jennychan12
  • jennychan12
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
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
bahrom7893
  • bahrom7893
total distance is integral of velocity but u gotta take into consideration which way it's going
jennychan12
  • jennychan12
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
bahrom7893
  • bahrom7893
so you found where it changes direction right?

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bahrom7893
  • bahrom7893
the thing is...|dw:1357322365854:dw|
bahrom7893
  • bahrom7893
if u take just the integral, the area on the bottom gets subtracted
bahrom7893
  • bahrom7893
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
bahrom7893
  • bahrom7893
|dw:1357322510296:dw| see if u start at pt A, displacement, which is the regular integral, is 0, but distance traveled, is 5
jennychan12
  • jennychan12
i thought to find the distance u have to take integral of the velocity.
bahrom7893
  • bahrom7893
yes you do.. integral of velocity is indeed the area under the velocity curve, which is the displacement
bahrom7893
  • bahrom7893
so in order to get the DISTANCE.. you want to add all the areas that end up below y- axis, instead of subtracting them
bahrom7893
  • bahrom7893
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.
bahrom7893
  • bahrom7893
okay listen, where did the particle change direction?
jennychan12
  • jennychan12
rad(2pi) and 0
bahrom7893
  • bahrom7893
0?
jennychan12
  • jennychan12
yeah those were the solutions but only rad(2pi) works
bahrom7893
  • bahrom7893
no it was just t = 2.507
jennychan12
  • jennychan12
yeah that's rad(2pi)
bahrom7893
  • bahrom7893
so integrate that function from 0 to 2.507, then from 2.507 to 3
bahrom7893
  • bahrom7893
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|
bahrom7893
  • bahrom7893
that's all he's doing by integrating the absolute value of V over 0 < t < 3
jennychan12
  • jennychan12
how'd u integrate the function tho? like algebraically?
bahrom7893
  • bahrom7893
isn't that the calculator part?
bahrom7893
  • bahrom7893
i dont think u can integrate exponents inside trig functions yet.. lol
jennychan12
  • jennychan12
or do u just plug it into the graphing calculator? ohhh -_- ok.
bahrom7893
  • bahrom7893
i mean i dont think there is a way that anyone has found how to do that
bahrom7893
  • bahrom7893
lol yea
jennychan12
  • jennychan12
ohhh yeah i got the answer when i plugged it into the calculator
bahrom7893
  • bahrom7893
You mean to tell me.... I WASTED MY TIME???!!!!!!
jennychan12
  • jennychan12
nooooooo...... like after you told me all that...
bahrom7893
  • bahrom7893
k good lol
jennychan12
  • jennychan12
sorry..if anyone's still viewing this for part d) why did they add 1 to rad(2pi) ?

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