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AP Calculus AB 2003 #2 http://www.collegeboard.com/prod_downloads/ap/students/calculus/calculus_ab_frq_03.pdf I don't understand how to do part c).

Mathematics
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total distance is integral of velocity but u gotta take into consideration which way it's going
I looked at the sample respoinses and they did like absolute value of v(t). But I don't understand why. http://apcentral.collegeboard.com/apc/public/repository/ap03_calculusab_q2_27980.pdf
so you found where it changes direction right?

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the thing is...|dw:1357322365854:dw|
if u take just the integral, the area on the bottom gets subtracted
but that's for displacement. So if u want to get the distance you need to add that area instead of subtracting it.. So you have to integrate from start till when it changed direction, then from when it changed direction till when it changed direction again, etc.. and finally take the absolute value of all integrals and add them up
|dw:1357322510296:dw| see if u start at pt A, displacement, which is the regular integral, is 0, but distance traveled, is 5
i thought to find the distance u have to take integral of the velocity.
yes you do.. integral of velocity is indeed the area under the velocity curve, which is the displacement
so in order to get the DISTANCE.. you want to add all the areas that end up below y- axis, instead of subtracting them
if you were looking for DISPLACEMENT, you would just take the integral of that function from 0 to 3. But, since you're looking for DISTANCE, take the integral of that function from 0 to a, where a is the time at which particle changed direction.
okay listen, where did the particle change direction?
rad(2pi) and 0
0?
yeah those were the solutions but only rad(2pi) works
no it was just t = 2.507
yeah that's rad(2pi)
so integrate that function from 0 to 2.507, then from 2.507 to 3
once you get the results, say Int from 0 to 2.507 = A, and Int from 2.507 to 3 = B; just add up the absolute values: |A| + |B|
that's all he's doing by integrating the absolute value of V over 0 < t < 3
how'd u integrate the function tho? like algebraically?
isn't that the calculator part?
i dont think u can integrate exponents inside trig functions yet.. lol
or do u just plug it into the graphing calculator? ohhh -_- ok.
i mean i dont think there is a way that anyone has found how to do that
lol yea
ohhh yeah i got the answer when i plugged it into the calculator
You mean to tell me.... I WASTED MY TIME???!!!!!!
nooooooo...... like after you told me all that...
k good lol
sorry..if anyone's still viewing this for part d) why did they add 1 to rad(2pi) ?

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