. Show the computations that lead to your answer.
To find the rate at t = 6 you have to use the Mean Value Theorem:
ab
afbf
)()(
According to what was given in the chart. Use t = 5 and t = 7 because 6 is between those two numbers.
42
8
2
1321
57
)5()7(
ff
E’(6) =
At t = 6 there are 4 hundred entries per
hour
Part B:Use a trapezoidal sum with the four subintervals given by the table
to approximate the value of
P
8
0
)(8
1dttE
Using correct units, explain the meaning of
8
0
)(8
1dttE
in terms of number of entries
t(hours) 0 2 5 7 8E(t) (hundreds of entries)
0 4 13 21 23
The four subintervals are (0,2), (2,5), (5,7), (7,8)
8
0
)(8
1dttE Is the average number of hundreds of entries in the
box between noon and 8 P.M.
Ahhh!
Part B : Continue…Trapezoidal rule : ))((
2
1hb Note: The base would be the amount
of entries added together and the height would be the difference between the t values.
)22)(1()17)(2()2
17)(3()2)(2(
8
1
)2
2321)(1()
2
2113)(2()
2
134)(3()
2
40)(2(
8
1
Giving you:
= 10.687 or 10.688 PLUG IT IN!
Part C:At 8 P.M., volunteers began to process the entries. They processed the entries at a rate modeled by the function P, where 97629830)( 23 ttttP
hundreds of entries per hour for 8≤t≤12. According to the model, how many entries had not yet been processed by midnight (t = 12)?
It is given in the chart that at E(8) = 23.Take the amount of entries processed which is E(8) subtract the integral of P(t) from 8≤t≤12. To determine how many entries were not processed.
12
8
23 )97629830(23 dtttt
Note: Integration is necessary because the function given models the rate.
???
Part C: Continue…Step 1: type in 23 –Step 2: go to MATH then scroll down to the 9th one and press enterStep 3: type in the equationStep4: then press the following buttons: , X , 8 , 12Step 5: press enter to solve.
Oh yeah. We
got this.
= 7 hundred entries
Part D:According to the model from part (c), at what time were the entries being processed most quickly? Justify your answer
The problem asked for what the entries are processed most quickly which means to find the maximum of the function.
298603)('
97629830)(2
23
tttP
ttttP Note: Find the derivative of the function in order to find the maximum of the function.
P’(t) = 0
To find the zeros of the function graph the function.
Graphing it.Plug it into the calculator to graph
The graph
Press 2nd trace and scroll down to ZERO.Set the bounds and get t = 9.183503 and t = 10.816497
That’s right
Part D: ContinueAfter obtaining the values plug it into the original equation (P(t)) to determine the maximum value.
t P(t)
89.1835010.8164912
05.0886622.911338
8
97629830)( 23 ttttP
At t = 12 the entries are processed most rapidly
THE END! Let’s dance!
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Works Cited
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