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costargh

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Exam 2 Question
« on: November 07, 2007, 06:07:54 pm »
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Exam 2 Question

Question 8 VCAA 2006 exam 2
The average value of the function y=cos (x)  over the interval [0, pie/2] is
A. 1/pie

B. pie/4

C. 0.5

D. 2/pie

E. pie/2


Does this have to be done by hand, can it be done using cas or graphically?
Itute gives some wierd explanation that i dont even think is for the same question lol
Q8 y = log3 (x); after reflection in the x-axis, y = − log3 (x);
after translation by 5 units up and translation 2 units right,
log3 ( 2) 5 y = − x − + .
(Will we need to know how to do this by hand also?)

Collin Li

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Exam 2 Question
« Reply #1 on: November 07, 2007, 06:11:49 pm »
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[ cos(pi/2) - cos(0) ] / (pi/2 - 0)

enwiabe

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Exam 2 Question
« Reply #2 on: November 07, 2007, 06:13:03 pm »
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That's not the answer for that question :S

costargh

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Exam 2 Question
« Reply #3 on: November 07, 2007, 06:19:06 pm »
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Yeh I think Itute made a mistake. Ok so just sub in the intervals and minus them? ok cool =)

costargh

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Exam 2 Question
« Reply #4 on: November 07, 2007, 06:27:16 pm »
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Ummm maybe Im doing something wrong but isnt [cos(pi/2)- cos(0)] /2 = -1/2?

And the VCAA answer is  D. 2/pi

=S

Collin Li

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Exam 2 Question
« Reply #5 on: November 07, 2007, 06:47:40 pm »
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Sorry, wasn't thinking properly. I edited my post to provide the correct answer.

reg

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Exam 2 Question
« Reply #6 on: November 07, 2007, 06:48:47 pm »
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VCAA is right.

First, we integrate cos(x) across the interval, and we get 1 as a result.
We then divide by the interval,

1 / (pi / 2) =
2 / pi

cara.mel

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Exam 2 Question
« Reply #7 on: November 07, 2007, 06:48:53 pm »
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You should be able to do translations by hand. It's a similar thing for all graphs.

Q2 I don't know how to explain it but I can do it. It's the integral of cos x (total area) divided by how wide it is (so then it would be 'how high does it get if you spread the area out evenly if that makes any sense). Like I said I don't know how to explain it, I just know it works

so integral cos(x) => sin(pi/2)-sin(0) = 1
Then divide this by (pi/2 - 0) = 2/pi.

Someone explain that better ty :K

costargh

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Exam 2 Question
« Reply #8 on: November 07, 2007, 06:58:42 pm »
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Thanks

Collin Li

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Exam 2 Question
« Reply #9 on: November 07, 2007, 07:03:01 pm »
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Hmm... that's weird.

I would have done that question last year and got it right (80/80 on exam 2). I was thinking of the average slope.

Yes, the integration method is the correct way. The sum of all possible values from 0, pi/2 is represented by the integral from 0 to pi/2, and then divided by the width: (pi/2 - 0) will give you the average value.

Galelleo

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Exam 2 Question
« Reply #10 on: November 07, 2007, 07:09:23 pm »
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yeah... cos(pi/2) - cos(0) /2 would be the average value if it was a triangle :P
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Ahmad

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Exam 2 Question
« Reply #11 on: November 07, 2007, 07:15:59 pm »
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To find the average value of a continuous function you extend the discrete definition to an infinite sum, evaluating it at more and more values. If the riemann sum is set up you get exactly what Coblin has in his last post :)
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Exam 2 Question
« Reply #12 on: November 07, 2007, 10:04:35 pm »
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I dunno where costargh is reading from, but I can't see that question on either 2006 exam. :S